Acids-bases chemistry

Class 11 Chemistry, unit 9, book pages 188–211

Practise 69 MCQs Take the unit quiz Review 38 cards

This unit studies acids, bases and salts in aqueous solution. It starts with conjugate acid-base pairs and the Lewis concept, then measures acid and base strength with $K_{w}$ , pH, $K_{a}$ and $K_{b}$ , and uses titration and indicators to find unknown concentrations. It then explains buffer solutions, the hydrolysis of salts, the solubility product with precipitation, and the common ion effect, all of which are applications of chemical equilibrium.

Learning outcomes (17)

Define conjugate acid-base pairs.
Identify conjugate acid-base pairs in reactions.
Apply the concept of conjugate acid and conjugate base on salt hydrolysis.
Define mathematically the terms pH, $K_{a}$ , $p K_{a}$ and $K_{w}$ and use them in calculations ( $K_{b}$ and the equation $K_{w} = K_{a} \times K_{b}$ will not be tested).
Calculate $[H X^{+}]$ and pH values for: (a) strong acid, (b) strong alkalis, (c) weak acids, (d) weak alkalis.
Distinguish that Lewis acids accept a lone pair, and Lewis bases donate a lone pair to make a coordinate covalent bond.
Calculate the pH of buffer solutions in given appropriate data.
Demonstrate the ability to comprehend and effectively apply the concept of solubility product ( $K_{s p}$ ).
Construct an expression for $K_{s p}$ .
Calculate $K_{s p}$ from concentrations and vice versa.
Apply the concept of the common ion effect to describe why the solubility of a substance changes when it is dissolved in a solution containing a common ion.
Perform calculations using $K_{s p}$ values and the concentration of a common ion.
Use the concept of hydrolysis to explain why aqueous solutions of some salts are acidic or basic.
Calculate the $[H X_{3} O X^{+}]$ given the $K_{a}$ and the molar concentration of weak acids.
Calculate concentrations of ions of slightly soluble salts.
Perform acid-base titrations to calculate the molarity and strength of given sample solutions.
Select suitable indicators for acid-alkali titrations, appropriate data ( $p K_{a}$ values will not be used).

Acids, bases and salts

Acids were first identified as substances with a sour taste. Vinegar tastes sour because it is a dilute solution of acetic acid, and citric acid is responsible for the sour taste of lemon. Bases, sometimes called alkalis, are characterised by their bitter taste and slippery feel. Commercial preparations for unclogging drains are highly basic. A salt is an ionic substance formed by neutralising an acid with a base.

Acid-base chemistry is important in many everyday applications. The effect of acids on living things has become particularly important in recent years because of acid rain.

Acids are classified as (i) mineral acids and (ii) organic acids. Organic acids are much weaker than mineral acids and are mostly found in vegetables, fruits and other natural things. Some common organic acids and where they occur are given below.

Some common organic acids
Organic acid	Where it is found
Lactic acid	Sour milk
Citric acid	Citrus fruits such as lemons and oranges
Formic acid	Insect bites
Tartaric acid	Grape juice
Maleic acid	Apples and pears

Acids and water

The presence of water is essential for the formation of $H X^{+}$ . For example, $HCl$ is covalent in nature and does not form $H X^{+}$ ions by itself. It forms $H X_{3} O X^{+}$ (hydronium) ions in the presence of $H X_{2} O$ .

HCl + H X_{2} O H X_{3} O X^{+} + Cl X^{-}

Bases and alkalis

Bases form a class of chemical substances that includes metal oxides and hydroxides. A soluble base is called an alkali and forms $OH X^{-}$ ions when it dissolves in $H X_{2} O$ .

CaO + H X_{2} O Ca (OH) X_{2}

$Ca (OH) X_{2}$ is an alkali and a base. Alkalis are important in the manufacture of soap and detergents, and caustic soda ( $NaOH$ ) is used for this purpose.

Salts

The ionic substances obtained as a result of the neutralisation of acids and bases are called salts.

acid HCl + base NaOH ⟶ salt NaCl + H X_{2} O

A salt may be neutral, acidic or basic. Why this happens is explained in the section on salt hydrolysis.

9.1Conjugate acid-base pairsOften examined

A conjugate acid is a species formed as a result of the acceptance of a proton by a base. A conjugate base is the species that remains after donating a proton from an acid. In an acid-base reaction, the acid forms a (conjugate) base and the base, having accepted a proton, forms a (conjugate) acid. The acid-base reaction is as shown below.

A X_{1} + B X_{2} B X_{1} + A X_{2}

$A X_{1}$ : Brønsted acid, $B X_{2}$ : Brønsted base, $B X_{1}$ : conjugate base, $A X_{2}$ : conjugate acid

The conjugate acid-base pairs are species on opposite sides of an equation that differ by a proton. So $A X_{1}$ and $B X_{1}$ form one pair, and $B X_{2}$ and $A X_{2}$ form the other. The weaker acids have stronger conjugate bases and the stronger acids have weaker conjugate bases.

Examples

In each equation below, identify the species that loses a proton (the acid) and the one that gains it (the base). Each product is the conjugate of one reactant.

Ionisation of

HCl

in water:

HCl

(acid) and

Cl X^{-}

(conjugate base) are one pair,

H X_{2} O

(base) and

H X_{3} O X^{+}

(conjugate acid) are the other.

Three examples from the book

Ionisation of $HCl$ in water. $HCl$ gives a proton to $H X_{2} O$ , so $HCl$ is the acid and $H X_{2} O$ is the base. Their conjugates are $Cl X^{-}$ (conjugate base) and $H X_{3} O X^{+}$ (conjugate acid).
$HCl + H X_{2} O H X_{3} O X^{+} + Cl X^{-}$
Ionisation of $NH X_{4} OH$ (ammonia in water). $NH X_{3}$ accepts a proton from water, so $NH X_{3}$ is the base and $H X_{2} O$ is the acid. The conjugate acid of $NH X_{3}$ is $NH X_{4} X^{+}$ and the conjugate base of $H X_{2} O$ is $OH X^{-}$ .
$NH X_{3} + H X_{2} O NH X_{4} X^{+} + OH X^{-}$
Ionisation of water. One water molecule gives a proton (acid) and another accepts it (base). The products are $H X_{3} O X^{+}$ (conjugate acid) and $OH X^{-}$ (conjugate base).
$H X_{2} O + H X_{2} O H X_{3} O X^{+} + OH X^{-}$

In the last two examples water acts as an acid in one and as a base in the other. Water is therefore amphoteric in nature, that is, it is acidic as well as basic.

Particular examples of conjugate acid-base pairs

Each row is $A X_{1} + B X_{2} B X_{1} + A X_{2}$ ; $A X_{1}$ / $B X_{1}$ and $B X_{2}$ / $A X_{2}$ are conjugate acid-base pairs
Acid ( $A X_{1}$ )	Base ( $B X_{2}$ )	Conjugate base ( $B X_{1}$ )	Conjugate acid ( $A X_{2}$ )
$HCl$	$H X_{2} O$	$Cl X^{-}$	$H X_{3} O X^{+}$
$HCl$	$NH X_{3}$	$Cl X^{-}$	$NH X_{4} X^{+}$
$HNO X_{3}$	$H X_{2} O$	$NO X_{3} X^{-}$	$H X_{3} O X^{+}$
$NH X_{4} X^{+}$	$H X_{2} O$	$NH X_{3}$	$H X_{3} O X^{+}$
$H X_{2} O$	$H X_{2} O$	$OH X^{-}$	$H X_{3} O X^{+}$
$H X_{2} O$	$CO X_{3} X^{2 -}$	$OH X^{-}$	$HCO X_{3} X^{-}$
$HClO X_{4}$	$H X_{2} O$	$ClO X_{4} X^{-}$	$H X_{3} O X^{+}$

9.2Lewis concept of acids and bases

Certain substances such as $SO X_{2}$ , $CO X_{2}$ , $CaO$ and $BF X_{3}$ behave as acids or bases although they do not have the ability to donate or accept protons. The nature of such substances cannot be explained by the Arrhenius theory or the Brønsted-Lowry theory.

In 1923, G. N. Lewis proposed an acid-base theory that focuses on the reaction. This concept is more general than either the Arrhenius theory or the Brønsted-Lowry theory.

Lewis acid: A substance that can accept a pair of electrons to form a coordinate covalent bond.
Lewis base: A substance that can donate a pair of electrons to form a coordinate covalent bond.

In a Lewis acid-base reaction, a coordinate covalent bond is formed between the acid and the base. Consider the following reaction of $HCl$ with water.

The lone pair on the oxygen of

H X_{2} O

(electron pair donor) forms a bond with the

H

HCl

(electron pair acceptor). The

H - Cl

bond pair goes to

Cl

, which leaves as

Cl X^{-}

The book asks four questions about this reaction.

Question	Answer	Reason
(i) Which species donates an electron pair?	$H X_{2} O$	The oxygen atom has lone pairs and gives one of them to $H$ to form a coordinate bond in $H X_{3} O X^{+}$ .
(ii) Which species accepts an electron pair?	$HCl$	Its hydrogen atom takes the pair from oxygen (the $H - Cl$ bond pair then leaves with $Cl X^{-}$ ).
(iii) Which species is a Lewis acid?	$HCl$	It is the electron pair acceptor.
(iv) Which species is a Lewis base?	$H X_{2} O$	It is the electron pair donor.

This theory can also explain gas-phase neutralisation reactions, in which there are no ions in solution and no solvent. In the reaction of $NH X_{3}$ gas with $HCl$ gas the nitrogen atom of ammonia donates an electron pair to the $H$ atom of $HCl$ (the $H$ end carries a partial positive charge, $δ +$ ).

NH X_{3} + HCl NH X_{4} X^{+} + Cl X^{-}

The book asks which species is the Lewis acid, $HCl$ or $NH X_{3}$ . It is $HCl$ , because it accepts the electron pair. $NH X_{3}$ is the Lewis base because it donates the pair.

Example 9.1: classifying substances as Lewis acids or Lewis bases

Identify the Lewis acid and the Lewis base in the reaction of

NH X_{3}

with

BF X_{3}

to form

H X_{3} N - BF X_{3}

Problem solving strategy from the book: (1) draw the electronic structures of both species; (2) look for the species with a lone pair of electrons or a negative charge. It has a complete octet, so it does not need electrons but can donate a pair: it is a Lewis base; (3) look for the species that can accommodate an electron pair. It has an incomplete octet, so it can accept a pair: it is a Lewis acid.

Draw the electronic structures. $NH X_{3}$ : N has five valence electrons, uses three in the three $N - H$ bonds and keeps one lone pair. $BF X_{3}$ : B has three valence electrons, all used in the three $B - F$ bonds, so B has only six electrons (three pairs) around it.
Look for the lone pair. $NH X_{3}$ has a lone pair on the N atom, so it is an electron pair donor.
$NH X_{3} is a Lewis base$
Look for the incomplete octet. B in $BF X_{3}$ has six electrons, so it needs one more electron pair to complete its octet. $BF X_{3}$ is an electron pair acceptor.
$BF X_{3} is a Lewis acid$
The lone pair of N is shared with B and a coordinate covalent bond $N B$ forms, giving $H X_{3} N BF X_{3}$ .

Figure for Example 9.1: the nitrogen lone pair of

NH X_{3}

(Lewis base) is donated to boron in

BF X_{3}

(Lewis acid), forming a coordinate covalent bond.

NH X_{3} : Lewis base, BF X_{3} : Lewis acid

Concept assessment exercise 9.1

Identify the Lewis acid and the Lewis base in the following examples.

Concept assessment exercise 9.1 (1): chloride ion and aluminium chloride

Cl X^{-} + AlCl X_{3} [AlCl X_{4}] X^{-}

$Cl X^{-}$ has four lone pairs (a complete octet) and a negative charge, so it can donate an electron pair.
$Cl X^{-} is a Lewis base$
Al in $AlCl X_{3}$ has only six electrons (three bonding pairs), an incomplete octet, so it can accept an electron pair.
$AlCl X_{3} is a Lewis acid$
The chloride ion donates a pair to Al, forming a coordinate bond and the ion $[AlCl X_{4}] X^{-}$ .

Cl X^{-} : Lewis base, AlCl X_{3} : Lewis acid

Concept assessment exercise 9.1 (2): hydrogen ion and hydroxide ion

H X^{+} + OH X^{-} H X_{2} O

$OH X^{-}$ has lone pairs on oxygen and a negative charge, so it donates a pair of electrons to form the $O - H$ bond.
$OH X^{-} is a Lewis base$
$H X^{+}$ has no electrons at all (an empty orbital), so it accepts the pair.
$H X^{+} is a Lewis acid$

H X^{+} : Lewis acid, OH X^{-} : Lewis base

9.3Strength of acids and bases

Different Brønsted acids donate protons to different extents. An acid that can donate a proton to a higher degree than another acid is said to be a relatively strong acid. For example, hydrochloric acid is a relatively stronger acid than acetic acid. Also, acetic acid is relatively stronger than water. The ability of an acid to donate a proton is called the strength of the acid or the acid strength.

Similarly, bases also differ in their ability to accept a proton. A base that can accept a proton to a higher degree than another base is a relatively stronger base. Thus ammonia is a relatively stronger base than water, because ammonia can accept a proton to a higher degree than water.

Strength is a comparison of how far each species donates or accepts a proton
	Acid strength	Base strength
Meaning	Ability (degree) to donate a proton	Ability (degree) to accept a proton
Example from the book	\ce{HCl} > \ce{CH3COOH} > \ce{H2O}	\ce{NH3} > \ce{H2O}

9.4Ionisation constant of water and calculation of pHOften examined

Water is a unique compound because of its ability to accept or donate a proton under different environments. Water acts as a Brønsted acid in the presence of ammonia, and as a Brønsted base in the presence of hydrochloric acid. In fact, water itself undergoes ionisation to a small extent, as shown in the following equation.

\ce{H2O + H2O <=> H3O+ + OH-}

This reaction is regarded as the auto-ionisation of water. There is an equilibrium between water molecules (on the left side) and hydronium ions and hydroxide ions (on the right side).

Derivation Ionic product constant of water, K_w

Aim. Show that K_w=[\ce{H+}][\ce{OH-}] and find its value for pure water at 25^\circ\mathrm{C}.

Write the equilibrium constant K for the auto-ionisation.
K=\dfrac{[\ce{H3O+}][\ce{OH-}]}{[\ce{H2O}]^2}
Water is a weak electrolyte and is also the solvent. Its concentration [\ce{H2O}] is in large excess, so it remains constant. Multiply both sides by [\ce{H2O}]^2.
K[\ce{H2O}]^2=[\ce{H3O+}][\ce{OH-}]
K[\ce{H2O}]^2 is the product of two constants, so it is itself a constant. It is represented by K_w.
K[\ce{H2O}]^2=K_w=[\ce{H3O+}][\ce{OH-}]
[\ce{H3O+}] is the concentration of hydrated protons at equilibrium, so it is written simply as [\ce{H+}].
K_w=[\ce{H+}][\ce{OH-}]
In pure water at 25^\circ\mathrm{C} the measured value is
K_w=[\ce{H+}][\ce{OH-}]=1.0\times10^{-14}

ResultK_w=[\ce{H+}][\ce{OH-}]=1.0\times10^{-14}\ \text{(pure water, }25^\circ\mathrm{C})

K_w is called the ionic product constant of water. At a given temperature, in any aqueous solution, the product [\ce{H+}][\ce{OH-}] is fixed, so if one concentration rises the other must fall.

Neutral, acidic and basic solutions

Whenever [\ce{H+}]=[\ce{OH-}], the aqueous solution is neutral, neither acidic nor basic. In pure water at 25^\circ\mathrm{C}:

\begin{aligned}[\ce{H+}]&=1.0\times10^{-7}\,M\\ [\ce{OH-}]&=1.0\times10^{-7}\,M\\ [\ce{H+}][\ce{OH-}]&=[\ce{H+}]^2=1.0\times10^{-14}\end{aligned}

Solution	Condition
Acidic	[\ce{H+}]>[\ce{OH-}]
Basic	[\ce{OH-}]>[\ce{H+}]
Neutral	[\ce{H+}]=[\ce{OH-}]

Because K_w is an equilibrium constant, it is temperature dependent. For example, at 40^\circ\mathrm{C}, K_w=3.8\times10^{-14}. Since [\ce{H+}]=[\ce{OH-}] in pure water, this corresponds to

[\ce{H+}]=1.9\times10^{-7}\,M\quad\text{and}\quad[\ce{OH-}]=1.9\times10^{-7}\,M

pH

The concentrations [\ce{H+}] and [\ce{OH-}] are usually very small numbers and inconvenient to work with, so a more practical measure called pH was proposed. It is defined as the negative logarithm of the hydrogen ion concentration (in \mathrm{mol\,dm^{-3}}).

