Chemical equilibria

Class 12 Chemistry, unit 3, book pages 49–67

Practise 41 MCQs Take the unit quiz Review 52 cards

This unit applies the idea of dynamic equilibrium to solutions. It uses the extent of ionization and the constants $K_{a}$ and $K_{b}$ to tell strong acids and bases from weak ones, explains how buffers (including the $H X_{2} CO X_{3}$ / $HCO X_{3} X^{-}$ buffer in blood) resist changes in pH, and uses the solubility product $K_{s p}$ and the common ion effect to calculate ion concentrations and predict precipitation. It ends with the partition coefficient $K_{p c}$ , which describes how a solute shares itself between two immiscible solvents and is important in solvent extraction and drug design.

Learning outcomes (8)

Explain common ion effects giving suitable examples.
Use the extent of ionization and the acid dissociation constant, $K_{a}$ , to distinguish between strong and weak acids.
Use the extent of ionization and the base dissociation constant, $K_{b}$ , to distinguish between strong and weak bases.
Explain what is meant by a chemical buffer and how a buffer system works: define a buffer solution; explain how a buffer solution can be made; explain how buffer solutions control pH, using chemical equations; describe and explain the uses of buffer solutions, including the role of $HCO X_{3} X^{-}$ in controlling pH in blood.
Calculate the concentration of slightly soluble salts.
State what is meant by the term partition coefficient, $K_{p c}$ .
Calculate a partition coefficient for a system in which the solute is in the same physical state in the two solvents.
Explain the factors affecting the value of a partition coefficient in terms of the polarities of the solute and the solvent used.

Chemical equilibrium: the ideas this unit uses

Many reactions do not go to completion. They are reversible: the products can react to form the reactants again. A double arrow, , shows this. The unit opener shows the reaction $A + B C + D$ in a closed beaker.

At the start only A and B are present, so the forward reaction is fast and there is no reverse reaction. As A and B are used up, the forward rate falls. As C and D build up, the reverse rate rises. After some time the two rates become equal. From then on the concentrations stay constant, although both reactions are still going on. This state is called dynamic chemical equilibrium.

Concentration against time

0.2

0.4

0.6

0.8

t

c

equilibrium

reactants

products

Concentration

c

against time

t

(arbitrary units). Reactant concentration falls and product concentration rises until both stop changing: equilibrium is established.

Reaction rate against time

0.2

0.4

0.6

0.8

t

rate

rates equal

forward

reverse

Rate against time

t

. The forward rate falls and the reverse rate rises until they are equal. Both reactions continue, so the equilibrium is dynamic.

Dynamic equilibrium: The state of a reversible reaction in which the forward and reverse reactions go on at the same rate, so the concentrations of reactants and products stay constant.
Law of mass action: The rate of a reaction is proportional to the product of the molar concentrations of the reactants, each raised to the power of its coefficient in the balanced equation.
Equilibrium constant, $K_{c}$: The product of the equilibrium concentrations of the products divided by the product of the equilibrium concentrations of the reactants, each raised to the power of its coefficient in the balanced equation.

a A + b B c C + d D K_{c} = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}}

Square brackets mean the molar concentration at equilibrium, in $mol d m^{- 3}$ (also written M). Three rules are used again and again in this unit:

Constant concentrations are left out. A pure solid, a pure liquid and the solvent of a dilute solution (water) have constant concentrations, so they are joined into the constant. This is why $[H X_{2} O]$ disappears from $K_{a}$ and $K_{b}$ , and $[CaF X_{2}]$ disappears from $K_{s p}$ .
Only temperature changes the value of $K$ . Changing a concentration or the pressure, or adding a catalyst, does not change the value of $K$ .
The size of $K$ shows how far the reaction goes. A large $K$ (for example $1 0^{6}$ ) means mostly products at equilibrium; a small $K$ (for example $1 0^{- 5}$ ) means mostly reactants.

Le Chatelier's principle

Le Chatelier's principle: if a system at equilibrium is disturbed by a change in concentration, pressure or temperature, the equilibrium shifts in the direction that tends to cancel the change.

Effect of changing conditions on a system at equilibrium
Change made	Direction of shift	Does $K$ change?
Add a reactant or remove a product	Forward (to the right), to use up the added reactant	No
Add a product or remove a reactant	Backward (to the left)	No
Increase the pressure (gases)	Towards the side with fewer gas molecules	No
Increase the temperature	In the endothermic direction	Yes
Add a catalyst	No shift; equilibrium is only reached sooner	No

The constants of this unit

The five equilibrium constants used in Unit 3
Constant	Equilibrium it describes	Expression	A large value means
$K_{a}$	Weak acid in water: $HX + H X_{2} O H X_{3} O X^{+} + X X^{-}$	$[H X_{3} O X^{+}] [X X^{-}] / [HX]$	a stronger acid
$K_{b}$	Weak base in water: $B + H X_{2} O BH X^{+} + OH X^{-}$	$[BH X^{+}] [OH X^{-}] / [B]$	a stronger base
$K_{w}$	Ionization of water: $H X_{2} O H X^{+} + OH X^{-}$	$[H X^{+}] [OH X^{-}]$	fixed: $1.0 \times 1 0^{- 14}$ at $25^{\circ} C$
$K_{s p}$	Slightly soluble salt: $CaF X_{2} (s) Ca X^{2 +} + 2 F X^{-}$	$[Ca X^{2 +}] [F X^{-}]^{2}$	a more soluble salt (same formula type)
$K_{p c}$	Solute shared by two solvents: $X (water) X (ether)$	$[X]_{ether} / [X]_{water}$	the solute prefers the organic solvent

pH in one line

Buffer questions need the pH scale, which Unit 4 treats in full. In this unit you only need these relations:

pH = - lo g [H X^{+}], pOH = - lo g [OH X^{-}], pH + pOH = 14 (25^{\circ} C)

A solution with pH below 7 is acidic, pH 7 is neutral and pH above 7 is basic. A fall of 1 pH unit means that $[H X^{+}]$ has become 10 times larger.

3.1Strength of acids and bases

In the Bronsted–Lowry theory an acid is a proton ( $H X^{+}$ ) donor and a base is a proton acceptor. Acids differ in how readily they give away protons, and bases differ in how readily they take them.

Acid strength: The degree to which a Bronsted acid gives off protons. A relatively strong acid can give off more protons than another acid.
Base strength: The degree to which a base accepts protons. A strong base accepts protons to a greater extent than another base.

Strength is always compared with something. Hydrochloric acid is strong compared with acetic acid, and acetic acid is relatively strong compared with water. Similarly, ammonia is a stronger base than water because it accepts protons to a higher degree than water does.

When an acid $HX$ gives a proton to water it becomes $X X^{-}$ . The pair $HX$ / $X X^{-}$ is a conjugate acid–base pair. In the same way, a base $B$ gains a proton and becomes its conjugate acid $BH X^{+}$ .

acid HX + base H X_{2} O conjugate acid H X_{3} O X^{+} + conjugate base X X^{-}

Two measures are used to compare strengths: the extent (degree) of ionization, and the dissociation constant ( $K_{a}$ for acids, $K_{b}$ for bases). The next two sections use both, exactly as the learning outcomes ask.

3.1.1Strong and weak acidsOften examined

A strong acid ionizes almost completely in water. In a solution of $HCl$ nearly every molecule has given its proton to water, so the solution contains $H X_{3} O X^{+}$ and $Cl X^{-}$ ions and practically no $HCl$ molecules. A weak acid ionizes only partly. In a solution of $HF$ or $CH X_{3} COOH$ most molecules stay un-ionized and only a few per cent form ions.

HCl (aq) + H X_{2} O (l) H X_{3} O X^{+} (aq) + Cl X^{-} (aq) HF (aq) + H X_{2} O (l) H X_{3} O X^{+} (aq) + F X^{-} (aq) (complete: strong acid) (partial: weak acid)

A strong acid is almost completely ionized; a weak acid is mostly present as un-ionized molecules (ions are shown in the accent colour).

Extent (degree) of ionization, $α$: The fraction of the dissolved acid (or base) that has ionized at equilibrium: $α = \frac{amount ionized}{amount dissolved}$ . Multiplied by 100 it gives the percentage ionization.

The acid dissociation constant, $K_{a}$

The strength of an acid is generally expressed by its acid ionization constant, $K_{a}$ . Consider the ionization of a general acid $HX$ in water.

Derivation Expression for $K_{a}$

Aim. Write an equilibrium constant that measures how far a weak acid $HX$ ionizes in water.

Assumptions.

The solution is dilute, so water is present in large excess.
The temperature is constant.

Write the ionization equilibrium of the acid in water.
$HX (aq) + H X_{2} O (l) H X_{3} O X^{+} (aq) + X X^{-} (aq)$
Apply the law of mass action to get the equilibrium constant $K$ .
$K = \frac{[ H X _{3} O X ^{+} ] [ X X ^{-} ]}{[ HX ] [ H X _{2} O ]}$
Multiply both sides by $[H X_{2} O]$ to collect the water term on the left.
$K [H X_{2} O] = \frac{[ H X _{3} O X ^{+} ] [ X X ^{-} ]}{[ HX ]}$
Water is the solvent and is present in large excess (about $55.5 mol d m^{- 3}$ ). So little of it reacts that $[H X_{2} O]$ is practically constant. A constant times a constant is a new constant, called $K_{a}$ .
$K [H X_{2} O] = K_{a}$

Result

K_{a} = \frac{[ H X _{3} O X ^{+} ] [ X X ^{-} ]}{[ HX ]} or K_{a} = \frac{[ H X ^{+} ] [ X X ^{-} ]}{[ HX ]}

$K_{a}$ compares the ionized part of the acid (top) with the un-ionized part (bottom). A large $K_{a}$ means the acid is mostly ionized, so it is strong; a small $K_{a}$ means it is mostly un-ionized, so it is weak.

$K_{a}$ is called the acid dissociation constant. It measures the extent to which an acid is ionized or dissociated at equilibrium. $K_{a}$ depends on temperature, so its value is always quoted with the temperature at which it was measured. For example, $K_{a}$ of acetic acid in water at $25^{\circ} C$ is $1.8 \times 1 0^{- 5}$ . Comparing $K_{a}$ values is a way to compare the strengths of acids:

$p K_{a}$

$K_{a}$ values are inconvenient numbers (from $1 0^{10}$ down to $1 0^{- 16}$ ), so they are converted into $p K_{a}$ values:

p K_{a} = - lo g K_{a}

Because of the minus sign, the lower the $p K_{a}$ , the stronger the acid: a lower $p K_{a}$ corresponds to a higher $K_{a}$ . A change of 1 in $p K_{a}$ is a 10-fold change in $K_{a}$ .