\mathrm{pH}=-\log[\ce{H+}]

Being a logarithmic value, pH has no units. At 25^\circ\mathrm{C} the different types of solution show the following behaviour.

Solution	[\ce{H+}] at 25^\circ\mathrm{C}	pH
Acidic	>1.0\times10^{-7}\,M	<7.00
Basic	<1.0\times10^{-7}\,M	>7.00
Neutral	=1.0\times10^{-7}\,M	=7.00

The pH scale at 25^\circ\mathrm{C}: below 7 the solution is acidic, at 7 it is neutral and above 7 it is basic.

pOH and the relation pH + pOH = 14

Derivation pH + pOH = 14

Aim. Prove that \mathrm{pH}+\mathrm{pOH}=14 at 25^\circ\mathrm{C}.

Start from the ionic product of water at 25^\circ\mathrm{C}.
K_w=[\ce{H+}][\ce{OH-}]=1.0\times10^{-14}
Take the logarithm of both sides. The logarithm of a product is the sum of the logarithms.
\log(1.0\times10^{-14})=\log[\ce{H+}]+\log[\ce{OH-}]
Since \log(1.0\times10^{-14})=-14,
-14=\log[\ce{H+}]+\log[\ce{OH-}]
Multiply by -1 and write the right-hand side as two negative logarithms.
14=(-\log[\ce{H+}])+(-\log[\ce{OH-}])
The first bracket is the pH. The second is defined as the pOH, \mathrm{pOH}=-\log[\ce{OH-}].
14=\mathrm{pH}+\mathrm{pOH}

Result\mathrm{pH}+\mathrm{pOH}=14\quad(25^\circ\mathrm{C})

If you know the pH you can find the pOH, and the reverse. A strongly basic solution has a high pH and a low pOH.

Worked examples

Example 9.2: [H+] of household ammonia

The concentration of \ce{OH-} ions in a household ammonia solution is 0.005\,M. Calculate the concentration of \ce{H+} in it.

Given

[\ce{OH-}] = 0.005\,M
K_w = 1.0\times10^{-14}

To find

[\ce{H+}]

Write the ionic product of water.
K_w=[\ce{H+}][\ce{OH-}]
Substitute K_w and [\ce{OH-}].
1.0\times10^{-14}=[\ce{H+}]\times0.005
Divide by 0.005.
[\ce{H+}]=\dfrac{1.0\times10^{-14}}{0.005}=2.0\times10^{-12}\,M

[\ce{H+}]=2.0\times10^{-12}\,M

Example 9.3: pH of 0.001 M hydrochloric acid

Calculate the pH of 0.001\,M aqueous hydrochloric acid solution.

Given

[\ce{HCl}] = 0.001\,M

To find

\mathrm{pH}

Hydrochloric acid ionises in water completely, so each mole of \ce{HCl} gives one mole of \ce{H3O+}.
\ce{HCl + H2O -> H3O+ + Cl-}
[\ce{H3O+}] is the same as [\ce{H+}], so [\ce{H+}]=0.001\,M.
Apply the definition of pH.
\mathrm{pH}=-\log(0.001)=-\log10^{-3}=3.00

\mathrm{pH}=3.00

Example 9.4: pH of 0.062 M sodium hydroxide

Calculate the pH of 0.062\,M \ce{NaOH} solution.

Given

[\ce{NaOH}] = 0.062\,M

To find

\mathrm{pH}

\ce{NaOH} is a strong base and ionises completely, so [\ce{OH-}] equals the concentration of \ce{NaOH}.
\ce{NaOH(aq) -> Na+(aq) + OH-(aq)}\qquad [\ce{OH-}]=0.062\,M
Find the pOH.
\mathrm{pOH}=-\log[\ce{OH-}]=-\log(0.062)=1.21
Use \mathrm{pH}+\mathrm{pOH}=14.
\mathrm{pH}=14-\mathrm{pOH}=14-1.21=12.79

\mathrm{pH}=12.79

Worked example: pH of a strong base, and [H+] from a pH

(a) Find the pH of 0.0025\,M barium hydroxide, \ce{Ba(OH)2}, at 25^\circ\mathrm{C}. (b) A solution has pH 4.30. Find [\ce{H+}]. (c) How does [\ce{H+}] compare in solutions of pH 4.30 and pH 3.30?

(a) \ce{Ba(OH)2} is a strong base and ionises completely. Each formula unit gives two hydroxide ions.
\ce{Ba(OH)2 -> Ba^2+ + 2OH-}\;\Rightarrow\;[\ce{OH-}]=2\times0.0025=5.0\times10^{-3}\,M
Find pOH, then use \mathrm{pH}+\mathrm{pOH}=14.
\mathrm{pOH}=-\log(5.0\times10^{-3})=2.30,\qquad \mathrm{pH}=14-2.30=11.70
(b) Reverse the definition \mathrm{pH}=-\log[\ce{H+}]: take the antilogarithm of -\mathrm{pH}.
[\ce{H+}]=10^{-4.30}=10^{0.70}\times10^{-5}=5.0\times10^{-5}\,M
(c) For pH 3.30 the same step gives [\ce{H+}]=5.0\times10^{-4}\,M.
\dfrac{5.0\times10^{-4}}{5.0\times10^{-5}}=10
One pH unit is a factor of ten in [\ce{H+}], because pH is a logarithm. The solution of pH 3.30 is ten times more acidic in [\ce{H+}] than the one of pH 4.30.

\mathrm{pH}=11.70;\quad[\ce{H+}]=5.0\times10^{-5}\,M;\quad\text{ten times larger at pH }3.30

Concept assessment exercise 9.2

Concept assessment exercise 9.2: pH of a sulphuric acid solution

What is the pH of a solution containing 1.95\,\mathrm{g} pure \ce{H2SO4} per \mathrm{dm^3} of solution?

Given

m = 1.95\,\mathrm{g}
V = 1\,\mathrm{dm^3}
M_r(\ce{H2SO4}) = 98\,\mathrm{g\,mol^{-1}}

To find

\mathrm{pH}

Convert the mass to moles. In 1\,\mathrm{dm^3} the number of moles is the molarity.
M=\dfrac{1.95}{98}=0.0199\,\mathrm{mol\,dm^{-3}}\approx0.02\,M
Sulphuric acid is a strong acid and gives two \ce{H+} ions per molecule (both ionise fully in this calculation).
\ce{H2SO4 -> 2H+ + SO4^2-}\qquad [\ce{H+}]=2\times0.0199=0.0398\,M
Apply the definition of pH.
\mathrm{pH}=-\log(0.0398)=1.40

\mathrm{pH}=1.4

9.5TitrationOften examined

A titration is defined as a method to find the volume of the standard solution required to react completely with a known volume of another solution under analysis.

Acid-base titrations are conducted using burettes and volumetric pipettes. Generally the acid solution is placed in the burette. A fixed volume of base is placed into a conical flask along with a few drops of a suitable acid-base indicator. Acid from the burette is added to the base until the indicator changes colour. This change of colour indicates the end point of the titration.

Apparatus for an acid-base titration: acid from the burette is added to a fixed volume of base and indicator in the conical flask until the colour changes (the end point).

In the neutralisation of a strong acid with a strong base, phenolphthalein is used as the indicator. It imparts a pink colour to the base solution. At the end point the solution just becomes colourless.

Calculating the molarity

The molarity of the acid solution under test is determined with the help of the following equation.

\dfrac{M_1V_1}{n_1}=\dfrac{M_2V_2}{n_2}

Symbol	Meaning
M_1	Molarity of the base
V_1	Volume of the base taken in the flask
M_2	Molarity of the acid
V_2	Volume of the acid used from the burette
n_1	Number of moles of the base in the balanced equation
n_2	Number of moles of the acid in the balanced equation

Knowing five of the six parameters, the sixth can be calculated. The strength of a solution (in \mathrm{g\,dm^{-3}}) follows from the molarity:

\text{strength}=\text{molarity}\times\text{molar mass}

Concept assessment exercise 9.3

Concept assessment exercise 9.3: concentration and strength of HCl

In a titration it is found that 25\,\mathrm{cm^3} of 0.12\,M \ce{NaOH} is neutralised with 30\,\mathrm{cm^3} of \ce{HCl} of unknown concentration. Calculate the concentration and strength of the \ce{HCl} solution.

Given

M_\text{base} = 0.12\,M NaOH
V_\text{base} = 25\,\mathrm{cm^3}
V_\text{acid} = 30\,\mathrm{cm^3} HCl
M_r(\ce{HCl}) = 36.5\,\mathrm{g\,mol^{-1}}

To find

M_\text{acid}
strength of the \ce{HCl} solution

Write the balanced equation to find n for each. Both coefficients are 1.
\ce{HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O}\qquad n_\text{acid}=1,\; n_\text{base}=1
Write the titration equation (base on one side, acid on the other). Volumes can stay in \mathrm{cm^3} because they appear on both sides.
\dfrac{M_\text{acid}V_\text{acid}}{n_\text{acid}}=\dfrac{M_\text{base}V_\text{base}}{n_\text{base}}
Substitute the values.
\dfrac{M_\text{acid}\times30}{1}=\dfrac{0.12\times25}{1}
Solve for the molarity of \ce{HCl}.
M_\text{acid}=\dfrac{0.12\times25}{30}=0.1\,M
Strength is molarity times molar mass.
\text{strength}=0.1\times36.5=3.65\,\mathrm{g\,dm^{-3}}

M=0.1\,M,\qquad \text{strength}=3.65\,\mathrm{g\,dm^{-3}}

9.5.1Selection of indicatorsOften examined

To perform an acid-base titration it is necessary to choose the correct indicator. For an acid-base titration, choose an indicator whose pH range lies within the limits of the pH change of the reaction. The colour change of the indicator then happens exactly when the reaction passes its end point.

Table 9.1: Indicators and their pH range
Indicator	Acidic colour	pH range of colour change	Basic colour
Phenolphthalein	Colourless	8.3 to 10.0	Pink
Bromothymol blue	Yellow	6.0 to 7.6	Blue
Phenol red	Yellow	6.8 to 8.0	Red
Methyl orange	Red	3.2 to 4.4	Yellow

The four cases

Strong acid with strong base. The pH changes rapidly from about 3 to 11 at the end point. Phenolphthalein changes colour from pH 8 to 10, which lies inside this jump. Therefore phenolphthalein is the best choice for this titration.
Weak acid with strong base (for example oxalic acid against sodium hydroxide). The pH changes only slightly at the end point, from about 6.5 to 10. Phenolphthalein, whose range is 8.3 to 10, is the best choice. Methyl orange works below pH 5, so it changes colour long before the end point and is not suitable.
Strong acid with weak base (for example \ce{HCl} against \ce{Na2CO3}). At the equivalent point the pH changes from 3.5 to 7.5. Methyl orange is the best indicator for this titration. Phenolphthalein (range 8.3 to 10) lies above 7.5, so it cannot be used.
Weak acid with weak base. There is no sharp change in pH at the end point, so neither indicator works satisfactorily.

Choose the indicator whose bar lies inside the pH jump of the titration (accent bars). Phenolphthalein fits the first two jumps and methyl orange fits the third.

The book asks two questions about the titration of a strong acid with a weak base (\ce{HCl} against \ce{Na2CO3}).

Question	Answer
Can you use phenolphthalein?	No. Its range 8.3 to 10 is above the pH change 3.5 to 7.5, so it would change colour after the equivalent point has already been passed.
Can you use bromocresol green also?	The book leaves this question open and does not list bromocresol green in Table 9.1. Like methyl orange it changes colour on the acid side, so it can also be used here, but methyl orange is the book's best indicator for this titration.

9.6Strong and weak acids

The extent of ionisation and the acid dissociation constant K_a can be used to distinguish between strong and weak acids. The strength of an acid is generally expressed in terms of the acid ionisation constant, K_a.

Derivation Acid dissociation constant, K_a

Aim. Obtain the expression for K_a of a general acid \ce{HX} in water.

Write the equilibrium established when the acid \ce{HX} ionises in water.
\ce{HX(aq) + H2O(l) <=> H3O+(aq) + X-(aq)}
Write the equilibrium constant K for this ionisation.
K=\dfrac{[\ce{H3O+}][\ce{X-}]}{[\ce{HX}][\ce{H2O}]}
Multiply both sides by [\ce{H2O}].
K[\ce{H2O}]=\dfrac{[\ce{H3O+}][\ce{X-}]}{[\ce{HX}]}
Water is the solvent and is present in excess, so its concentration may be regarded as constant. K[\ce{H2O}] is therefore another constant, designated K_a.
K[\ce{H2O}]=K_a=\dfrac{[\ce{H3O+}][\ce{X-}]}{[\ce{HX}]}

ResultK_a=\dfrac{[\ce{H3O+}][\ce{X-}]}{[\ce{HX}]}

K_a is called the acid dissociation constant. It is a measure of the extent to which an acid is ionised or dissociated at the equilibrium state.

K_a depends on temperature. Therefore the value of K_a should be mentioned along with the temperature at which it was determined. The dissociation constant of acetic acid in water at 25^\circ\mathrm{C} is K_a=1.8\times10^{-5}. The comparison of K_a values of different acids provides a method to compare their strengths.

pK_a

The values of K_a are usually inconvenient numbers. For convenience they are converted to \mathrm{p}K_a values:

\mathrm{p}K_a=-\log K_a

Since \mathrm{p}K_a is the negative logarithm of K_a, the **smaller the value of \mathrm{p}K_a, the stronger the acid**, because a smaller \mathrm{p}K_a corresponds to a greater K_a. Table 9.2 lists the ionisation constants and \mathrm{p}K_a values of some common acids in water at 25^\circ\mathrm{C}.

Table 9.2: Ionisation constants and \mathrm{p}K_a of acids
Name of acid	Formula	K_a	\mathrm{p}K_a
Perchloric acid	\ce{HClO4}	1.0\times10^{10}	-10.0
Hydroiodic acid	\ce{HI}	1.0\times10^{10}	-10.0
Hydrobromic acid	\ce{HBr}	1.0\times10^{9}	-9.0
Hydrochloric acid	\ce{HCl}	1.0\times10^{6}	-6.0
Sulphuric acid	\ce{H2SO4}	1.0\times10^{3}	-3.0
Hydrofluoric acid	\ce{HF}	7.2\times10^{-4}	+3.1
Formic acid	\ce{HCOOH}	1.8\times10^{-4}	+3.75
Benzoic acid	\ce{C6H5COOH}	6.3\times10^{-5}	+4.2
Acetic acid	\ce{CH3COOH}	1.8\times10^{-5}	+4.7
Phenol	\ce{C6H5OH}	1.3\times10^{-10}	+8.9 (see note)
Water	\ce{H2O}	1.8\times10^{-16}	+15.7

The book asks which acid is strongest and which is weakest. Perchloric acid (\ce{HClO4}) and hydroiodic acid (\ce{HI}) are the strongest in the table (largest K_a=1.0\times10^{10}, lowest \mathrm{p}K_a=-10.0). Water is the weakest (smallest K_a=1.8\times10^{-16}, highest \mathrm{p}K_a=+15.7).

The book also asks: which acid is stronger, \ce{HCl} or \ce{HF}? It is \ce{HCl}, because its K_a (1.0\times10^{6}) is far greater than that of \ce{HF} (7.2\times10^{-4}), that is, its \mathrm{p}K_a (-6.0) is much lower than that of \ce{HF} (+3.1).

Strong and weak acids compared (using the book's table)
	Strong acid	Weak acid
Ionisation in water	Almost complete	Partial (small extent), an equilibrium is set up
K_a	Large	Small
\mathrm{p}K_a	Small (even negative)	Large
Examples from Table 9.2	\ce{HClO4}, \ce{HI}, \ce{HBr}, \ce{HCl}, \ce{H2SO4}	\ce{HF}, \ce{HCOOH}, \ce{C6H5COOH}, \ce{CH3COOH}, \ce{C6H5OH}

Example 9.5: [H+] in 1.0 M hydrofluoric acid

Calculate the concentration of \ce{H+} ions of a solution that contains 1.0\,M \ce{HF} (K_a=7.2\times10^{-4}).

Given

[\ce{HF}]_\text{initial} = 1.0\,M
K_a = 7.2\times10^{-4}

To find

[\ce{H+}]

Write the ionisation equilibrium and let x mol \mathrm{dm^{-3}} of \ce{HF} ionise. Each \ce{HF} that ionises gives one \ce{H+} and one \ce{F-}. Set out the concentrations (in \mathrm{mol\,dm^{-3}}) as in the table below.
\ce{HF(aq) <=> H+(aq) + F-(aq)}

	\ce{HF}	\ce{H+}	\ce{F-}
Initial conc.	1.0	0	0
Equilibrium conc.	1.0-x	x	x

Write the expression for K_a and substitute the equilibrium values.
K_a=\dfrac{[\ce{H+}][\ce{F-}]}{[\ce{HF}]}\quad\Rightarrow\quad 7.2\times10^{-4}=\dfrac{x\cdot x}{1.0-x}
Since x is very small compared with 1.0, the term in the denominator can be approximated: 1.0-x\approx1.0.
7.2\times10^{-4}=\dfrac{x^2}{1.0}
Solve for x by taking the square root.
x=\sqrt{7.2\times10^{-4}}=2.68\times10^{-2}=0.0268\,M
Since x=[\ce{H+}],
[\ce{H+}]=0.0268\,M

[\ce{H+}]=0.0268\,M\;(\text{the book prints }0.268\,M)

Worked example: pH of a weak acid and comparing acid strengths

(a) Calculate the pH of 0.10\,M acetic acid, \ce{CH3COOH}, with K_a=1.8\times10^{-5}. (b) Acid X has K_a=4.0\times10^{-6} and acid Y has \mathrm{p}K_a=2.85. Which is the stronger acid?

(a) Write the ionisation. Let x be the concentration of \ce{H+} formed. Then [\ce{CH3COO-}]=x and [\ce{CH3COOH}]=0.10-x.
\ce{CH3COOH <=> H+ + CH3COO-}\qquad K_a=\dfrac{x\cdot x}{0.10-x}
The acid is weak, so x is very small compared with 0.10. Take 0.10-x\approx0.10.
1.8\times10^{-5}=\dfrac{x^{2}}{0.10}\;\Rightarrow\;x^{2}=1.8\times10^{-6}
Solve and find the pH. Check: x=1.3\times10^{-3} is only about 1\% of 0.10, so the approximation was fair.
x=[\ce{H+}]=1.34\times10^{-3}\,M,\qquad \mathrm{pH}=-\log(1.34\times10^{-3})=2.87
(b) Convert to the same quantity. Acid X: \mathrm{p}K_a=-\log(4.0\times10^{-6})=5.40. Acid Y: K_a=10^{-2.85}=1.4\times10^{-3}.
A larger K_a, or a smaller \mathrm{p}K_a, means a stronger acid. Acid Y has \mathrm{p}K_a=2.85, smaller than 5.40 for X, and its K_a (1.4\times10^{-3}) is greater than 4.0\times10^{-6}.
\mathrm{p}K_a(\text{Y})=2.85<\mathrm{p}K_a(\text{X})=5.40\;\Rightarrow\;\text{Y is stronger}

\mathrm{pH}=2.87;\quad\text{acid Y is stronger}

9.7Strong and weak bases

The strength of a base is its ability to accept a proton from a solvent. Hydroxides of alkali metals such as sodium hydroxide and potassium hydroxide are strong bases and ionise completely in aqueous solution.

\ce{NaOH(aq) -> Na+(aq) + OH-(aq)}

\ce{KOH(aq) -> K+(aq) + OH-(aq)}

The \ce{OH-} ion thus formed is a Brønsted base because it can accept a proton \ce{H+}.

\ce{OH- + H+ -> H2O}

Base ionisation constant, K_b

The ability of a base to accept a proton from an acid, usually water, is termed the strength of the base. For a base \ce{B}, the equilibrium reaction with water can be represented by

\ce{B(aq) + H2O(l) <=> BH+(aq) + OH-(aq)}

The equilibrium constant K_b is called the base ionisation constant. It is derived in the same way as K_a for acids: [\ce{H2O}] is constant and is absorbed into K_b.

K_b=\dfrac{[\ce{BH+}][\ce{OH-}]}{[\ce{B}]}

K_b will be large if the degree of ionisation of the base \ce{B} is high, that is, if the base is strong. K_b will be small for a weak base. For convenience, a parameter \mathrm{p}K_b has been devised to express K_b in convenient numbers. It is defined as the negative logarithm of K_b.

\mathrm{p}K_b=-\log K_b

Table 9.3: K_b and \mathrm{p}K_b values of some common bases
Name of base	Formula	K_b	\mathrm{p}K_b
Diethylamine	\ce{(C2H5)2NH}	9.6\times10^{-4}	3.02
Ethylamine	\ce{C2H5NH2}	5.6\times10^{-4}	3.25
Methylamine	\ce{CH3NH2}	4.5\times10^{-4}	3.34
Ammonia	\ce{NH3}	1.7\times10^{-5}	4.76
Pyridine	\ce{C5H5N}	5.6\times10^{-9}	8.25
Aniline	\ce{C6H5NH2}	4.3\times10^{-10}	9.37

According to these values, ammonia is a stronger base than pyridine and aniline but weaker than methylamine and ethylamine. Also, diethylamine is the strongest base among all those listed in the table.

9.8Buffer solutions and their applicationsOften examined

A buffer solution is a solution, the pH of which does not change significantly when a small amount of acid or base is added to it. Such a solution has a constant pH that does not change on keeping it for a long time.

9.8.1Types of buffer solutionsOften examined

A buffer solution can be made in two ways.

Types of buffer solutions
Type	Made by mixing	pH	Example
Acidic buffer	A weak acid and a salt of it with a strong base	Less than 7	\ce{CH3COOH + CH3COONa}
Basic buffer	A weak base and a salt of it with a strong acid	More than 7	\ce{NH4OH + NH4Cl}

In the acidic buffer the weak acid is \ce{CH3COOH} and the salt supplies its conjugate base, the acetate ion. In the basic buffer the weak base is \ce{NH4OH} (aqueous ammonia) and the salt \ce{NH4Cl} supplies its conjugate acid, the ammonium ion.

9.8.2Calculation of pH of buffer solutionOften examined

The concentration of the conjugate base in the reaction mixture is predominantly supplied by the salt, which is a strong electrolyte. Therefore, assuming that the concentration of the conjugate base is equal to that of the salt, and that the equilibrium concentration of the acid is its original concentration, the pH of a buffer can be calculated. The pH of a buffer is calculated by using the Henderson-Hasselbalch equation.

Henderson-Hasselbalch equations
Buffer	Equation
Acidic buffer	\mathrm{pH}=\mathrm{p}K_a+\log\dfrac{[\text{salt}]}{[\text{acid}]}
Basic buffer	\mathrm{pOH}=\mathrm{p}K_b+\log\dfrac{[\text{salt}]}{[\text{base}]}

Example 9.6: pH of an acetic acid and sodium acetate buffer

What is the pH of a buffer if the concentration of \ce{CH3COOH} is 0.1\,M and of \ce{CH3COONa} is 1.0\,M? \mathrm{p}K_a for \ce{CH3COOH} is 4.76.

Given

[\text{acid}] = 0.1\,M \ce{CH3COOH}
[\text{salt}] = 1.0\,M \ce{CH3COONa}
\mathrm{p}K_a = 4.76

To find

\mathrm{pH}

This is an acidic buffer (weak acid and its salt), so use the Henderson-Hasselbalch equation.
\mathrm{pH}=\mathrm{p}K_a+\log\dfrac{[\text{salt}]}{[\text{acid}]}
Substitute the values. The ratio is \dfrac{1.0}{0.1}=\dfrac{10}{1}.
\mathrm{pH}=4.76+\log\left(\dfrac{1.0}{0.1}\right)=4.76+\log10
Since \log10=1,
\mathrm{pH}=4.76+1.00=5.76

\mathrm{pH}=5.76

Worked example: a buffer resists added acid

A buffer contains 0.20\,M \ce{CH3COOH} and 0.10\,M \ce{CH3COONa} (K_a=1.8\times10^{-5}, so \mathrm{p}K_a=4.74). (a) Find its pH. (b) Then 0.010\,\mathrm{mol} of \ce{HCl} is added to 1\,\mathrm{dm^3} of this buffer (take the volume as unchanged). Find the new pH and compare it with 0.010\,M \ce{HCl} in pure water.

pH of an acetic acid and acetate buffer (\mathrm{p}K_a=4.74) against \log([\text{salt}]/[\text{acid}]). The line has slope 1 and passes through \mathrm{p}K_a where the two concentrations are equal.

(a) Use the Henderson-Hasselbalch equation for an acidic buffer. The salt gives the acetate ion, so [\text{salt}]=0.10 and [\text{acid}]=0.20.
\mathrm{pH}=\mathrm{p}K_a+\log\dfrac{[\text{salt}]}{[\text{acid}]}=4.74+\log\dfrac{0.10}{0.20}=4.74-0.30=4.44
(b) The added \ce{H+} is used up by the acetate ion, the conjugate base. So 0.010\,M of acetate is lost and 0.010\,M of acetic acid is formed.
\ce{CH3COO- + H+ -> CH3COOH}
New amounts: acid =0.20+0.010=0.21\,M, salt =0.10-0.010=0.09\,M.
\mathrm{pH}=4.74+\log\dfrac{0.09}{0.21}=4.74-0.37=4.37
Compare. In pure water, 0.010\,M \ce{HCl} has [\ce{H+}]=0.010, so \mathrm{pH}=2.00: a fall of 5 units from the neutral 7. The buffer changed by only 0.07, from 4.44 to 4.37.
\Delta\mathrm{pH}=4.44-4.37=0.07
The buffer holds its pH because it contains a large supply of both the weak acid and its conjugate base, so the ratio changes only a little (the two points close together on the graph).

\mathrm{pH}=4.44\ \text{before},\quad4.37\ \text{after adding HCl}

Concept assessment exercise 9.4

Concept assessment exercise 9.4: pH of an acetate buffer

Calculate the pH of a buffer solution in which 0.11 molar \ce{CH3COONa} and 0.09 molar \ce{CH3COOH} solutions are present. K_a for \ce{CH3COOH} is 1.8\times10^{-5}.

Given

[\text{salt}] = 0.11\,M
[\text{acid}] = 0.09\,M
K_a = 1.8\times10^{-5}

To find

\mathrm{pH}

The equation needs \mathrm{p}K_a, so convert K_a first.
\mathrm{p}K_a=-\log(1.8\times10^{-5})=4.74
Write the Henderson-Hasselbalch equation and substitute.
\mathrm{pH}=\mathrm{p}K_a+\log\dfrac{[\text{salt}]}{[\text{acid}]}=4.74+\log\dfrac{0.11}{0.09}
Evaluate the logarithm: \dfrac{0.11}{0.09}=1.222 and \log1.222=0.087.
\mathrm{pH}=4.74+0.087=4.83

\mathrm{pH}=4.83

pH of some biological systems
Biological system	pH range
Human blood	7.35 to 7.45
Tears	7.40
Stomach	1.65 to 1.75
Milk	6.7 to 6.8
Egg white (protein)	8.0 to 8.1

9.9Salt hydrolysisOften examined

Consider the following observations.

An aqueous solution of \ce{NH4Cl} turns blue litmus red.
An aqueous solution of \ce{K2CO3} turns red litmus blue.
An aqueous solution of \ce{NaCl} has no action on litmus solutions.

These observations can be explained on the basis of the Brønsted-Lowry acid-base theory. When a salt (\ce{MX}) is dissolved in water, it splits up into its \ce{M+} and \ce{X-} ions. These ions may react with water and give the following reactions.

\ce{M+ + H-OH(l) <=> MOH(aq) + H+(aq)}

\ce{X-(aq) + H-OH(l) <=> HX(aq) + OH-(aq)}

Since \ce{H+} and \ce{OH-} ions are produced in these reactions, the solution of the salt may be acidic or basic. In salts the anions are derived from acids and the cations from bases.

Anions of weak acids: basic solutions

The anions of weak acids are strong conjugate bases. Such anions react with water, producing basic solutions. Examples are \ce{CH3COO-}, \ce{CN-}, \ce{HCO3-} and \ce{CO3^2-}.

\ce{CH3COO-(aq) + H-OH(l) <=> CH3COOH(aq) + OH-(aq)}

\ce{CN-(aq) + H-OH(l) <=> HCN(aq) + OH-(aq)}

\ce{CO3^2-(aq) + 2H-OH(l) <=> H2CO3(aq) + 2OH-(aq)}

\ce{HCO3-(aq) + H-OH(l) <=> H2CO3(aq) + OH-(aq)}

Anions like \ce{Cl-}, \ce{NO3-} and \ce{SO4^2-} are such weak conjugate bases that they do not react with water.

Cations of weak bases: acidic solutions

Cations of weak bases are strong conjugate acids. Such cations react with water, producing acidic solutions. Examples are \ce{Cu^2+}, \ce{Al^3+} and \ce{NH4+}.

\ce{Cu^2+(aq) + 2H-OH(l) <=> Cu(OH)2(aq) + 2H+(aq)}

\ce{NH4+(aq) + H-OH(l) <=> NH4OH(aq) + H+(aq)}

Cations like \ce{Na+}, \ce{K+}, \ce{Ca^2+} and \ce{Mg^2+} are so weak as conjugate acids that they do not react with water.

Which ions hydrolyse (conjugate acid-base idea)
Ion	Comes from	Conjugate nature	Reaction with water	Solution
\ce{CH3COO-}, \ce{CN-}, \ce{CO3^2-}, \ce{HCO3-}	A weak acid	Strong conjugate base	Reacts, gives \ce{OH-}	Basic
\ce{Cl-}, \ce{NO3-}, \ce{SO4^2-}	A strong acid	Very weak conjugate base	None	No effect
\ce{Cu^2+}, \ce{Al^3+}, \ce{NH4+}	A weak base	Strong conjugate acid	Reacts, gives \ce{H+}	Acidic
\ce{Na+}, \ce{K+}, \ce{Ca^2+}, \ce{Mg^2+}	A strong base	Very weak conjugate acid	None	No effect

Hydrolysis: The chemical reaction of cations and anions of salts with water.

These reactions are called hydrolysis reactions. Hydrolysis is different from hydration. In hydrolysis the \ce{H-OH} bond is broken, whereas in hydration a water molecule adds on to a substance without bond breakage.

Four types of salts

There are four types of salts on the basis of their reactivity with water.

Salts of strong acids and strong bases do not hydrolyse (\mathrm{pH}=7). Examples: \ce{NaCl}, \ce{Na2SO4}, \ce{KNO3}.
Salts of weak acids and strong bases hydrolyse, producing basic solutions (\mathrm{pH}>7). Examples: \ce{CH3COONa}, \ce{NaCN}, \ce{Na2S}.
Salts of strong acids and weak bases hydrolyse, producing acidic solutions (\mathrm{pH}<7). Examples: \ce{CuSO4}, \ce{NH4Cl}, \ce{NH4NO3}.
Salts of weak acids and weak bases hydrolyse, but the resulting solution is either neutral, acidic or basic. This depends upon the relative values of K_a and K_b of the cations and anions of the salt.

The important aspects of salt hydrolysis are summarised in Table 9.5.

Table 9.5: Salt hydrolysis
Acid	Base	Common example	Ions which impart hydrolysis	Solution pH (nature)
Strong	Strong	\ce{NaCl}, \ce{KBr}	None	=7.0 (neutral)
Strong	Weak	\ce{NH4NO3}, \ce{NH4Cl}	Cations	<7.0 (acidic)
Weak	Strong	\ce{NaCN}, \ce{K2CO3}	Anions	>7.0 (basic)
Weak	Weak	\ce{NH4CN}, \ce{NH4NO2}	Anions and cations	May be equal to, smaller or greater than 7.0

9.10Solubility product and precipitation reactionsOften examined

This section discusses some of the important equilibria which have analytical importance. They apply the idea of chemical equilibrium to ionic compounds that dissolve only slightly in water.