Table 3.1: Ionization constants and $p K_{a}$ of acids in water at $25^{\circ} C$ . Acid strength decreases down the table.
Name of acid	Formula	$K_{a}$	$p K_{a}$
Perchloric acid	$HClO X_{4}$	$1.0 \times 1 0^{10}$	$- 10.0$
Hydroiodic acid	$HI$	$1.0 \times 1 0^{10}$	$- 10.0$
Hydrobromic acid	$HBr$	$1.0 \times 1 0^{9}$	$- 9.0$
Hydrochloric acid	$HCl$	$1.0 \times 1 0^{6}$	$- 6.0$
Sulphuric acid	$H X_{2} SO X_{4}$	$1.0 \times 1 0^{3}$	$- 3.0$
Hydrofluoric acid	$HF$	$7.2 \times 1 0^{- 4}$	$+ 3.1$
Formic acid	$HCOOH$	$1.8 \times 1 0^{- 4}$	$+ 3.75$
Benzoic acid	$C X_{6} H X_{5} COOH$	$6.3 \times 1 0^{- 5}$	$+ 4.2$
Acetic acid	$CH X_{3} COOH$	$1.8 \times 1 0^{- 5}$	$+ 4.7$
Phenol	$C X_{6} H X_{5} OH$	$1.3 \times 1 0^{- 10}$	$+ 9.9$ (book prints 8.9)
Water	$H X_{2} O$	$1.8 \times 1 0^{- 16}$	$+ 15.7$

formic acid

acetic acid

benzoic acid

phenol

Strongest and weakest acid in Table 3.1. Perchloric acid and hydroiodic acid are the strongest ( $K_{a} = 1.0 \times 1 0^{10}$ , $p K_{a} = - 10$ ). Water is the weakest ( $K_{a} = 1.8 \times 1 0^{- 16}$ , $p K_{a} = 15.7$ ); among the true acids in the list, phenol is the weakest.

HCl or HF: which is stronger? HCl is far stronger. Its $K_{a}$ ( $1.0 \times 1 0^{6}$ ) is about $1 0^{9}$ times larger than that of HF ( $7.2 \times 1 0^{- 4}$ ); in $p K_{a}$ terms, $- 6.0$ is much lower than $+ 3.1$ . HF is weak mainly because the H–F bond is very strong.

Extent of ionization and $K_{a}$ together

For a weak acid we can connect its $K_{a}$ to its degree of ionization $α$ and to $[H X^{+}]$ . This gives a quick way to calculate $[H X^{+}]$ , as in Example 3.1.

Derivation $[H X^{+}]$ and degree of ionization of a weak acid

Aim. Express $α$ and $[H X^{+}]$ for a weak acid $HA$ of initial concentration $c$ in terms of $K_{a}$ .

Assumptions.

The acid is weak, so $α$ is small (less than about 0.05).
The $H X^{+}$ coming from water itself is negligible.

Write the equilibrium. If a fraction $α$ of the acid ionizes, $c α$ of $HA$ is used up and $c α$ each of $H X^{+}$ and $A X^{-}$ are formed.
$initial at equilibrium [HA] c c (1 - α) [H X^{+}] 0 c α [A X^{-}] 0 c α$
Substitute the equilibrium concentrations into $K_{a}$ .
$K_{a} = \frac{( c α ) ( c α )}{c ( 1 - α )} = \frac{c α ^{2}}{1 - α}$
For a weak acid $α$ is very small, so $1 - α \approx 1$ .
$K_{a} \approx c α^{2} ⟹ α = \frac{K _{a}}{c}$
Multiply $α$ by $c$ to get the hydrogen ion concentration.
$[H X^{+}] = c α = c \frac{K _{a}}{c} = K_{a} c$

Result

α = \frac{K _{a}}{c}, [H X^{+}] = K_{a} c

At the same concentration, the acid with the larger $K_{a}$ has the larger $α$ and the larger $[H X^{+}]$ , so it is the stronger acid. The relation $α = K_{a} / c$ (Ostwald's dilution law) also shows that a weak acid ionizes more when it is diluted, while its $K_{a}$ stays the same.

Percentage ionization of two weak acids

0.2

0.4

0.6

0.8

c

α (%)

2.7%

(0.1, 8.1%)

(0.1, 1.3%)

HF (K_{a} = 7.2 \times 1 0^{- 4})

CH X_{3} COOH (K_{a} = 1.8 \times 1 0^{- 5})

Percentage ionization

α

against concentration

c

mol d m^{- 3}

. At every concentration HF (larger

K_{a}

) is more ionized than acetic acid. Both ionize more when diluted, but their

K_{a}

values stay the same.

Example 3.1:

[H X^{+}]

in a weak acid

Calculate the concentration of

H X^{+}

ions in a solution that contains

1.0 M

HF (

K_{a} = 7.2 \times 1 0^{- 4}

Write the ionization and the concentrations. Let $x mol d m^{- 3}$ of HF ionize.
$initial equilibrium HF 1.0 1.0 - x H X^{+} 0 x F X^{-} 0 x$
Write $K_{a}$ and substitute.
$K_{a} = \frac{[ H X ^{+} ] [ F X ^{-} ]}{[ HF ]} ⟹ 7.2 \times 1 0^{- 4} = \frac{x \cdot x}{1.0 - x}$
HF is a weak acid, so $x$ is very small compared with 1.0 and $1.0 - x \approx 1.0$ .
$7.2 \times 1 0^{- 4} = \frac{x ^{2}}{1.0}$
Take the square root.
$x = 7.2 \times 1 0^{- 4} = 2.68 \times 1 0^{- 2} mol d m^{- 3}$
Check the approximation: $0.0268$ is only 2.7% of $1.0$ , so neglecting $x$ was justified. The pH is $- lo g (0.0268) = 1.57$ .