Solubility product (9.10.1): the equilibrium between an insoluble solid and its ions in a saturated solution, and the constant K_{sp} that describes it.
Precipitation reactions (9.10.2): predicting whether a solid will form when two solutions are mixed, by comparing the ion product with K_{sp}.

9.10.1Solubility productOften examined

When an excess of a sparingly soluble ionic compound is mixed with water, some of it dissolves and the rest of the compound settles to the bottom. A dynamic equilibrium is reached between the insoluble solid compound and its ions in a saturated solution. For example, when calcium fluoride is mixed with water the following equilibrium occurs.

\ce{CaF2(s) <=> Ca^2+(aq) + 2F-(aq)}

A saturated solution of \ce{CaF2}: the solid and its ions are in dynamic equilibrium, so the rate of dissolving equals the rate of precipitation.

Derivation Solubility product constant, K_{sp}

Aim. Obtain the expression for K_{sp} of \ce{CaF2} and then of a general slightly soluble salt \ce{A_mB_n}.

Write K_c for the equilibrium of \ce{CaF2}.
K_c=\dfrac{[\ce{Ca^2+}][\ce{F-}]^2}{[\ce{CaF2}]}
\ce{CaF2} is a slightly soluble salt, so the concentration of the solid almost remains constant. Multiply both sides by [\ce{CaF2}].
K_c[\ce{CaF2}]=[\ce{Ca^2+}][\ce{F-}]^2
The left-hand side is a product of constants, so it is a new constant, K_{sp}.
K_{sp}=[\ce{Ca^2+}][\ce{F-}]^2
For any slightly soluble ionic compound \ce{A_mB_n}, the equilibrium is \ce{A_mB_n(s) <=> mA^{n+}(aq) + nB^{m-}(aq)}. The same argument gives
K_{sp}=[\ce{A^{n+}}]^m[\ce{B^{m-}}]^n

ResultK_{sp}=[\ce{A^{n+}}]^m[\ce{B^{m-}}]^n

K_{sp} is the solubility product constant: the product of the equilibrium concentrations of the ions, each raised to a power equal to the number of such ions in the formula unit of the compound (the coefficient of the ion in the balanced equation).

9.10.2Precipitation reactionsOften examined

In the previous section we considered solids dissolving in solutions. Now we consider the reverse process, the formation of a solid from solution. An aqueous reaction that takes place when two or more solutions are mixed together, yielding a solid insoluble substance, is called a precipitation reaction.

To predict whether a precipitate will form when two solutions are mixed, the term ion product (Q') is used. It is obtained by substituting the initial concentrations instead of the equilibrium concentrations in the expression for K_{sp}. For example, the ion product expression for solid \ce{CaF2} is

Q'=[\text{initial conc. of }\ce{Ca^2+}]\,[\text{initial conc. of }\ce{F-}]^2

If we add a solution containing \ce{Ca^2+} ions to a solution containing \ce{F-} ions, a precipitate may or may not form. To predict whether precipitation will occur, we compare Q' and K_{sp}. There are two possibilities.

Predicting precipitation
Condition	Result
Q'>K_{sp}	Precipitation occurs and continues until the concentrations satisfy K_{sp}
Q'<K_{sp}	Precipitation does not occur

Example 9.7: Ksp of silver bromide from its solubility

The solubility of \ce{AgBr} is 7.1\times10^{-7}\,M at 25^\circ\mathrm{C}. Calculate its K_{sp}.

Given

\text{solubility} = 7.1\times10^{-7}\,M \ce{AgBr} at 25^\circ\mathrm{C}

To find

K_{sp}

Write the dissolving equilibrium. 7.1\times10^{-7}\,M of dissolved \ce{AgBr} produces equal concentrations of \ce{Ag+} and \ce{Br-} ions.
\ce{AgBr(s) <=> Ag+(aq) + Br-(aq)}\qquad [\ce{Ag+}]=[\ce{Br-}]=7.1\times10^{-7}\,M
Write the K_{sp} expression.
K_{sp}=[\ce{Ag+}][\ce{Br-}]
Substitute the ion concentrations.
K_{sp}=(7.1\times10^{-7})(7.1\times10^{-7})=5.041\times10^{-13}

K_{sp}=5.041\times10^{-13}

Worked example: Ksp from solubility and solubility from Ksp

(a) The solubility of silver chromate, \ce{Ag2CrO4}, is 6.5\times10^{-5}\,\mathrm{mol\,dm^{-3}}. Calculate K_{sp}. (b) K_{sp} of magnesium hydroxide, \ce{Mg(OH)2}, is 5.6\times10^{-12}. Calculate its solubility in \mathrm{mol\,dm^{-3}} and in \mathrm{g\,dm^{-3}} (molar mass 58.3\,\mathrm{g\,mol^{-1}}).

(a) Write the dissolving equation and the K_{sp} expression. Let the solubility be s. Each formula unit gives 2 silver ions and 1 chromate ion.
\ce{Ag2CrO4(s) <=> 2Ag+(aq) + CrO4^2-(aq)},\quad [\ce{Ag+}]=2s,\ [\ce{CrO4^2-}]=s
Substitute into K_{sp}=[\ce{Ag+}]^2[\ce{CrO4^2-}]. The power 2 applies to the whole 2s.
K_{sp}=(2s)^{2}(s)=4s^{3}=4\times(6.5\times10^{-5})^{3}=1.1\times10^{-12}
(b) Write the equation. Each formula unit gives 1 magnesium ion and 2 hydroxide ions.
\ce{Mg(OH)2(s) <=> Mg^2+(aq) + 2OH-(aq)},\quad K_{sp}=s\,(2s)^{2}=4s^{3}
Solve for s.
s^{3}=\dfrac{5.6\times10^{-12}}{4}=1.4\times10^{-12}\;\Rightarrow\;s=1.1\times10^{-4}\,\mathrm{mol\,dm^{-3}}
Convert to \mathrm{g\,dm^{-3}} by multiplying by the molar mass.
1.1\times10^{-4}\times58.3=6.5\times10^{-3}\,\mathrm{g\,dm^{-3}}
Both salts are of the \ce{AB2} or \ce{A2B} type, so K_{sp}=4s^{3} for each. Compare \ce{AgBr} (Example 9.7), where K_{sp}=s^{2}: the formula depends on how many ions the salt gives.

\begin{aligned}K_{sp}(\ce{Ag2CrO4})&=1.1\times10^{-12}\\ s(\ce{Mg(OH)2})&=1.1\times10^{-4}\,\mathrm{mol\,dm^{-3}}=6.5\times10^{-3}\,\mathrm{g\,dm^{-3}}\end{aligned}

Concept assessment exercise 9.5

Concept assessment exercise 9.5 (i): Ksp expressions

Write K_{sp} expressions for (a) iron(II) hydroxide and (b) calcium sulphate.

(a) Iron(II) hydroxide is \ce{Fe(OH)2}. Write the dissolving equation, then copy the coefficients as powers.
\ce{Fe(OH)2(s) <=> Fe^2+(aq) + 2OH-(aq)}\qquad K_{sp}=[\ce{Fe^2+}][\ce{OH-}]^2
(b) Calcium sulphate is \ce{CaSO4}.
\ce{CaSO4(s) <=> Ca^2+(aq) + SO4^2-(aq)}\qquad K_{sp}=[\ce{Ca^2+}][\ce{SO4^2-}]

K_{sp}(\ce{Fe(OH)2})=[\ce{Fe^2+}][\ce{OH-}]^2,\qquad K_{sp}(\ce{CaSO4})=[\ce{Ca^2+}][\ce{SO4^2-}]

Concept assessment exercise 9.5 (ii): solubility of lead(II) sulphate

Lead(II) sulphate is used as a white pigment. What is the solubility of \ce{PbSO4}? K_{sp}=1.96\times10^{-8} at 25^\circ\mathrm{C}.

Given

K_{sp} = 1.96\times10^{-8} 25^\circ\mathrm{C}

To find

solubility s of \ce{PbSO4}

Write the equilibrium. Let the solubility be s mol \mathrm{dm^{-3}}. Each formula unit gives one \ce{Pb^2+} and one \ce{SO4^2-}.
\ce{PbSO4(s) <=> Pb^2+(aq) + SO4^2-(aq)}\qquad [\ce{Pb^2+}]=[\ce{SO4^2-}]=s
Write the K_{sp} expression and substitute.
K_{sp}=[\ce{Pb^2+}][\ce{SO4^2-}]=s^2=1.96\times10^{-8}
Take the square root.
s=\sqrt{1.96\times10^{-8}}=1.4\times10^{-4}\,M

s=1.4\times10^{-4}\,M

Concept assessment exercise 9.5 (iii): does calcium phosphate precipitate?

Phosphate in natural water often precipitates as insoluble \ce{Ca3(PO4)2}. In the Indus river the concentration of \ce{Ca^2+} and \ce{PO4^3-} ions is 1.0\times10^{-9}\,M each. Will calcium phosphate precipitate? K_{sp}=1.2\times10^{-29} at 25^\circ\mathrm{C}.

Given

[\ce{Ca^2+}] = 1.0\times10^{-9}\,M
[\ce{PO4^3-}] = 1.0\times10^{-9}\,M
K_{sp} = 1.2\times10^{-29}

To find

will \ce{Ca3(PO4)2} precipitate?

Write the dissolving equilibrium and the ion product with the coefficients as powers.
\ce{Ca3(PO4)2(s) <=> 3Ca^2+(aq) + 2PO4^3-(aq)}\qquad Q'=[\ce{Ca^2+}]^3[\ce{PO4^3-}]^2
Substitute the given concentrations.
Q'=(1.0\times10^{-9})^3(1.0\times10^{-9})^2=1.0\times10^{-45}
Compare with K_{sp}. Since 1.0\times10^{-45} is far smaller than 1.2\times10^{-29}, Q'<K_{sp}, so precipitation does not occur.
Q'=1.0\times10^{-45}<K_{sp}=1.2\times10^{-29}

Q'<K_{sp}\;\Rightarrow\;\text{no precipitate}

Worked example: will a precipitate form on mixing?

Equal volumes of silver nitrate and sodium chloride solutions are mixed (K_{sp} of \ce{AgCl}=1.8\times10^{-10}). Decide whether \ce{AgCl} precipitates when (a) 2.0\times10^{-4}\,M \ce{AgNO3} is mixed with 2.0\times10^{-4}\,M \ce{NaCl}, and (b) 2.0\times10^{-6}\,M of each is mixed. Then find [\ce{Ag+}] needed to start precipitation from 0.010\,M \ce{Cl-}.

Mixing equal volumes doubles the total volume, so each concentration is halved before Q' is worked out. Forgetting this is the commonest mistake.
(a) After mixing: [\ce{Ag+}]=[\ce{Cl-}]=1.0\times10^{-4}\,M. Write Q' for \ce{AgCl(s) <=> Ag+ + Cl-} and compare with K_{sp}.
Q'=[\ce{Ag+}][\ce{Cl-}]=(1.0\times10^{-4})^{2}=1.0\times10^{-8}>1.8\times10^{-10}\;\Rightarrow\;\text{precipitate forms}
(b) After mixing: [\ce{Ag+}]=[\ce{Cl-}]=1.0\times10^{-6}\,M.
Q'=(1.0\times10^{-6})^{2}=1.0\times10^{-12}<1.8\times10^{-10}\;\Rightarrow\;\text{no precipitate}

The ion product Q' of each mixture on a log scale. Case (a) lies to the right of K_{sp} (accent), so \ce{AgCl} precipitates. Case (b) lies to the left, so the solution stays clear.

(c) Precipitation starts when Q' just reaches K_{sp}. Put [\ce{Cl-}]=0.010\,M into K_{sp}=[\ce{Ag+}][\ce{Cl-}] and solve.
[\ce{Ag+}]=\dfrac{K_{sp}}{[\ce{Cl-}]}=\dfrac{1.8\times10^{-10}}{0.010}=1.8\times10^{-8}\,M
Any [\ce{Ag+}] above 1.8\times10^{-8}\,M in this solution makes Q'>K_{sp} and starts the precipitation of \ce{AgCl}.

\text{(a) precipitate};\quad\text{(b) none};\quad[\ce{Ag+}]=1.8\times10^{-8}\,M

9.11Common ion effect

An interesting situation arises when a weak electrolyte and a salt containing a common ion are present simultaneously in an aqueous solution. For example, in a solution of the weak acid hydrofluoric acid (K_a=7.2\times10^{-4}), its salt sodium fluoride produces the common ion \ce{F-}.

\ce{HF(aq) <=>[H2O] H+(aq) + F-(aq)}

\ce{NaF(s) ->[H2O] Na+(aq) + F-(aq)}

Why the ionisation of HF falls

A strong electrolyte breaks up completely into its ions. \ce{HF} is a weak electrolyte, so it dissociates only slightly. \ce{NaF} is a strong electrolyte and dissociates completely.
The common ion \ce{F-} produced by \ce{NaF} disturbs the \ce{HF} equilibrium. It increases the concentration of \ce{F-} ions.
According to Le Chatelier's principle the equilibrium shifts to the left to use up some of the \ce{F-} ions. This reduces the dissociation of \ce{HF}.
\ce{HF <=> H+ + F-}\quad(\text{shifts left})
So the dissociation of \ce{HF} is reduced in the presence of dissolved \ce{NaF}, and the \ce{HF} concentration increases as a result of the equilibrium change.

Similarly, when a highly soluble salt is added to a saturated solution of a less soluble salt containing a common ion, the degree of dissociation of the less soluble salt decreases. Therefore its solubility is reduced. The term common ion effect is used to describe the behaviour of a solution in which two different compounds produce the same ion.

Common ion effect: The phenomenon in which the degree of ionisation or solubility of an electrolyte is reduced by the addition of a highly soluble electrolyte containing a common ion.

Example 9.8: potassium perchlorate and potassium chloride

Potassium perchlorate \ce{KClO4} is moderately soluble in water. When highly soluble \ce{KCl} is added to a saturated solution of \ce{KClO4}, it causes an increase in the concentration of \ce{K+} ions. Explain what happens.

Write the two dissolving equations. \ce{K+} is common to both.
\ce{KClO4(s) <=> K+(aq) + ClO4-(aq)}\qquad \ce{KCl(s) -> K+(aq) + Cl-(aq)}
The added \ce{KCl} ionises completely and raises [\ce{K+}].
According to Le Chatelier's principle the \ce{K+} ions react with \ce{ClO4-} ions to form \ce{KClO4}, so the first equilibrium shifts to the left. The ionisation of \ce{KClO4} is suppressed, and \ce{KClO4} precipitates out.

\text{the solubility of }\ce{KClO4}\text{ decreases and it precipitates}

Example 9.9: hydrogen chloride and brine

When \ce{HCl} gas is passed through a saturated solution of \ce{NaCl} (brine), it increases the concentration of \ce{Cl-} ions. Explain what happens.

Write the equilibria. \ce{Cl-} is common to both.
\ce{NaCl(s) <=> Na+(aq) + Cl-(aq)}\qquad \ce{HCl(aq) -> H+(aq) + Cl-(aq)}
The \ce{HCl} ionises completely, so [\ce{Cl-}] rises.
According to Le Chatelier's principle the \ce{Cl-} ions combine with \ce{Na+} ions to form a precipitate of pure \ce{NaCl}.

\text{pure }\ce{NaCl}\text{ precipitates}

Concept assessment exercise 9.6

Concept assessment exercise 9.6 (i): ammonium chloride added to ammonium hydroxide

Ammonium chloride, \ce{NH4Cl}, is a water soluble salt. What will happen if this salt is added to a solution containing ammonium hydroxide? \ce{NH4OH(aq) <=> NH4+(aq) + OH-(aq)}

\ce{NH4Cl} is a strong electrolyte and ionises completely, so it adds \ce{NH4+} ions, which are common to the equilibrium.
\ce{NH4Cl -> NH4+ + Cl-}
By Le Chatelier's principle the equilibrium shifts to the left to use up the extra \ce{NH4+}.
\ce{NH4OH <=> NH4+ + OH-}\quad(\text{shifts left})
The ionisation of \ce{NH4OH} is suppressed (common ion effect), so [\ce{OH-}] decreases and the solution becomes less basic (its pH falls).