[H X^{+}] = 2.68 \times 1 0^{- 2} mol d m^{- 3} (0.0268 M)

Strong and weak acids compared

Property	Strong acid (e.g. $HCl$ )	Weak acid (e.g. $CH X_{3} COOH$ , $HF$ )
Extent of ionization	Almost 100%	Small, usually a few per cent or less
Equilibrium	Lies completely to the right (written with )	Lies to the left (written with )
$K_{a}$	Very large (greater than 1, e.g. $1 0^{6}$ )	Small (less than 1, e.g. $1 0^{- 4}$ to $1 0^{- 10}$ )
$p K_{a}$	Low, often negative	Higher and positive
$[H X^{+}]$ at concentration $c$	Equal to $c$	Much smaller: about $K_{a} c$
pH of a $0.1 M$ solution	1.0	About 2.9 (acetic acid)
Electrical conductivity	High (many ions)	Low (few ions)
Conjugate base	Very weak ( $Cl X^{-}$ )	Weak but useful base ( $CH X_{3} COO X^{-}$ )
Examples	$HClO X_{4}$ , $HI$ , $HBr$ , $HCl$ , $H X_{2} SO X_{4}$ , $HNO X_{3}$	$HF$ , $HCOOH$ , $C X_{6} H X_{5} COOH$ , $CH X_{3} COOH$ , $C X_{6} H X_{5} OH$

3.1.2Strong and weak basesOften examined

The strength of a base is its ability to accept a proton from a solvent (usually water). Hydroxides of alkali metals, such as sodium hydroxide and potassium hydroxide, are strong bases: they ionize completely in aqueous solution.

\begin{aligned}&\ce{NaOH(aq) -> Na+(aq) + OH-(aq)}\\&\ce{KOH(aq) -> K+(aq) + OH-(aq)}\end{aligned}

The \ce{OH-} ion formed is a Bronsted base because it can accept a proton:

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

A weak base, such as ammonia or an amine, reacts with water only partly. It takes a proton from a water molecule and leaves an \ce{OH-} ion behind. For a base B:

\begin{aligned}&\ce{B + H2O(l) <=> BH+(aq) + OH-(aq)}\\&\ce{NH3(aq) + H2O(l) <=> NH4+(aq) + OH-(aq)}\end{aligned}

Derivation Expression for K_b

Aim. Write a constant that measures how far a weak base B reacts with water.

Apply the law of mass action to the equilibrium above.
K=\dfrac{[\ce{BH+}][\ce{OH-}]}{[\ce{B}][\ce{H2O}]}
Water is the solvent and is in large excess, so its concentration is constant. Multiply both sides by [\ce{H2O}].
K[\ce{H2O}]=\dfrac{[\ce{BH+}][\ce{OH-}]}{[\ce{B}]}
The product of two constants is a new constant, the base ionization constant K_b.
K[\ce{H2O}]=K_b

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

K_b is large when the degree of ionization of the base is high, that is, when the base is strong. It is small for a weak base. Like K_a, it depends only on temperature.

For convenience, \mathrm{p}K_b is defined as the negative logarithm of K_b. The lower the \mathrm{p}K_b, the stronger the base.

\mathrm{p}K_b=-\log K_b

Table 3.2: K_b and \mathrm{p}K_b values of some common bases. Base strength decreases down the table.
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

diethylamine

ethylamine

methylamine

pyridine

aniline

According to these values, ammonia is a stronger base than pyridine and aniline but weaker than methylamine and ethylamine. Diethylamine is the strongest base in Table 3.2 and aniline the weakest:

\ce{(C2H5)2NH}>\ce{C2H5NH2}>\ce{CH3NH2}>\ce{NH3}>\ce{C5H5N}>\ce{C6H5NH2}

Exactly the same steps as for a weak acid (previous section) give the degree of ionization and [\ce{OH-}] of a weak base of concentration c:

\alpha=\sqrt{\dfrac{K_b}{c}},\qquad[\ce{OH-}]=\sqrt{K_b\,c}

Comparing two bases by their K_b values

Two bases A and B have K_b values of 1.8\times10^{-5} and 1.0\times10^{-2} respectively. Identify the stronger base and justify your answer on the basis of extent of ionization.

Base B is the stronger base. Its K_b is about 560 times larger than that of A. A larger K_b means that, at equilibrium, a larger fraction of the base has taken protons from water to form \ce{BH+} and \ce{OH-} ions.

Compare the two bases at the same concentration, say 0.10\,\mathrm{mol\,dm^{-3}}. For A, \alpha is small, so the short formula works.
\alpha_A=\sqrt{\dfrac{1.8\times10^{-5}}{0.10}}=1.3\times10^{-2}\quad(1.3\%)
For B, K_b is large, so 1-\alpha cannot be taken as 1. Solve K_b=\dfrac{c\alpha^{2}}{1-\alpha} exactly.
0.10\,\alpha^{2}+0.010\,\alpha-0.010=0\quad\Longrightarrow\quad\alpha_B=0.27\quad(27\%)

At the same concentration B is about 20 times more ionized than A and produces about 20 times more \ce{OH-} ions, so B is the stronger base. Neither is as strong as NaOH, which is completely ionized.

\text{B }(K_b=1.0\times10^{-2})\text{ is the stronger base}

Link between K_a and K_b

Derivation K_a\times K_b=K_w for a conjugate pair

Aim. Relate K_a of a weak acid to K_b of its conjugate base.