\text{ionisation of }\ce{NH4OH}\text{ is suppressed; }[\ce{OH-}]\text{ falls}

Concept assessment exercise 9.6 (ii): sodium carbonate added to carbonic acid

Carbonic acid is a weak acid. It ionises in water as follows: \ce{H2CO3(aq) <=> 2H+(aq) + CO3^2-(aq)}. What will happen if a strong electrolyte such as \ce{Na2CO3} is added to a solution containing carbonic acid?

\ce{Na2CO3} is a strong electrolyte and ionises completely, so it adds \ce{CO3^2-} ions, which are common to the equilibrium.
\ce{Na2CO3 -> 2Na+ + CO3^2-}
By Le Chatelier's principle the equilibrium shifts to the left to use up the extra \ce{CO3^2-}.
\ce{H2CO3 <=> 2H+ + CO3^2-}\quad(\text{shifts left})
The ionisation of \ce{H2CO3} is suppressed (common ion effect), so [\ce{H+}] falls and the concentration of undissociated \ce{H2CO3} increases.

\text{ionisation of }\ce{H2CO3}\text{ is suppressed; }[\ce{H+}]\text{ falls}

Formula sheet

Ionic product of water

K_w=[\ce{H+}][\ce{OH-}]=1.0\times10^{-14}

[\ce{H+}] and [\ce{OH-}] are the molar concentrations of hydrogen and hydroxide ions in an aqueous solution. The value is for 25^\circ\mathrm{C}; K_w changes with temperature (for example 3.8\times10^{-14} at 40^\circ\mathrm{C}).

\mathrm{mol^2\,dm^{-6}} (usually written without a unit)

\mathrm{pH}=-\log[\ce{H+}]

[\ce{H+}] is the hydrogen ion concentration in \mathrm{mol\,dm^{-3}}. At 25^\circ\mathrm{C} acidic means \mathrm{pH}<7, neutral \mathrm{pH}=7 and basic \mathrm{pH}>7.

no unit

pOH

\mathrm{pOH}=-\log[\ce{OH-}]

[\ce{OH-}] is the hydroxide ion concentration in \mathrm{mol\,dm^{-3}}.

no unit

Relation between pH and pOH

\mathrm{pH}+\mathrm{pOH}=14

Obtained by taking the logarithm of K_w=1.0\times10^{-14}; valid at 25^\circ\mathrm{C}.

no unit

Titration equation

\dfrac{M_1V_1}{n_1}=\dfrac{M_2V_2}{n_2}

M_1,V_1,n_1: molarity, volume and number of moles (coefficient in the balanced equation) of the base; M_2,V_2,n_2: the same for the acid. V_1 and V_2 must be in the same unit.

Units: M in \mathrm{mol\,dm^{-3}}; V in \mathrm{cm^3} or \mathrm{dm^3}

Strength of a solution

\text{strength}=\text{molarity}\times\text{molar mass}

Molarity in \mathrm{mol\,dm^{-3}} and molar mass in \mathrm{g\,mol^{-1}}.

Unit: \mathrm{g\,dm^{-3}}

Acid dissociation constant

K_a=\dfrac{[\ce{H3O+}][\ce{X-}]}{[\ce{HX}]}

For the ionisation \ce{HX + H2O <=> H3O+ + X-} at equilibrium. A larger K_a means a stronger acid.

\mathrm{mol\,dm^{-3}} (usually written without a unit)

pK_a

\mathrm{p}K_a=-\log K_a

A smaller \mathrm{p}K_a means a stronger acid.

no unit

Base ionisation constant

K_b=\dfrac{[\ce{BH+}][\ce{OH-}]}{[\ce{B}]}

For the ionisation \ce{B + H2O <=> BH+ + OH-}. A larger K_b means a stronger base.

\mathrm{mol\,dm^{-3}} (usually written without a unit)

pK_b

\mathrm{p}K_b=-\log K_b

A smaller \mathrm{p}K_b means a stronger base.

no unit

Henderson-Hasselbalch equation (acidic buffer)

\mathrm{pH}=\mathrm{p}K_a+\log\dfrac{[\text{salt}]}{[\text{acid}]}

[\text{salt}] is the concentration of the salt (conjugate base) and [\text{acid}] that of the weak acid.

no unit

Henderson-Hasselbalch equation (basic buffer)

\mathrm{pOH}=\mathrm{p}K_b+\log\dfrac{[\text{salt}]}{[\text{base}]}

[\text{salt}] is the concentration of the salt (conjugate acid) and [\text{base}] that of the weak base. Then \mathrm{pH}=14-\mathrm{pOH}.

no unit

Solubility product

\ce{A_mB_n(s) <=> mA^{n+}(aq) + nB^{m-}(aq)}\qquad K_{sp}=[\ce{A^{n+}}]^m[\ce{B^{m-}}]^n

Concentrations are those of the saturated solution at equilibrium. The solid does not appear in the expression.

depends on m+n (usually written without a unit)

Ion product and precipitation

Q'>K_{sp}:\ \text{precipitation occurs};\qquad Q'<K_{sp}:\ \text{no precipitation}

Q' is the expression for K_{sp} written with the initial concentrations.

no unit

Key terms

Conjugate acid: A species formed as a result of the acceptance of a proton by a base.
Conjugate base: The species that remains after an acid has donated a proton.
Conjugate acid-base pair: Two species on opposite sides of an equation that differ by a proton.
Amphoteric: Acting both as an acid and as a base, like water.
Lewis acid: A substance that can accept a pair of electrons to form a coordinate covalent bond.
Lewis base: A substance that can donate a pair of electrons to form a coordinate covalent bond.
Strength of an acid: The ability of an acid to donate a proton (acid strength); a stronger acid donates a proton to a higher degree.
Auto-ionisation of water: The ionisation of water to a small extent, \ce{H2O + H2O <=> H3O+ + OH-}.
Ionic product constant of water: K_w=[\ce{H+}][\ce{OH-}], equal to 1.0\times10^{-14} in pure water at 25^\circ\mathrm{C}.
pH: The negative logarithm of the hydrogen ion concentration, \mathrm{pH}=-\log[\ce{H+}].
Titration: A method to find the volume of the standard solution required to react completely with a known volume of another solution under analysis.
End point: The stage of a titration at which the indicator changes colour.
Acid dissociation constant: K_a, a measure of the extent to which an acid is ionised at equilibrium; the greater the K_a, the stronger the acid.
pK_a: The negative logarithm of K_a; the smaller the \mathrm{p}K_a, the stronger the acid.
Base ionisation constant: K_b, the equilibrium constant for the ionisation of a base in water; \mathrm{p}K_b=-\log K_b.
Buffer solution: A solution, the pH of which does not change significantly when a small amount of acid or base is added to it.
Acidic buffer: A mixture of a weak acid and a salt of it with a strong base (pH less than 7), for example \ce{CH3COOH + CH3COONa}.
Basic buffer: A mixture of a weak base and a salt of it with a strong acid (pH more than 7), for example \ce{NH4OH + NH4Cl}.
Hydrolysis: The chemical reaction of cations and anions of salts with water.
Salt: An ionic substance obtained as a result of the neutralisation of an acid and a base.
Solubility product constant: K_{sp}, the product of the equilibrium concentrations of ions, each raised to a power which is the coefficient of the ion in the balanced chemical equation.
Precipitation reaction: An aqueous reaction that takes place when two or more solutions are mixed together, yielding a solid insoluble substance.
Ion product: Q', the expression for K_{sp} written with the initial concentrations; precipitation occurs if Q'>K_{sp}.
Common ion effect: The phenomenon in which the degree of ionisation or solubility of an electrolyte is reduced by the addition of a highly soluble electrolyte containing a common ion.

Multiple choice questionsp. 208–210

Choose the correct answer. The book prints no answer key for these questions, so every key was worked out from the unit text. Item (i) is not printed in the book: the list starts at (ii).

Water cannot act as:

ALewis acidAnswer
BLewis base
CConjugate acid
DConjugate base

Why

A Lewis acid must have an incomplete octet, so it can accept an electron pair. The oxygen atom in water already has a complete octet with two lone pairs, so water cannot accept a pair. Water does donate a lone pair, so it acts as a Lewis base (B).

Water is also the conjugate acid of \ce{OH-} (\ce{OH- + H+ -> H2O}) and the conjugate base of \ce{H3O+}, so options C and D are possible.

iii

\mathrm{pH} of 0.01\,\mathrm{M} \ce{HCl} solution is:

A10^{-2}
B10^{+2}
C2.0Answer
D1.0

Why

\ce{HCl} is a strong acid and ionises completely, so [\ce{H+}]=0.01\,\mathrm{mol\,dm^{-3}}=10^{-2}\,\mathrm{M}.

\mathrm{pH}=-\log[\ce{H+}]=-\log(10^{-2})=2.0

Option A is the hydrogen ion concentration, not the pH. B is a value above 7, which an acid cannot have, and D is the pH of a 0.1 M solution.

iv2026 Annual I, Set D, MCQ 12

An aqueous solution of ammonium chloride is:

ABasic
BAcidicAnswer
CNeutral
DAmphoteric

Why

\ce{NH4Cl} is a salt of the strong acid \ce{HCl} and the weak base \ce{NH3}. The cation is the conjugate acid of a weak base, so it hydrolyses: \ce{NH4+ + H2O <=> NH4OH + H+}. The \ce{H+} produced makes the solution acidic (pH below 7), which is why it turns blue litmus red. The chloride ion does not react with water.

Basic (A) would need an anion from a weak acid, and neutral (C) would need both ions from strong acid and base. Amphoteric (D) describes a species such as water, not a salt.

An aqueous solution of which compound is basic?

AAmmonium nitrate
BCalcium chloride
CAmmonium acetate
DPotassium carbonateAnswer

Why

\ce{K2CO3} is a salt of the strong base \ce{KOH} and the weak acid \ce{H2CO3}. The carbonate ion is the conjugate base of a weak acid and takes a proton from water: \ce{CO3^2- + H2O <=> HCO3- + OH-}. The \ce{OH-} ions make the solution basic.

Ammonium nitrate (A) is acidic because \ce{NH4+} hydrolyses. Calcium chloride (B) is neutral since neither ion hydrolyses. Ammonium acetate (C) is almost neutral because the cation and anion hydrolyse to about the same extent.

Which statement about acids is NOT correct? An acid:

AContains hydrogen ions in solution.
BAlways contains oxygen.Answer
CHas a pH of less than 7.
DGives off carbon dioxide from a carbonate.

Why

The false statement is B. Acids do not need to contain oxygen: \ce{HCl} and \ce{HF} are acids with no oxygen. Acids in water give \ce{H+} (or \ce{H3O+}) ions, so pH is below 7, and they react with carbonates to release carbon dioxide.

vii

If a liquid has a pH of 7;

AIt must be colourless.
BIt has boiling point of 100^\circ\mathrm{C}.
CIt must be a solution.
DIt must be neutral.Answer

Why

At 25^\circ\mathrm{C}, \mathrm{pH}=7 means [\ce{H+}]=1.0\times10^{-7}\,\mathrm{M}. Since [\ce{H+}]=[\ce{OH-}] in a neutral solution, a pH of 7 means the liquid is neutral (D).

Colour (A) and boiling point (B) do not follow from pH. A liquid with pH 7 need not be a solution (C): pure water has pH 7 at 25^\circ\mathrm{C}.

viii

When air is bubbled through pure water, the pH is lowered from 7.0 to 5.6; which gas in the air is responsible for this change?

AArgon
BCarbon dioxideAnswer
CNitrogen
DOxygen

Why

Carbon dioxide dissolves in water to form carbonic acid, \ce{CO2 + H2O <=> H2CO3}, and the acid ionises to give \ce{H+} ions. This lowers the pH. Argon, nitrogen and oxygen do not form acids when they dissolve in water, so they do not change the pH.

The book describes carbonic acid as a weak acid (Concept Assessment 9.6), so its ionisation gives only a small drop in pH, from 7.0 to 5.6.

If 25\,\mathrm{cm^3} of 1\,\mathrm{mol\,dm^{-3}} nitric acid is added to 50\,\mathrm{cm^3} of 0.5 potassium hydroxide solution, what would be the pH of the resulting solution?

A5
B7Answer
C9
D14

Why

Moles of acid: n(\ce{HNO3})=1.0\times0.025=0.025\,\mathrm{mol}. Moles of base: n(\ce{KOH})=0.5\times0.050=0.025\,\mathrm{mol}.

The moles are equal, so the acid and base neutralise each other exactly. The product is the salt \ce{KNO3}, from a strong acid and a strong base, which does not hydrolyse. The solution is neutral with pH 7 (B).

Note. The book prints '0.5 potassium hydroxide solution' with no unit. The answer reads it as 0.5 mol dm^-3, the only reading that gives a pH.

If dry citric acid crystals are placed on dry litmus paper it will;

ATurn yellow
BTurn green
CTurn red
DRemain unchangedAnswer

Why

The book states that the presence of water is essential for the formation of \ce{H+} ions. Dry citric acid does not ionise, so it releases no \ce{H+} and the litmus paper does not change colour (D).

Red litmus turns red only in the presence of \ce{H+} ions in solution, so the colour changes (A, B, C) are not produced.

A base is a substance which will neutralize an acid, which of these substances is not a base?

AAqueous ammonia
BCopper oxide
CPotassium chlorideAnswer
DSodium carbonate

Why

Aqueous ammonia (A) accepts protons and forms \ce{OH-}. Copper oxide (B) is a metal oxide, a base that reacts with acid. Sodium carbonate (D) is a base because its carbonate ion accepts protons, \ce{CO3^2- + 2H+ -> H2O + CO2}.

Potassium chloride (C) is a salt of a strong acid and a strong base. It does not accept protons, so it does not neutralise an acid.

xii

A strong acid;

AIs always partially ionized when in solution.
BIs always fully ionized when in solution.Answer
CAlways decomposes carbonates.
DAlways contains oxygen.

Why

A strong acid is defined by its complete ionisation in solution, so B is correct. A describes a weak acid, which is only partly ionised (equilibrium lies far to the left, so K_a is small).

C is wrong as a definition: many acids react with carbonates, but this does not make an acid strong. D is wrong because \ce{HCl}, a strong acid, contains no oxygen.

xiii

Which one of the following oxides dissolves in water to form acidic solution?

A\ce{MgO}
B\ce{Na2O}
C\ce{SO2}Answer
D\ce{SiO2}

Why

Sulphur dioxide dissolves and forms an acid: \ce{SO2 + H2O -> H2SO3}, so the solution is acidic. \ce{MgO} and \ce{Na2O} are basic oxides: they form \ce{Mg(OH)2} and \ce{NaOH}, which make the solution basic. \ce{SiO2} does not dissolve in water.

xiv

When crystals of copper sulphate are heated, the colour changes from blue to white. This is caused by;

ALoss of water onlyAnswer
BLoss of water and \ce{SO2}
CReaction with \ce{CO2} in the air
DLoss of water, sulphur dioxide and oxygen

Why

Blue hydrated copper sulphate loses its five molecules of water of crystallisation on heating, giving white anhydrous copper sulphate: \ce{CuSO4.5H2O ->[\Delta] CuSO4 + 5H2O}. The colour change is caused by water alone (A).

Decomposition into \ce{SO2} and oxygen needs far higher temperatures, and \ce{CO2} does not change the colour, so B, C and D are wrong.

Short questionsp. 210

Give short answers. Items (i) to (ix) of the book's second group. Each answer covers every point the book makes on the question.

i3 marks

What are conjugate acid-base pairs? Explain with examples.

Solution

A conjugate acid-base pair is two species, on opposite sides of an acid-base equation, that differ by one proton (\ce{H+}). The acid gives up a proton and is left as its conjugate base. The base accepts the proton and becomes its conjugate acid. In general, for \ce{A1 + B2 -> B1 + A2}, the pairs are \ce{A1/B1} and \ce{B2/A2}.

\ce{HCl + H2O <=> H3O+ + Cl-}\qquad\text{pairs: }\ce{HCl/Cl-},\quad\ce{H2O/H3O+}

\ce{NH3 + H2O <=> NH4+ + OH-}\qquad\text{pairs: }\ce{NH4+/NH3},\quad\ce{H2O/OH-}

Water is amphoteric. With \ce{HCl} it acts as a base, accepting a proton to form \ce{H3O+}; with \ce{NH3} it acts as an acid, donating a proton to form \ce{OH-}. A stronger acid has a weaker conjugate base: \ce{Cl-}, the conjugate base of the strong acid \ce{HCl}, is far weaker as a base than \ce{CH3COO-}, the conjugate base of the weak acid \ce{CH3COOH}.

ii3 marks

Define Lewis acid and Lewis bases. Give one example in each case.

Solution

A Lewis acid is a substance that can accept a pair of electrons to form a coordinate covalent bond. Its central atom has an incomplete octet, so it has a deficiency of electron pairs. Example: boron trifluoride, \ce{BF3}, which has only six electrons around boron.

A Lewis base is a substance that can donate a pair of electrons to form a coordinate covalent bond. Its central atom has a complete octet and at least one unshared (lone) pair. Example: ammonia, \ce{NH3}, whose nitrogen lone pair is donated to boron.

\ce{NH3 + BF3 -> H3N-BF3}\qquad\text{(N donates a pair to B, a coordinate covalent bond)}

In every Lewis acid-base reaction, the coordinate covalent bond forms between the acid and the base.

iii3 marks

Write briefly about the ionisation of water.

Solution

Water ionises to a very small extent. This is called autoionisation: one water molecule acts as an acid and another as a base.

\ce{H2O + H2O <=> H3O+ + OH-}\qquad K=\dfrac{[\ce{H3O+}][\ce{OH-}]}{[\ce{H2O}]^2}

Water is in large excess, so [\ce{H2O}] is constant and K[\ce{H2O}]^2=K_w=[\ce{H+}][\ce{OH-}]. At 25^\circ\mathrm{C}, K_w=1.0\times10^{-14}, so in pure water [\ce{H+}]=[\ce{OH-}]=1.0\times10^{-7}\,\mathrm{M}. Because K_w is an equilibrium constant it changes with temperature: it is 3.8\times10^{-14} at 40^\circ\mathrm{C}.

iv3 marks

Define pH. What are the values of pH for acidic, basic and neutral solutions?

Solution

pH is the negative logarithm (to base 10) of the hydrogen ion concentration, in mol dm^-3: \mathrm{pH}=-\log[\ce{H+}]. The pH has no units.

Solution	[\ce{H+}] at 25^\circ\mathrm{C}	pH value
Acidic	[\ce{H+}]>1.0\times10^{-7}\,\mathrm{M}	\mathrm{pH}<7.00
Neutral	[\ce{H+}]=1.0\times10^{-7}\,\mathrm{M}	\mathrm{pH}=7.00
Basic	[\ce{H+}]<1.0\times10^{-7}\,\mathrm{M}	\mathrm{pH}>7.00

At 25^\circ\mathrm{C}, \mathrm{pH}+\mathrm{pOH}=14, where \mathrm{pOH}=-\log[\ce{OH-}].

v3 marks

What are K_a and \mathrm{p}K_a?

Solution

For a weak acid \ce{HX} in water, \ce{HX + H2O <=> H3O+ + X-}. The acid dissociation constant is K_a=\dfrac{[\ce{H3O+}][\ce{X-}]}{[\ce{HX}]}, because the concentration of water is constant and is absorbed into the constant (book's K[\ce{H2O}]=K_a).

K_a measures the extent to which the acid ionises. The greater the value of K_a, the stronger the acid. K_a depends on temperature, so the temperature must be stated. Acetic acid has K_a=1.8\times10^{-5} at 25^\circ\mathrm{C}.

\mathrm{p}K_a=-\log K_a\qquad\text{for acetic acid: }\mathrm{p}K_a=-\log(1.8\times10^{-5})\approx4.7

Because \mathrm{p}K_a is the negative logarithm, a smaller \mathrm{p}K_a means a larger K_a and a stronger acid.

vi3 marks

What are K_b and \mathrm{p}K_b?

Solution

For a base \ce{B} in water, \ce{B + H2O <=> BH+ + OH-}. The base ionisation constant is K_b=\dfrac{[\ce{BH+}][\ce{OH-}]}{[\ce{B}]}. A large K_b means the base ionises to a high degree, so it is a strong base. The value of K_b is small for a weak base.

\mathrm{p}K_b=-\log K_b\qquad\text{for ammonia: }\mathrm{p}K_b=4.76\ (K_b=1.7\times10^{-5})

A smaller \mathrm{p}K_b means a stronger base. Ammonia is a stronger base than pyridine (\mathrm{p}K_b=8.25) and aniline (\mathrm{p}K_b=9.37), but weaker than methylamine (\mathrm{p}K_b=3.34).

vii3 marks

What is the relationship between K_a and K_b?

Solution

For a conjugate acid-base pair \ce{HA/A-}, the product of the two constants is the ionic product of water:

\begin{aligned}&\ce{HA + H2O <=> H3O+ + A-} &&K_a=\dfrac{[\ce{H3O+}][\ce{A-}]}{[\ce{HA}]}\\ &\ce{A- + H2O <=> HA + OH-} &&K_b=\dfrac{[\ce{HA}][\ce{OH-}]}{[\ce{A-}]}\end{aligned}

K_aK_b=\dfrac{[\ce{H3O+}][\ce{A-}]}{[\ce{HA}]}\times\dfrac{[\ce{HA}][\ce{OH-}]}{[\ce{A-}]}=[\ce{H3O+}][\ce{OH-}]=K_w

So K_a\times K_b=K_w=1.0\times10^{-14} at 25^\circ\mathrm{C}, and \mathrm{p}K_a+\mathrm{p}K_b=14. A strong acid therefore has a very weak conjugate base, and a weak acid has a relatively strong conjugate base.

Note. The unit's SLO says that K_b and the relation K_w=K_a\times K_b will not be tested. The question is still printed in the exercise, so the relation is given here in full.

viii3 marks

Give two examples of a buffer solution.

Solution

A buffer solution is one whose pH does not change significantly when a small amount of acid or base is added to it. Two examples given in the book are:

A weak acid with a salt of that acid and a strong base (acidic buffer, pH below 7): \ce{CH3COOH + CH3COONa}.
A weak base with a salt of that base and a strong acid (basic buffer, pH above 7): \ce{NH4OH + NH4Cl}.

For example, a solution containing 0.1\,\mathrm{M} \ce{CH3COOH} and 1.0\,\mathrm{M} \ce{CH3COONa} has a pH of 5.76 (Example 9.6).

ix3 marks

How does temperature impact solubility?

Solution

For most sparingly soluble solids, the dissolving process absorbs heat (it is endothermic). When the temperature is raised, the equilibrium shifts in the direction that absorbs heat, so more solid dissolves and the solubility increases.

This follows from Le Chatelier's principle: a system at equilibrium responds to a change in conditions by shifting to counteract it. The same principle explains the common ion effect in this unit, where adding an ion pushes the equilibrium back and reduces solubility.

Like K_w and K_a, the solubility product K_{sp} is an equilibrium constant, so its value depends on temperature and must be quoted with the temperature.

Note. The unit text does not discuss the effect of temperature on solubility. The answer applies Le Chatelier's principle, which the book uses for the common ion effect.

Long questionsp. 210–211

Questions 3 to 6 and 10 to 15 of the book's exercise, answered in full. The book gives no heading for these groups.

Elaborate the ionization equation of water. How does it lead to the ion-product constant of water?

Solution

Ionisation equation of water

Water ionises to a small extent. One molecule acts as an acid and the other as a base, so the two ions formed are the hydronium ion and the hydroxide ion:

\ce{H2O + H2O <=> H3O+ + OH-}

Water is therefore amphoteric: it acts as an acid and as a base.

From the equilibrium constant to K_w

Derivation Ionic product of water

Aim. Show how the ionisation of water gives the ion-product constant K_w.

Assumptions.

Water is a very weak electrolyte, so the ionisation is small.
Water is the solvent and is in large excess, so its concentration is effectively constant.
The temperature is fixed at 25^\circ\mathrm{C}.

Write the equilibrium for the ionisation of water.
\ce{H2O + H2O <=> H3O+ + OH-}
Write the equilibrium constant. Water appears on the left side, so its concentration is squared.
K=\dfrac{[\ce{H3O+}][\ce{OH-}]}{[\ce{H2O}]^2}
Water is in large excess and its concentration barely changes, so [\ce{H2O}]^2 is a constant. Multiply both sides by [\ce{H2O}]^2.
K[\ce{H2O}]^2=[\ce{H3O+}][\ce{OH-}]
The product of the two constants K and [\ce{H2O}]^2 is a new constant, named K_w, the ionic product of water.
K_w=K[\ce{H2O}]^2=[\ce{H3O+}][\ce{OH-}]
The hydronium ion is the hydrated proton, so [\ce{H3O+}] is written as [\ce{H+}] for convenience.
K_w=[\ce{H+}][\ce{OH-}]
At 25^\circ\mathrm{C} the value is fixed by experiment.
K_w=[\ce{H+}][\ce{OH-}]=1.0\times10^{-14}

ResultK_w=[\ce{H+}][\ce{OH-}]=1.0\times10^{-14}\ \text{at}\ 25^\circ\text{C}

The product of the hydrogen and hydroxide ion concentrations is fixed for any aqueous solution at a given temperature. If [\ce{H+}] rises, [\ce{OH-}] falls in the same proportion.

Acidic, basic and neutral solutions

Neutral: [\ce{H+}]=[\ce{OH-}]=1.0\times10^{-7}\,\mathrm{M} and \mathrm{pH}=7.00.
Acidic: [\ce{H+}]>[\ce{OH-}], so [\ce{H+}]>1.0\times10^{-7}\,\mathrm{M} and \mathrm{pH}<7.00.
Basic: [\ce{OH-}]>[\ce{H+}], so [\ce{H+}]<1.0\times10^{-7}\,\mathrm{M} and \mathrm{pH}>7.00.

Because K_w is an equilibrium constant it depends on temperature. At 40^\circ\mathrm{C}, K_w=3.8\times10^{-14}, which gives [\ce{H+}]=[\ce{OH-}]=1.9\times10^{-7}\,\mathrm{M}. The solution is still neutral, because the two ion concentrations are equal.

What are buffer solutions? Elaborate with suitable examples, their significance in acid-base reactions. Write three common applications of buffer solutions.

Solution

What a buffer solution is

A buffer solution is a solution whose pH does not change significantly when a small amount of acid or base is added to it. It keeps a nearly constant pH for a long time.

Types of buffer solution

Acidic buffer: a weak acid with a salt of a strong base, for example \ce{CH3COOH + CH3COONa}. The pH is below 7.
Basic buffer: a weak base with a salt of a strong acid, for example \ce{NH4OH + NH4Cl}. The pH is above 7.

How a buffer works and why it matters

A weak acid is only slightly ionised, so it holds most of its molecules unionised. The salt supplies a large amount of the conjugate base. When a small amount of acid (an extra \ce{H+}) is added, it combines with the large reserve of conjugate base, and the pH barely moves. When base is added, it is neutralised by the weak acid. In this way a buffer resists change in pH during an acid-base reaction, so the reaction is kept at a steady pH.

Example: pH of a buffer

The pH of a buffer is found with the Henderson-Hasselbalch equation: \mathrm{pH}=\mathrm{p}K_a+\log\dfrac{[\text{salt}]}{[\text{acid}]}.

Data: [\ce{CH3COOH}]=0.1\,\mathrm{M}, [\ce{CH3COONa}]=1.0\,\mathrm{M}, \mathrm{p}K_a=4.76.
Substitute the salt and acid concentrations into the equation.
\mathrm{pH}=4.76+\log\dfrac{1.0}{0.1}=4.76+\log10
\log 10=1.
\mathrm{pH}=4.76+1.00=5.76

The pH of the buffer is 5.76.

Three applications

The book gives the pH ranges of several biological systems that must be kept nearly constant. Three examples are:

Human blood: the pH is kept in the range 7.35 to 7.45.
Tears: the pH is about 7.40.
Milk: the pH is in the range 6.7 to 6.8, and egg white is 8.0 to 8.1.

Note. The book's text gives the pH ranges of these biological systems (Do You Know, page 202) but does not name the buffers in them. Those three applications are taken from that table; the book names no others on these pages.

Write detailed notes on each of the followings:

Solution

(a) Conjugate acid base pairs

Definition

A conjugate acid-base pair is two species, on opposite sides of an acid-base equation, that differ by a single proton. A conjugate acid is formed when a base accepts a proton. A conjugate base is what remains when an acid donates a proton.

Examples

Equation	Acid	Conjugate base	Base	Conjugate acid
\ce{HCl + H2O <=> H3O+ + Cl-}	\ce{HCl}	\ce{Cl-}	\ce{H2O}	\ce{H3O+}
\ce{NH3 + H2O <=> NH4+ + OH-}	\ce{H2O}	\ce{OH-}	\ce{NH3}	\ce{NH4+}

Rule

A weak acid has a strong conjugate base, and a strong acid has a weak conjugate base. Water is the base in the first reaction and the acid in the second, so it is amphoteric.

(b) \mathrm{p}K_a

Definition

\mathrm{p}K_a=-\log K_a

K_a is the acid dissociation constant. It is inconvenient to work with such small numbers, so the negative logarithm is used. A smaller \mathrm{p}K_a means a larger K_a and a stronger acid.

Values

Acid	K_a	\mathrm{p}K_a
Hydrochloric acid, \ce{HCl}	1.0\times10^{6}	-6.0
Hydrofluoric acid, \ce{HF}	7.2\times10^{-4}	+3.1
Acetic acid, \ce{CH3COOH}	1.8\times10^{-5}	+4.7
Water, \ce{H2O}	1.8\times10^{-16}	+15.7

\mathrm{p}K_a values can be negative for strong acids. Perchloric acid and hydroiodic acid have \mathrm{p}K_a=-10.0, the strongest acids in the table, while water, with \mathrm{p}K_a=+15.7, is the weakest.

Definition

\mathrm{p}K_b=-\log K_b

K_b is the base ionisation constant. As with \mathrm{p}K_a, the logarithm gives convenient numbers. A smaller \mathrm{p}K_b means a larger K_b and a stronger base.

Values

Base	K_b	\mathrm{p}K_b
Diethylamine, (\ce{C2H5})2\ce{NH}	9.6\times10^{-4}	3.02
Ammonia, \ce{NH3}	1.7\times10^{-5}	4.76
Pyridine, \ce{C5H5N}	5.6\times10^{-9}	8.25
Aniline, \ce{C6H5NH2}	4.3\times10^{-10}	9.37

Diethylamine is the strongest base in the table. Ammonia is a stronger base than pyridine and aniline, but weaker than methylamine and ethylamine.

Note. Acetic acid is given as \mathrm{p}K_a=4.7 in Table 9.2 and as 4.76 in Example 9.6. From K_a=1.8\times10^{-5} the value is 4.74. The three values differ only by rounding.

What is hydrolysis? Discuss in detail, the behaviour of each of the following salts in their aqueous solutions.

Solution

What hydrolysis is

Hydrolysis is the chemical reaction of the cations and anions of a salt with water. Water is a reactant: an H-OH bond is broken. Salts whose ions react with water make the solution acidic or basic, while salts whose ions do not react leave the solution neutral.

(a) \ce{K2CO3}

Ions of the salt

\ce{K2CO3 -> 2K+ + CO3^2-}

\ce{K2CO3} is a salt of the strong base \ce{KOH} and the weak acid \ce{H2CO3}. \ce{K+} is the conjugate acid of a strong base and does not react with water. \ce{CO3^2-} is the conjugate base of a weak acid and does react.

Hydrolysis

\begin{aligned}&\ce{CO3^2- + H2O <=> HCO3- + OH-}\\ \text{overall: }&\ce{CO3^2- + 2H2O <=> H2CO3 + 2OH-}\end{aligned}

The hydroxide ions make the solution basic. The book's observation is that aqueous \ce{K2CO3} turns red litmus blue, so the solution is basic with pH above 7.

(b) \ce{NH4Cl}

Ions of the salt

\ce{NH4Cl -> NH4+ + Cl-}

\ce{NH4Cl} is a salt of the strong acid \ce{HCl} and the weak base \ce{NH3}. \ce{Cl-} is the conjugate base of a strong acid and does not react with water. \ce{NH4+} is the conjugate acid of a weak base and does react.

Hydrolysis

\ce{NH4+ + H2O <=> NH4OH + H+}

The \ce{H+} ions make the solution acidic with pH below 7. The book's observation is that an aqueous solution of \ce{NH4Cl} turns blue litmus red.