Take the ammonium ion (an acid) and write its K_a.
\ce{NH4+ <=> NH3 + H+}\qquad K_a=\dfrac{[\ce{NH3}][\ce{H+}]}{[\ce{NH4+}]}
Write K_b for its conjugate base, ammonia.
\ce{NH3 + H2O <=> NH4+ + OH-}\qquad K_b=\dfrac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]}
Multiply the two expressions. [\ce{NH3}] and [\ce{NH4+}] cancel, leaving the ionic product of water.
K_aK_b=[\ce{H+}][\ce{OH-}]=K_w
Take -\log of both sides.
\mathrm{p}K_a+\mathrm{p}K_b=\mathrm{p}K_w=14\quad(25\,^\circ\mathrm{C})

ResultK_a\times K_b=K_w=1.0\times10^{-14}\ \ (25\,^\circ\mathrm{C})

The stronger an acid, the weaker its conjugate base, and the other way round. For example, K_b of the acetate ion is 1.0\times10^{-14}/(1.8\times10^{-5})=5.6\times10^{-10}, so acetate is a weak base.

Strong and weak bases compared

Property	Strong base (e.g. \ce{NaOH}, \ce{KOH})	Weak base (e.g. \ce{NH3}, amines)
Extent of ionization	Complete (100%) in water	Partial, usually a few per cent
Equation	\ce{NaOH -> Na+ + OH-}	\ce{NH3 + H2O <=> NH4+ + OH-}
K_b	Very large	Small, e.g. 1.7\times10^{-5} for \ce{NH3}
\mathrm{p}K_b	Very low	Higher (3 to 10 in Table 3.2)
[\ce{OH-}] at concentration c	Equal to c	About \sqrt{K_b c}
pH of a 0.1\,\mathrm{M} solution	13	About 11.1 (\ce{NH3})
Conjugate acid	Negligible acid strength (\ce{Na+})	Weak acid (\ce{NH4+})

3.2Buffer solutions and buffer actionOften examined

Pure water has pH 7, but one drop of concentrated HCl added to a litre of water lowers its pH to about 3. Blood, enzymes, many industrial processes and many products cannot tolerate such changes. They are protected by buffers.

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

Types of buffer solutions

A buffer solution can be made in two ways:

Acidic buffer: mix a weak acid with a salt of that acid with a strong base. Such solutions have a pH less than 7. Example: \ce{CH3COOH + CH3COONa}.
Basic buffer: mix a weak base with a salt of that base with a strong acid. Such solutions have a pH more than 7. Example: \ce{NH4OH + NH4Cl}.

Buffer action of an acidic buffer

Consider a buffer of \ce{CH3COOH} and \ce{CH3COONa}. The common ion effect explains how it works. Acetic acid is a weak electrolyte and ionizes very little. Sodium acetate is a strong electrolyte and ionizes completely. The large amount of \ce{CH3COO-} from the salt suppresses the ionization of \ce{CH3COOH} even further (common ion \ce{CH3COO-}).

\begin{aligned}&\ce{CH3COOH(l) + H2O(l) <=> CH3COO-(aq) + H3O+(aq)}&&\text{(very little)}\\&\ce{CH3COONa(s) -> CH3COO-(aq) + Na+(aq)}&&\text{(complete)}\end{aligned}

The buffer therefore holds two large reserves: un-ionized \ce{CH3COOH} molecules, which can react with any added base, and \ce{CH3COO-} ions, which can react with any added acid.

Buffer action of \ce{CH3COOH}/\ce{CH3COONa}: each reserve removes one kind of added ion, so [\ce{H+}] hardly changes.

(i) Adding a small amount of strong acid (HCl). The \ce{H+} ions from HCl are used up by the \ce{CH3COO-} ions and turned into acetic acid, which is only weakly ionized. [\ce{H+}] stays almost unchanged, so the addition of HCl does not change the pH significantly.

\begin{aligned}&\ce{HCl(aq) -> H+(aq) + Cl-(aq)}\\&\ce{CH3COO-(aq) + H+(aq) <=> CH3COOH(aq)}\end{aligned}

(ii) Adding a small amount of strong base (NaOH). The \ce{OH-} ions are neutralised by the \ce{H+} ions of the acid. More \ce{CH3COOH} then ionizes to replace the \ce{H+} used up (Le Chatelier's principle). The overall result is that the acetic acid removes the added \ce{OH-}, and the pH does not change significantly.

\begin{aligned}&\ce{NaOH(aq) -> Na+(aq) + OH-(aq)}\\&\ce{H+(aq) + OH-(aq) <=> H2O(l)}\\&\text{overall: }\ce{CH3COOH(aq) + OH-(aq) -> CH3COO-(aq) + H2O(l)}\end{aligned}

Buffer action of a basic buffer

The \ce{NH4OH}/\ce{NH4Cl} buffer works in the same way. \ce{NH4Cl} supplies a large amount of \ce{NH4+}, and \ce{NH4OH} is only slightly ionized, so both reserves are large.

\begin{aligned}&\text{added acid: }\ce{NH4OH(aq) + H+(aq) -> NH4+(aq) + H2O(l)}\\&\text{added base: }\ce{NH4+(aq) + OH-(aq) -> NH4OH(aq)}\end{aligned}

Adding HCl to water and to a buffer

pH against n, the moles of HCl added to 1\,\mathrm{dm^3} of solution. The buffer of Example 3.2 (1.0\,\mathrm{mol} each of \ce{CH3COOH} and \ce{CH3COONa} in 1\,\mathrm{dm^3}) stays near pH 4.7, while pure water drops from pH 7 to 2 with only 0.01\,\mathrm{mol} of HCl.