Ions of the salt

\ce{NaNO3 -> Na+ + NO3-}

\ce{NaNO3} is a salt of the strong base \ce{NaOH} and the strong acid \ce{HNO3}. Neither \ce{Na+} nor \ce{NO3-} reacts with water, so there is no hydrolysis.

[\ce{H+}]=[\ce{OH-}] and the solution is neutral with pH 7. It has no action on litmus.

Note. The book writes the carbonate reaction as \ce{CO3^2- + 2H-OH <=> H2CO3 + 2OH-}. Both the stepwise and the overall form are shown.

What is hydrolysis? Write the equations of hydrolysis equilibrium for each of the followings:

Solution

Hydrolysis is the reaction of the cations or anions of a salt with water, in which an H-OH bond is broken. A cation or anion that is the conjugate of a weak base or weak acid takes part; the conjugate of a strong acid or strong base does not.

(i) \ce{K+}

\ce{K+} is so weak a conjugate acid that it does not react with water. No hydrolysis equilibrium can be written: \ce{K+} leaves the solution neutral.

(ii) \ce{NH4+}

\ce{NH4+ + H2O <=> NH4OH + H+}

\ce{NH4+} donates a proton to water. The \ce{H+} formed makes the solution acidic.

(iii) \ce{CN-}

\ce{CN- + H2O <=> HCN + OH-}

\ce{CN-} is the conjugate base of the weak acid \ce{HCN}. It accepts a proton from water and the \ce{OH-} formed makes the solution basic.

Note. K+ does not hydrolyse, so no equation exists for part (i). The book states that such cations do not react with water.

Compare and contrast the concepts of strong acids and weak acids.

Solution

Comparison

Property	Strong acid	Weak acid
Ionisation in water	Complete: every molecule gives \ce{H+}	Partial: an equilibrium is set up
Position of equilibrium	Lies far to the right	Lies to the left, mostly unionised
K_a and \mathrm{p}K_a	Large K_a; negative \mathrm{p}K_a (e.g. -6.0 for HCl)	Small K_a; positive \mathrm{p}K_a (e.g. +4.7 for acetic acid)
Conjugate base	Very weak	Relatively strong
[\ce{H+}] and pH	[\ce{H+}] equals the acid concentration, so pH is low	[\ce{H+}] is found from K_a, so pH is higher for the same concentration
Examples	\ce{HCl}, \ce{H2SO4}, \ce{HClO4}	\ce{CH3COOH}, \ce{HF}, \ce{HCOOH}

Calculation: strong acid

\ce{HCl} ionises completely, so [\ce{H+}]=0.001\,\mathrm{M}.
Find the pH.
\mathrm{pH}=-\log(0.001)=3.00

Calculation: weak acid

For 1.0\,\mathrm{M} \ce{HF} (K_a=7.2\times10^{-4}), the equilibrium concentrations are [\ce{HF}]=1.0-x and [\ce{H+}]=[\ce{F-}]=x.

Write the expression for K_a.
K_a=\dfrac{[\ce{H+}][\ce{F-}]}{[\ce{HF}]}=\dfrac{x\cdot x}{1.0-x}
x is much smaller than 1.0, so 1.0-x\approx1.0.
7.2\times10^{-4}=\dfrac{x^2}{1.0}
Solve for x.
x=\sqrt{7.2\times10^{-4}}=0.0268\,\mathrm{M}
The pH follows from [\ce{H+}]=0.0268\,\mathrm{M}.
\mathrm{pH}=-\log(0.0268)=1.57

At the same concentration of 1.0\,\mathrm{M}, the strong acid \ce{HCl} would give [\ce{H+}]=1.0\,\mathrm{M} (pH 0), while the weak acid \ce{HF} gives only 0.0268\,\mathrm{M} (pH 1.57). The weak acid is partly unionised, so its [\ce{H+}] is much smaller.

Note. Book Example 9.5 prints [\ce{H+}]=0.268\,\mathrm{M}. Since x^2=7.2\times10^{-4}, x=0.0268\,\mathrm{M}; the book's value is a misprint by a factor of ten (pH 1.57, not 0.57).

Analyse how hydrolysis contribute to the pH of a solution. Provide examples.

Solution

Principle

Hydrolysis changes the pH because the ions of a salt react with water and release \ce{H+} or \ce{OH-}. The type of salt decides which ion reacts and so the nature of the solution.

Types of salt

Acid	Base	Ions that hydrolyse	pH	Example
Strong	Strong	None	=7 (neutral)	\ce{NaCl}, \ce{KBr}
Strong	Weak	Cation	<7 (acidic)	\ce{NH4NO3}, \ce{NH4Cl}
Weak	Strong	Anion	>7 (basic)	\ce{NaCN}, \ce{K2CO3}
Weak	Weak	Both	Depends on K_a and K_b	\ce{NH4CN}, \ce{NH4NO2}

Analysis

Salt of a strong acid and weak base: the cation (\ce{NH4+}) gives \ce{H+}. Example: \ce{NH4Cl + H2O <=> NH4OH + H+} gives an acidic solution.
Salt of a weak acid and strong base: the anion (\ce{CN-}, \ce{CO3^2-}) takes a proton from water and releases \ce{OH-}. Example: \ce{CN- + H2O <=> HCN + OH-} gives a basic solution.
Salt of a strong acid and strong base: neither ion hydrolyses, so the pH stays at 7. Example: \ce{NaCl}.

Weak acid and weak base

When both ions hydrolyse, the pH depends on the relative values of K_a and K_b of the cation and anion. The solution can be neutral, acidic or basic, so each salt must be checked separately.

Discuss the significance of the dissociation constant (K_a) in predicting the behaviour of weak acids.

Solution

What K_a measures

K_a is the acid dissociation constant of a weak acid \ce{HX}: K_a=\dfrac{[\ce{H3O+}][\ce{X-}]}{[\ce{HX}]}. It measures the extent to which the acid ionises at equilibrium. A greater K_a means a larger fraction of the acid is ionised and a stronger acid.

Uses in prediction

Comparing acids: K_a (or \mathrm{p}K_a) ranks acids. Hydrofluoric acid (K_a=7.2\times10^{-4}) is a stronger acid than acetic acid (K_a=1.8\times10^{-5}). A smaller \mathrm{p}K_a predicts a stronger acid.
**Predicting the [\ce{H+}] and pH:** the equilibrium table uses K_a with the initial concentration to find x=[\ce{H+}], as in Example 9.5 for 1.0\,\mathrm{M} \ce{HF}.
Predicting the pH of a buffer: the Henderson equation \mathrm{pH}=\mathrm{p}K_a+\log\dfrac{[\text{salt}]}{[\text{acid}]} uses \mathrm{p}K_a to predict the pH of a buffer made from the weak acid.
Predicting the conjugate base: a weak acid has a relatively strong conjugate base, so its salt hydrolyses to give a basic solution.

Temperature

K_a depends on temperature, so the value must be quoted with the temperature at which it was measured. The value of K_a at 25^\circ\mathrm{C} is the one used in calculations in the book.

How does the choice of indicators impact the accuracy of titration results?

Solution

Why the indicator matters

In a titration the indicator must change colour at the end point, which should be as close as possible to the equivalence point (the point at which acid and base have reacted in the exact stoichiometric ratio). Near the equivalence point the pH changes very sharply. An indicator whose pH range lies inside that sharp change gives an end point close to the equivalence point, and so an accurate volume.

Indicator pH ranges

Table 9.1 of the book
Indicator	pH range of colour change
Methyl orange	3.2 to 4.4
Bromothymol blue	6.0 to 7.6
Phenol red	6.8 to 8.0
Phenolphthalein	8.3 to 10.0

Matching indicator to titration

Strong acid with strong base (e.g. HCl with NaOH): pH rises rapidly from about 3 to 11. Phenolphthalein (8.3 to 10.0) changes inside this range and is the best choice.
Weak acid with strong base (e.g. oxalic acid with NaOH): pH changes from about 6.5 to 10 at the end point. Phenolphthalein again works. Methyl orange changes below pH 5, so it would give a wrong end point.
Strong acid with weak base (e.g. HCl with Na2CO3): pH changes from about 3.5 to 7.5 at the equivalence point. Methyl orange (3.2 to 4.4) is the best indicator. Phenolphthalein changes too late, so it would give an inaccurate result.
Weak acid with weak base: there is no sharp change in pH at the end point, so neither indicator works satisfactorily.

Effect on accuracy

If the colour change occurs outside the sharp pH change, the volume of titrant read at the end point is wrong. The molarity calculated from this volume, by M_1V_1/n_1=M_2V_2/n_2, is then wrong too. A correct indicator gives the most accurate titre and the most accurate molarity and strength of the unknown solution.

Evaluate the role of water in acid-base chemistry.

Solution

Water as the medium

Acids form \ce{H+} (as \ce{H3O+}) only in the presence of water. Hydrogen chloride is covalent and does not form ions by itself, but \ce{HCl + H2O -> H3O+ + Cl-} in water. Acid-base strength values such as K_a and K_b are measured in water, and the pH scale is defined for aqueous solutions.

Water as an acid and a base

Water is amphoteric. It is a base towards \ce{HCl} (it accepts a proton) and an acid towards \ce{NH3} (it donates a proton). Water also ionises itself, \ce{H2O + H2O <=> H3O+ + OH-}, with K_w=1.0\times10^{-14} at 25^\circ\mathrm{C}. This self-ionisation fixes the scale of pH and defines neutral solutions.

Water as a reactant

In hydrolysis, water takes part in the reaction: an H-OH bond breaks and the salt ion gains or loses a proton. Without water, salts would not give acidic or basic solutions.

Evaluation

Strength: water is essential for ionisation and makes the pH concept usable.
Limit: neutrality is defined by [\ce{H+}]=[\ce{OH-}], not by pH 7 at every temperature. At 40^\circ\mathrm{C}, K_w=3.8\times10^{-14}, so [\ce{H+}]=1.9\times10^{-7}\,\mathrm{M} and \mathrm{pH}=6.72, and the water is still neutral.
Buffers: the buffer pH calculations in this unit are also carried out in water, so water is part of every buffer.

In conclusion, water is the medium, a reactant and a participant in its own ionisation. Its role is central to all acid-base chemistry in this unit.

Numerical problemsp. 211

Questions 7 to 9 of the book's exercise. The book prints an answer for each; the working below reproduces it.

72026 Annual I, Set B, Q.2 (xiii) OR

Calculate the pH of formic acid-sodium formate buffer solution containing 1.0 mole of each component.

Solution

Given

n_{\ce{HCOOH}} = 1.0\,\mathrm{mol}
n_{\ce{HCOONa}} = 1.0\,\mathrm{mol}
K_a(\ce{HCOOH}) = 1.8\times10^{-4}

To find

\mathrm{pH} of the buffer

The salt and the acid are both present in equal amounts, so their concentration ratio is 1.0/1.0=1, and \log1=0.
\dfrac{[\text{salt}]}{[\text{acid}]}=\dfrac{1.0}{1.0}=1
Find \mathrm{p}K_a from K_a.
\mathrm{p}K_a=-\log(1.8\times10^{-4})=4-\log1.8=3.7447
Apply the Henderson-Hasselbalch equation.
\mathrm{pH}=\mathrm{p}K_a+\log\dfrac{[\text{salt}]}{[\text{acid}]}=3.7447+\log1=3.7447

\mathrm{pH}=3.7447\approx3.74

Book answer. Ans: 3.7447

Note. Table 9.2 gives \mathrm{p}K_a of formic acid as +3.75, the same value rounded. The book's answer uses K_a=1.8\times10^{-4} and agrees with this working.

When solid \ce{PbCl2} is added to pure water at 25^\circ\mathrm{C}, the salt dissolves until the concentration of \ce{Pb^2+} reaches 1.6\times10^{-2}\,\mathrm{M}. After this concentration is reached, excess solid remains undissolved. What is K_{sp} for this salt?

Solution

Given

[\ce{Pb^2+}] = 1.6\times10^{-2}\,\mathrm{M}

To find

K_{sp} of \ce{PbCl2}

Write the dissolution equilibrium. Each formula unit gives one \ce{Pb^2+} and two \ce{Cl-}.
\ce{PbCl2(s) <=> Pb^2+(aq) + 2Cl-(aq)}
Write the solubility product expression.
K_{sp}=[\ce{Pb^2+}][\ce{Cl-}]^2
Since [\ce{Cl-}]=2\times[\ce{Pb^2+}], the chloride concentration is 2\times1.6\times10^{-2}=3.2\times10^{-2}\,\mathrm{M}.
[\ce{Cl-}]=3.2\times10^{-2}\,\mathrm{M}
Substitute the two concentrations.
K_{sp}=(1.6\times10^{-2})(3.2\times10^{-2})^2=(1.6\times10^{-2})(1.024\times10^{-3})
Multiply.
K_{sp}=1.6384\times10^{-5}

K_{sp}=1.6384\times10^{-5}

Book answer. Ans: 1.6384 x 10^-5

Note. Working agrees with the book's answer.

(a) Calculate the \ce{H+} ion concentration of an aqueous solution having pH 10.6. (b) An aqueous solution contain 1.0\times10^{-9}\,\mathrm{moles/dm^3} of hydronium ions. Calculate the pOH of this solution.

Solution

(a) [\ce{H+}] of a solution with pH 10.6

Given

\mathrm{pH} = 10.6

To find

[\ce{H+}]

Use \mathrm{pH}=-\log[\ce{H+}], so [\ce{H+}]=10^{-\mathrm{pH}}.
[\ce{H+}]=10^{-10.6}
Write 10^{-10.6} as 10^{0.4}\times10^{-11}, since -10.6=0.4-11.
[\ce{H+}]=10^{0.4}\times10^{-11}
10^{0.4}=2.512.
[\ce{H+}]=2.5\times10^{-11}\,\mathrm{mol\,dm^{-3}}

[\ce{H+}]=2.5\times10^{-11}\,\mathrm{mol\,dm^{-3}}

(b) pOH of a solution with [\ce{H3O+}]=1.0\times10^{-9}

Given

[\ce{H3O+}] = 1.0\times10^{-9}\,\mathrm{mol\,dm^{-3}}

To find

\mathrm{pOH}

Hydronium ion is the same as \ce{H+}, so \mathrm{pH}=-\log(1.0\times10^{-9})=9.0. The solution is basic, since \mathrm{pH}>7.
\mathrm{pH}=9.0
At 25^\circ\mathrm{C}, \mathrm{pH}+\mathrm{pOH}=14.
\mathrm{pOH}=14-\mathrm{pH}=14-9.0=5.0
Check: [\ce{OH-}]=K_w/[\ce{H+}]=10^{-14}/10^{-9}=10^{-5}, so \mathrm{pOH}=-\log(10^{-5})=5.0.

\mathrm{pOH}=5.0

Book answer. (a) Ans: 2.5 x 10^-11 mol/dm^3; (b) Ans: 5.0

Note. Both answers agree with the book.

FBISE 2025 and 2026 questions

Questions from the FBISE HSSC-I 2026 (1st Annual) and FBISE HSSC-I 2025 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2025 Annual I, Set B, MCQ 71 mark

Aqueous solution of which of the following salts will have pH less than 7?

A\ce{Na2SO4}
B\ce{KNO3}
C\ce{CuSO4}Answer
D\ce{Na2CO3}

Why

A salt gives an acidic solution when it comes from a strong acid and a weak base, because the cation of the weak base hydrolyses and produces \ce{H+}. \ce{CuSO4} is the salt of \ce{H2SO4} (strong) and \ce{Cu(OH)2} (weak), so \ce{Cu^2+} reacts with water.

\ce{Cu^2+(aq) + 2H-OH(l) <=> Cu(OH)2(aq) + 2H+(aq)}

\ce{Na2SO4} and \ce{KNO3} are salts of a strong acid and a strong base, so they do not hydrolyse and \mathrm{pH}=7. \ce{Na2CO3} is a salt of a weak acid and a strong base, so it is basic (\mathrm{pH}>7).

2026 Annual I, Set B, MCQ 121 mark

pH value of aqueous solution of \ce{CH3COONa} will be:

ApH>7Answer
BpH<7
CpH=7
DpH is undefined

Why

\ce{CH3COONa} is the salt of a weak acid (\ce{CH3COOH}) and a strong base (\ce{NaOH}). The acetate ion is a strong conjugate base and reacts with water, producing \ce{OH-} ions, so the solution is basic (\mathrm{pH}>7).

\ce{CH3COO-(aq) + H-OH(l) <=> CH3COOH(aq) + OH-(aq)}

\ce{Na+} is the ion of a strong base and does not react with water. The solution is not neutral, because the acetate ion does hydrolyse.

2025 Annual I, Set B, Q.2 (viii)1.5+1.5 = 3 marks

Why aqueous solution of \ce{Na2CO3} is basic whereas that of \ce{NH4Cl} is acidic? Justify by giving chemical equations.

Solution

Both solutions are explained by hydrolysis, the reaction of the ions of a salt with water. The ion that comes from the weak parent (acid or base) reacts with water and decides the pH.

Na2CO3 is basic (1.5 marks)

\ce{Na2CO3} is the salt of a weak acid (\ce{H2CO3}) and a strong base (\ce{NaOH}). The carbonate ion is a strong conjugate base, so it takes protons from water and leaves an excess of \ce{OH-}. \ce{Na+} does not react with water.

\ce{Na2CO3 -> 2Na+ + CO3^2-}

\ce{CO3^2-(aq) + 2H-OH(l) <=> H2CO3(aq) + 2OH-(aq)}\qquad(\mathrm{pH}>7)

NH4Cl is acidic (1.5 marks)

\ce{NH4Cl} is the salt of a strong acid (\ce{HCl}) and a weak base (\ce{NH4OH}). The ammonium ion is a strong conjugate acid, so it gives protons to water and leaves an excess of \ce{H+}. \ce{Cl-} does not react with water.

\ce{NH4Cl -> NH4+ + Cl-}

\ce{NH4+(aq) + H-OH(l) <=> NH4OH(aq) + H+(aq)}\qquad(\mathrm{pH}<7)

2026 Annual I, Set B, Q.2 (x) OR3 marks

Calculate solubility of \ce{PbSO4} when K_{sp}=1.96\times10^{-8} at 25^{\circ}C.

Solution

Given

K_{sp}(\ce{PbSO4}) = 1.96\times10^{-8}
T = 25^\circ\mathrm{C}

To find

\text{solubility }s

Write the dissolving equilibrium. Let the solubility be s\ \mathrm{mol\,dm^{-3}}. Each formula unit gives one \ce{Pb^2+} and one \ce{SO4^2-}.
\ce{PbSO4(s) <=> Pb^2+(aq) + SO4^2-(aq)}\qquad[\ce{Pb^2+}]=[\ce{SO4^2-}]=s
Write K_{sp} and substitute (the solid is left out).
K_{sp}=[\ce{Pb^2+}][\ce{SO4^2-}]=s\times s=s^2=1.96\times10^{-8}
Take the square root.
s=\sqrt{1.96\times10^{-8}}=1.4\times10^{-4}\,\mathrm{mol\,dm^{-3}}

s=1.4\times10^{-4}\,\mathrm{mol\,dm^{-3}}

Note. This is the book's Concept assessment exercise 9.5 (ii), which is not one of the end-of-unit exercise questions.

2026 Annual I, Set B, Q.2 (xiii) OR3 marks

Calculate the pH of formic acid sodium formate buffer solution containing 1 mole of each component (K_a=1.8\times10^{-4}).

Solution

Given

n_{\ce{HCOOH}} = 1\,\mathrm{mol}
n_{\ce{HCOONa}} = 1\,\mathrm{mol}
K_a = 1.8\times10^{-4}

To find

\mathrm{pH}

Formic acid and its sodium salt make an acidic buffer. Both are in the same volume, so the ratio of concentrations equals the ratio of moles.
\dfrac{[\text{salt}]}{[\text{acid}]}=\dfrac{1}{1}=1,\qquad\log1=0
Convert K_a to \mathrm{p}K_a.
\mathrm{p}K_a=-\log(1.8\times10^{-4})=4-\log1.8=3.74
Apply the Henderson-Hasselbalch equation.
\mathrm{pH}=\mathrm{p}K_a+\log\dfrac{[\text{salt}]}{[\text{acid}]}=3.74+0=3.74

\mathrm{pH}=3.74

2025 Annual I, Set B, Q.61+2+4 = 7 marks

What is solubility product? Derive an expression for the solubility product of \ce{CaF2}. How can the concept of solubility product be applied to predict the precipitation of \ce{CaF2} on mixing aqueous solutions of \ce{Ca^{+2}} and \ce{F^{-1}} ions?

Solution

Solubility product (1 mark)

When a slightly soluble ionic compound is in contact with its saturated solution, a dynamic equilibrium exists between the undissolved solid and its ions. The solubility product constant K_{sp} is the product of the equilibrium concentrations of the ions, each raised to a power equal to its coefficient in the balanced equation. The solid itself does not appear in the expression.

Derivation Solubility product expression of CaF2 (2 marks)

Aim. Obtain K_{sp} for the equilibrium \ce{CaF2(s) <=> Ca^2+(aq) + 2F-(aq)}.

Assumptions.

The solution is saturated and at a fixed temperature.
\ce{CaF2} is slightly soluble, so the concentration of the solid stays practically constant.

A saturated solution of \ce{CaF2}: the solid and its ions are in dynamic equilibrium.

Write the equilibrium between the solid and its ions in a saturated solution.
\ce{CaF2(s) <=> Ca^2+(aq) + 2F-(aq)}
Write K_c for it. The fluoride ion has coefficient 2, so [\ce{F-}] is squared.
K_c=\dfrac{[\ce{Ca^2+}][\ce{F-}]^2}{[\ce{CaF2}]}
Multiply both sides by [\ce{CaF2}]. It is constant, so K_c[\ce{CaF2}] is a new constant.
K_c[\ce{CaF2}]=[\ce{Ca^2+}][\ce{F-}]^2
Call this constant the solubility product.
K_{sp}=[\ce{Ca^2+}][\ce{F-}]^2

ResultK_{sp}=[\ce{Ca^2+}][\ce{F-}]^2

For a general salt \ce{A_mB_n} the same argument gives K_{sp}=[\ce{A^{n+}}]^m[\ce{B^{m-}}]^n.

Prediction of precipitation of CaF2 (4 marks)

On mixing a solution of \ce{Ca^2+} ions with a solution of \ce{F-} ions, a precipitate of \ce{CaF2} may or may not form. To predict this, the ion product Q' is calculated. It has the same form as K_{sp} but uses the initial concentrations (after mixing) instead of the equilibrium concentrations.

Q'=[\text{initial }\ce{Ca^2+}]\,[\text{initial }\ce{F-}]^2

Condition	Result
Q'>K_{sp}	The solution is supersaturated: \ce{CaF2} precipitates until the concentrations satisfy K_{sp}
Q'=K_{sp}	Saturated solution at equilibrium: no net precipitation
Q'<K_{sp}	Unsaturated: no precipitate forms

Illustration

Take K_{sp}(\ce{CaF2})=3.45\times10^{-11} (from the solubility 2.05\times10^{-4}\,\mathrm{M}). Mix 10\,\mathrm{cm^3} of 0.010\,\mathrm{M} \ce{CaCl2} with 10\,\mathrm{cm^3} of 0.010\,\mathrm{M} \ce{NaF}. Each solution is diluted to half by the mixing, so [\ce{Ca^2+}]=[\ce{F-}]=0.0050\,\mathrm{M}.

Q'=(0.0050)(0.0050)^2=1.25\times10^{-7}>K_{sp}=3.45\times10^{-11}

Since Q'>K_{sp}, a precipitate of \ce{CaF2} forms.

Note. The illustration is added to show the method; the numbers are chosen for the example and are not from the paper.

2025 Annual I, Set D, MCQ 81 mark

If solubility of a salt MX is 2\times10^{-2}\ \mathrm{mol\,dm^{-3}}, what will be the value of K_{sp} if salt dissociates into one cation (\ce{M+}) and one anion (\ce{X-})?

A1\times10^{-4}
B2\times10^{-4}
C4\times10^{-4}Answer
D8\times10^{-4}

Why

\ce{MX} gives one cation and one anion, so \ce{MX(s) <=> M+ + X-} and [\ce{M+}]=[\ce{X-}]=s=2\times10^{-2}\,\mathrm{M}.

K_{sp}=[\ce{M+}][\ce{X-}]=s^2=(2\times10^{-2})^2=4\times10^{-4}

The tempting options come from squaring wrongly: (2\times10^{-2})^2 is 4\times10^{-4}, not 2\times10^{-4} (squaring the power of ten but not the 2) and not 8\times10^{-4} (doubling the 2 instead of squaring it).

2026 Annual I, Set D, MCQ 121 mark

pH value of aqueous solution of \ce{NH4Cl}:

ApH>7
BpH<7Answer
CpH=7
DpH is undefined

Why

\ce{NH4Cl} is the salt of a strong acid (\ce{HCl}) and a weak base (\ce{NH4OH}). The ammonium ion is a strong conjugate acid and reacts with water to produce \ce{H+}, so the solution is acidic (\mathrm{pH}<7).

\ce{NH4+(aq) + H-OH(l) <=> NH4OH(aq) + H+(aq)}

The chloride ion is a very weak conjugate base and does not react with water, so nothing balances the extra \ce{H+}.

2026 Annual I, Set D, Q.2 (x) OR1+1+1 = 3 marks

The solubility of \ce{CaF2} is 2.05\times10^{-4}\ \mathrm{M} at 25^{\circ}C. Calculate its K_{sp}.

Solution

Given

s = 2.05\times10^{-4}\,\mathrm{M}
T = 25^\circ\mathrm{C}

To find

K_{sp}(\ce{CaF2})

Write the dissolving equilibrium. Each formula unit of \ce{CaF2} that dissolves gives one \ce{Ca^2+} and two \ce{F-} ions.
\ce{CaF2(s) <=> Ca^2+(aq) + 2F-(aq)}
Find the ion concentrations from the solubility s.
[\ce{Ca^2+}]=s=2.05\times10^{-4}\,\mathrm{M},\qquad[\ce{F-}]=2s=4.10\times10^{-4}\,\mathrm{M}
Write K_{sp} (the solid is left out) and substitute.
K_{sp}=[\ce{Ca^2+}][\ce{F-}]^2=s(2s)^2=4s^3
Evaluate.
K_{sp}=4(2.05\times10^{-4})^3=4\times8.62\times10^{-12}=3.45\times10^{-11}

K_{sp}=3.45\times10^{-11}

2025 Annual I, Set D, Q.2 (xiii)3 marks

Calculate concentration of \ce{H+} ion in 1.0 M solution of formic acid (\ce{HCOOH}) which ionizes as: \ce{HCOOH <=> H+ + HCOO-}, K_a = 1.8\times10^{-4}

Solution

Given

[\ce{HCOOH}]_\text{initial} = 1.0\,\mathrm{M}
K_a = 1.8\times10^{-4}

To find

[\ce{H+}]

Let x\ \mathrm{mol\,dm^{-3}} of \ce{HCOOH} ionise. Each one gives one \ce{H+} and one \ce{HCOO-}.
\begin{aligned}&\ce{HCOOH <=> H+ + HCOO-}\\ &[\ce{HCOOH}]=1.0-x\\ &[\ce{H+}]=[\ce{HCOO-}]=x\end{aligned}
Write K_a and substitute.
K_a=\dfrac{[\ce{H+}][\ce{HCOO-}]}{[\ce{HCOOH}]}\ \Rightarrow\ 1.8\times10^{-4}=\dfrac{x^2}{1.0-x}
x is very small compared with 1.0 (a weak acid), so 1.0-x\approx1.0.
1.8\times10^{-4}=\dfrac{x^2}{1.0}
Take the square root.
x=\sqrt{1.8\times10^{-4}}=1.34\times10^{-2}\,\mathrm{M}

Check: x is only 1.3\% of 1.0\,\mathrm{M}, so the approximation is good. This follows the book's Example 9.5 for \ce{HF}.

[\ce{H+}]=1.34\times10^{-2}\,\mathrm{mol\,dm^{-3}}

2026 Annual I, Set D, Q.2 (xiii) OR1+1+1 = 3 marks

Calculate the pH of a buffer solution in which 0.11\ M \ce{CH3COONa} and 0.09\ M \ce{CH3COOH} solution are present. (K_a=1.8\times10^{-5})

Solution

Given

[\text{salt}]\ (\ce{CH3COONa}) = 0.11\,\mathrm{M}
[\text{acid}]\ (\ce{CH3COOH}) = 0.09\,\mathrm{M}
K_a = 1.8\times10^{-5}

To find

\mathrm{pH}

Acetic acid and sodium acetate make an acidic buffer, so use the Henderson-Hasselbalch equation. First convert K_a.
\mathrm{p}K_a=-\log(1.8\times10^{-5})=5-\log1.8=4.74
Write the equation with the salt in the numerator and the acid in the denominator, and substitute.
\mathrm{pH}=\mathrm{p}K_a+\log\dfrac{[\text{salt}]}{[\text{acid}]}=4.74+\log\dfrac{0.11}{0.09}
Evaluate the logarithm: 0.11/0.09=1.222 and \log1.222=0.087.
\mathrm{pH}=4.74+0.087=4.83

\mathrm{pH}=4.83

Note. This is the book's Concept assessment exercise 9.4, which is not one of the end-of-unit exercise questions.

2025 Annual I, Set D, Q.2 (xiv) OR3 marks

Why is the ionization of \ce{HF} decreased when aqueous solution of its salt is added to it? Briefly explain by applying Le Chatelier's principle.

Solution

\ce{HF} is a weak acid and ionises only slightly, setting up an equilibrium. Its salt, for example \ce{NaF}, is a strong electrolyte and ionises completely, so it supplies the common ion \ce{F-}.

\ce{HF(aq) <=> H+(aq) + F-(aq)}\qquad\ce{NaF -> Na+ + F-}

The salt raises the concentration of \ce{F-}, which disturbs the equilibrium.
By Le Chatelier's principle, the system opposes the increase in [\ce{F-}] by using up some of the \ce{F-}, so the equilibrium shifts to the left.
\ce{HF <=> H+ + F-}\quad(\text{shifts left})
As a result the ionisation of \ce{HF} decreases, [\ce{H+}] falls and [\ce{HF}] rises.

This effect is called the common ion effect: the ionisation of a weak electrolyte is reduced by adding a soluble salt that has an ion in common with it.

Exam focus

Calculate the pH of a buffer (acid and its salt) with the Henderson-Hasselbalch equation. (numerical)
Asked in both 2026 sets as a 3-mark question (Set B Q.2 (xiii) OR: formic acid and sodium formate; Set D Q.2 (xiii) OR: acetic acid and sodium acetate), and as book numerical 7.
Go to answer
Calculate K_{sp} from solubility, or solubility from K_{sp}, for salts such as \ce{PbSO4}, \ce{CaF2} and \ce{MX}. (numerical)
Asked in 2026 (Set B Q.2 (x) OR: \ce{PbSO4}; Set D Q.2 (x) OR: \ce{CaF2}) and 2025 (Set D MCQ 8).
Go to answer
Define solubility product, derive the K_{sp} expression for \ce{CaF2} and use Q' against K_{sp} to predict precipitation. (long)
A 7-mark Section C question in 2025 (Set B Q.6, split 1+2+4).
Go to answer
State whether a salt solution is acidic, basic or neutral and write the hydrolysis equation (\ce{CH3COONa}, \ce{NH4Cl}, \ce{Na2CO3}, \ce{CuSO4}). (short)
Asked in 2025 (Set B MCQ 7 and Q.2 (viii)) and 2026 (Set B MCQ 12 and Set D MCQ 12); book exercise MCQs 4 and 5 and long questions 6 and 10.
Go to answer
Explain the common ion effect, for example the decrease of ionisation of \ce{HF} when \ce{NaF} is added. (short)
Asked in 2025 (Set D Q.2 (xiv) OR); also book Concept assessment 9.6.
Go to answer
Find [\ce{H+}] and pH of a weak acid from K_a (for example 1.0 M formic acid). (numerical)
Asked in 2025 (Set D Q.2 (xiii)); same method as book Example 9.5.
Go to answer
Define buffer solutions, give the two types with examples, and write three applications. (long)
Book long question 4 and short question 8; the buffer pH calculation of the same section was asked in both 2026 sets.
Go to answer
Define pH, the ionic product of water and the pH of strong acids and bases, and compare strong and weak acids. (short)
Book exercise MCQs 3 and 7, short questions 3 to 5, long question 3 and numerical 9; not asked directly in the 2025 or 2026 papers.
Go to answer