3.2.1Calculating the pH of a buffer (Henderson equation)Often examined

In a buffer, the conjugate base is supplied almost entirely by the salt, which is a strong electrolyte. The weak acid ionizes so little (and even less in the presence of the common ion) that its concentration hardly changes. So we can take the concentrations of the salt and the acid we started with as the equilibrium concentrations:

[\ce{A-}]_{\text{eq}}\approx[\text{salt}],\qquad[\ce{HA}]_{\text{eq}}\approx[\text{acid}]

Derivation Henderson equation for an acidic buffer

Aim. Express the pH of a buffer of a weak acid \ce{HA} and its salt \ce{NaA} in terms of \mathrm{p}K_a and the concentrations.

Assumptions.

The salt is completely ionized, so [\ce{A-}] equals the salt concentration.
The weak acid is so little ionized that its concentration stays equal to the starting value.

Write K_a for the weak acid.
\ce{HA <=> H+ + A-}\qquad K_a=\dfrac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}
Make [\ce{H+}] the subject.
[\ce{H+}]=K_a\,\dfrac{[\ce{HA}]}{[\ce{A-}]}
Take -\log of both sides. The log of a product is the sum of the logs.
-\log[\ce{H+}]=-\log K_a-\log\dfrac{[\ce{HA}]}{[\ce{A-}]}
Use \mathrm{pH}=-\log[\ce{H+}], \mathrm{p}K_a=-\log K_a and -\log\dfrac{a}{b}=\log\dfrac{b}{a}.
\mathrm{pH}=\mathrm{p}K_a+\log\dfrac{[\ce{A-}]}{[\ce{HA}]}
Replace [\ce{A-}] by the salt concentration and [\ce{HA}] by the acid concentration.
\mathrm{pH}=\mathrm{p}K_a+\log\dfrac{[\text{salt}]}{[\text{acid}]}

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

The pH of a buffer depends on \mathrm{p}K_a and on the ratio of salt to acid, not on their actual amounts. When [\text{salt}]=[\text{acid}], \log1=0 and \mathrm{pH}=\mathrm{p}K_a. A little added acid or base changes the ratio only slightly, so the pH changes only slightly.

Derivation Henderson equation for a basic buffer

Aim. Find the pH of a buffer of a weak base (\ce{NH4OH}) and its salt with a strong acid (\ce{NH4Cl}).

Write K_b for the weak base.
\ce{NH4OH <=> NH4+ + OH-}\qquad K_b=\dfrac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH4OH}]}
Make [\ce{OH-}] the subject; [\ce{NH4+}] is supplied by the salt.
[\ce{OH-}]=K_b\,\dfrac{[\text{base}]}{[\text{salt}]}
Take -\log of both sides, as for the acidic buffer.
\mathrm{pOH}=\mathrm{p}K_b+\log\dfrac{[\text{salt}]}{[\text{base}]}
Convert pOH into pH.
\mathrm{pH}=14-\mathrm{pOH}

Result\mathrm{pOH}=\mathrm{p}K_b+\log\dfrac{[\text{salt}]}{[\text{base}]},\qquad\mathrm{pH}=14-\mathrm{pOH}

For equal amounts of base and salt, \mathrm{pOH}=\mathrm{p}K_b. With K_b=1.8\times10^{-5} this gives \mathrm{pOH}=4.74 and \mathrm{pH}=9.26 (long question v).

Example 3.2: pH of an acetate buffer before and after adding HCl

(a) Calculate the pH of an acetic acid–sodium acetate buffer solution containing 1.0 mole of each component. (b) What will be the pH of this solution after the addition of 0.01 mole of hydrochloric acid to 1\,\mathrm{dm^3} of the solution? Assume that the volume of the solution remains unchanged. (K_a for acetic acid is 1.8\times10^{-5}.)

(a) pH of the buffer

Take the starting amounts (in 1\,\mathrm{dm^3}) as the equilibrium concentrations of the acid and its conjugate base.
[\ce{CH3COOH}]=1.0\,\mathrm{M},\qquad[\ce{CH3COO-}]=1.0\,\mathrm{M}
Substitute into the K_a expression for acetic acid.
K_a=\dfrac{[\ce{CH3COO-}][\ce{H+}]}{[\ce{CH3COOH}]}\;\Longrightarrow\;1.8\times10^{-5}=\dfrac{(1.0)[\ce{H+}]}{1.0}
So [\ce{H+}]=1.8\times10^{-5}\,\mathrm{M}. Find the pH.
\mathrm{pH}=-\log(1.8\times10^{-5})=4.745

(b) After adding 0.01 mol HCl

HCl ionizes completely, giving 0.01\,\mathrm{mol} of \ce{H+}.
\ce{HCl(aq) -> H+(aq) + Cl-(aq)}
The added \ce{H+} combines with 0.01\,\mathrm{mol} of acetate ions to form 0.01\,\mathrm{mol} more acetic acid.
\ce{CH3COO-(aq) + H+(aq) -> CH3COOH(aq)}
Find the new amounts of acid and conjugate base.
\begin{aligned}n(\ce{CH3COOH})&=1.0+0.01=1.01\,\mathrm{mol}\\n(\ce{CH3COO-})&=1.0-0.01=0.99\,\mathrm{mol}\end{aligned}
Rearrange K_a for [\ce{H+}] and substitute.
[\ce{H+}]=\dfrac{K_a[\ce{CH3COOH}]}{[\ce{CH3COO-}]}=\dfrac{1.8\times10^{-5}\times1.01}{0.99}=1.84\times10^{-5}\,\mathrm{M}
Find the new pH.
\mathrm{pH}=-\log(1.84\times10^{-5})=4.736

The pH changes only from 4.745 to 4.736, a difference of 0.009. The same 0.01\,\mathrm{mol} of HCl added to 1\,\mathrm{dm^3} of pure water would change the pH from 7 to 2, a change of 5 units. A buffer does a very good job of limiting the change in pH.

\text{(a) }\mathrm{pH}=4.745\qquad\text{(b) }\mathrm{pH}=4.736

Choosing the ratio for a required pH

What ratio of [\ce{CH3COONa}] to [\ce{CH3COOH}] gives a buffer of pH 5.00? (\mathrm{p}K_a of acetic acid =4.74)

Write the Henderson equation.
\mathrm{pH}=\mathrm{p}K_a+\log\dfrac{[\text{salt}]}{[\text{acid}]}
Substitute and isolate the log term.
5.00=4.74+\log\dfrac{[\text{salt}]}{[\text{acid}]}\;\Longrightarrow\;\log\dfrac{[\text{salt}]}{[\text{acid}]}=0.26
Take the antilog.
\dfrac{[\text{salt}]}{[\text{acid}]}=10^{0.26}=1.8

So dissolving 1.8 mol of sodium acetate for every 1 mol of acetic acid gives a buffer of pH 5.00. A pH above \mathrm{p}K_a needs more salt than acid.

[\ce{CH3COONa}]:[\ce{CH3COOH}]=1.8:1

3.2.2Applications of buffers and the bicarbonate buffer in bloodOften examined

Buffers play a key role in many everyday systems, where they keep conditions stable and prevent sudden changes in pH. The most important example is our own blood.

1. The buffer system in blood plasma

Blood pH must be kept within narrow limits so that enzymes and other body processes can work. The normal pH range of blood is about 7.35 to 7.45. A value higher than 7.8 or lower than 6.8 can lead to death. The main buffer in blood plasma is the bicarbonate buffer system, \ce{H2CO3}/\ce{HCO3-}, in which carbon dioxide and bicarbonate ions are exchanged reversibly.

(a) Carbon dioxide dissolves in the water of blood and forms carbonic acid:

\ce{CO2(g) + H2O(l) <=> H2CO3(aq)}

(b) Carbonic acid dissociates into a bicarbonate ion and a hydrogen ion:

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

This system is catalysed by the enzyme carbonic anhydrase, found in red blood cells. The levels of \ce{CO2}, \ce{H2CO3}, \ce{HCO3-} and \ce{H+} are linked, so any imbalance affects the pH of the blood. \ce{HCO3-} regulates blood pH in two ways.

The bicarbonate buffer system of blood plasma. Added acid is taken up by \ce{HCO3-}; added base is neutralised by \ce{H2CO3}.

Neutralization of acids. When there are many acids (\ce{H+} ions) in the blood, bicarbonate acts as a base. It binds the extra hydrogen ions and forms carbonic acid, which removes the excess acidity. The carbonic acid breaks down to \ce{CO2} and water, and the lungs breathe out the \ce{CO2}.

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

Neutralization of bases. When there are many bases in the blood, such as hydroxide ions, carbonic acid dissociates and releases hydrogen ions, which combine with the excess base.

\ce{OH-(aq) + H2CO3(aq) <=> HCO3-(aq) + H2O(l)}

Through these reversible reactions, bicarbonate ions keep the acid–base balance of the blood. The system prevents rapid changes in blood pH and provides a stable environment for enzymes and other biochemical processes. The respiratory system (lungs, which control \ce{CO2}) and the renal system (kidneys, which control \ce{HCO3-}) also play an important part in keeping blood pH in the normal range.

2 to 6. Other uses of buffers

Applications of buffer solutions (book section 3.2.2)
Area	How a buffer is used
2. Household cleaning products	Many cleaning agents and detergents contain buffers to keep a stable pH, so they clean well without harming the surface or the skin.
3. Personal care products	Buffers keep the pH of shampoos and soaps steady. This prevents skin irritation and keeps the products gentle.
4. Swimming pools	In water treatment, buffers keep the pH of pool water stable. This prevents corrosion of pool equipment and keeps the water comfortable for swimmers.
5. Food and beverage industry	In food processing, buffers control acidity, preserve flavours and keep some food products stable. In beverages they control pH, which keeps the taste consistent and prevents spoilage.
6. Photography	Photographic developer solutions contain buffers to keep a stable pH, so film or prints develop in a controlled and consistent way.

3.3, 3.3.1Solubility product, KspOften examined

Some equilibria are important in chemical analysis. One of them is the equilibrium between a slightly soluble ionic solid and its ions in a saturated solution.

When an excess of a slightly soluble ionic compound is mixed with water, some of it dissolves and the rest settles at the bottom. A dynamic equilibrium is set up between the undissolved solid and its ions in the saturated solution: ions leave the solid at the same rate as other ions return to it. For calcium fluoride:

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

Solubility equilibrium: in a saturated solution, dissolving and precipitation go on at the same rate.

Derivation The solubility product expression

Aim. Obtain K_{sp} for calcium fluoride from the equilibrium constant K_c.

Apply the law of mass action to the equilibrium.
K_c=\dfrac{[\ce{Ca^2+}][\ce{F-}]^{2}}{[\ce{CaF2}]}
\ce{CaF2} is a slightly soluble solid, so its concentration remains practically constant. Multiply both sides by [\ce{CaF2}].
K_c[\ce{CaF2}]=[\ce{Ca^2+}][\ce{F-}]^{2}
A constant times a constant is a new constant, the solubility product constant K_{sp}.
K_c[\ce{CaF2}]=K_{sp}

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

Only the ions appear in K_{sp}. Each ion concentration is raised to the power of its coefficient in the balanced equation, so [\ce{F-}] is squared because one \ce{CaF2} gives two \ce{F-} ions.

Solubility product constant, K_{sp}: The product of the molar concentrations of the ions in a saturated solution of a slightly soluble salt, each raised to a power equal to the coefficient of the ion in the balanced chemical equation. It is constant at a given temperature.

In general, the K_{sp} expression of any slightly soluble ionic compound \mathrm{A}_m\mathrm{B}_n is written as:

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

This means that the solubility product constant equals 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.

Concept Assessment Exercise 3.1

Write the dissociation reaction and the solubility product constant expression for each of the following solids: (1) \ce{PbSO4} (2) \ce{Al(OH)3} (3) \ce{BaSO4}

Solid	Dissociation reaction	K_{sp} expression
\ce{PbSO4}	\ce{PbSO4(s) <=> Pb^2+(aq) + SO4^2-(aq)}	K_{sp}=[\ce{Pb^2+}][\ce{SO4^2-}]
\ce{Al(OH)3}	\ce{Al(OH)3(s) <=> Al^3+(aq) + 3OH-(aq)}	K_{sp}=[\ce{Al^3+}][\ce{OH-}]^{3}
\ce{BaSO4}	\ce{BaSO4(s) <=> Ba^2+(aq) + SO4^2-(aq)}	K_{sp}=[\ce{Ba^2+}][\ce{SO4^2-}]

The solid never appears in the expression, and the power 3 on [\ce{OH-}] comes from the 3 hydroxide ions in one formula unit of \ce{Al(OH)3}.

Precipitation reactions: ionic product and K_{sp}

The ionic product, Q, has the same form as K_{sp} but uses the actual ion concentrations at any moment, not only at equilibrium. Comparing Q with K_{sp} tells us whether a precipitate forms:

Q>K_{sp}: the solution holds more ions than it can at equilibrium, so a precipitate forms until Q falls to K_{sp}.
Q=K_{sp}: the solution is exactly saturated, and nothing changes.
Q<K_{sp}: the solution is unsaturated; no precipitate forms and more solid can dissolve.

Will a precipitate form?

Equal volumes of 1.0\times10^{-4}\,\mathrm{M} \ce{AgNO3} and 1.0\times10^{-4}\,\mathrm{M} NaCl are mixed. Will AgCl precipitate? (K_{sp} of AgCl =1.8\times10^{-10})

Mixing equal volumes doubles the volume, so each concentration is halved.
[\ce{Ag+}]=[\ce{Cl-}]=5.0\times10^{-5}\,\mathrm{M}
Calculate the ionic product.
Q=[\ce{Ag+}][\ce{Cl-}]=(5.0\times10^{-5})^{2}=2.5\times10^{-9}
Compare Q with K_{sp}.
2.5\times10^{-9}>1.8\times10^{-10}

Q>K_{sp},\ \text{so AgCl precipitates}

3.3.2Calculating concentrations of ions from KspOften examined

Slightly soluble salts dissociate only to a small extent in water, so their ion concentrations are low. To calculate them we use the solubility product constant, because the K_{sp} expression relates the concentrations of the ions in equilibrium with the undissolved salt.

Let x (or s) be the molar solubility: the number of moles of the salt that dissolve in 1\,\mathrm{dm^3} of water to give a saturated solution.

Steps to calculate the concentration

Write the dissociation equation for the salt.
Set up the K_{sp} expression.
Solve for the ion concentrations using the stoichiometry (mole ratio) of the equation.

Ion concentrations in saturated AgBr

Given K_{sp} for AgBr =1.60\times10^{-10}, calculate the concentration of each ion in its saturated solution.

Let x mol of AgBr dissolve in 1\,\mathrm{dm^3} of water. Each mole gives one mole of each ion.
\ce{AgBr(s) <=> Ag+(aq) + Br-(aq)}\qquad[\ce{Ag+}]=[\ce{Br-}]=x
Set up K_{sp} and substitute.
K_{sp}=[\ce{Ag+}][\ce{Br-}]\;\Longrightarrow\;1.60\times10^{-10}=x\cdot x=x^{2}
Take the square root.
x=\sqrt{1.60\times10^{-10}}=1.26\times10^{-5}\,\mathrm{M}

[\ce{Ag+}]=[\ce{Br-}]=1.26\times10^{-5}\,\mathrm{mol\,dm^{-3}}

Example 3.3: solubility of silver chloride

Calculate the concentration of silver chloride (AgCl) in its aqueous solution. (K_{sp} for AgCl =1.8\times10^{-10} at 25\,^\circ\mathrm{C})

Dissociation.
\ce{AgCl(s) <=> Ag+(aq) + Cl-(aq)}
K_{sp} expression.
K_{sp}=[\ce{Ag+}][\ce{Cl-}]
Concentration calculation. Each mole of AgCl gives one mole of \ce{Ag+} and one mole of \ce{Cl-}. Let the solubility be x\,\mathrm{mol\,dm^{-3}}.
1.8\times10^{-10}=x^{2}\;\Longrightarrow\;x=\sqrt{1.8\times10^{-10}}=1.34\times10^{-5}\,\mathrm{mol\,dm^{-3}}

So the solubility of AgCl, which is also the concentration of \ce{Ag+} and of \ce{Cl-} in water, is about 1.34\times10^{-5}\,\mathrm{mol\,dm^{-3}}.

s=[\ce{Ag+}]=[\ce{Cl-}]=1.34\times10^{-5}\,\mathrm{mol\,dm^{-3}}

Derivation Molar solubility of any salt \mathrm{A}_m\mathrm{B}_n

Aim. Find a general link between K_{sp} and the molar solubility s.

Each mole of \mathrm{A}_m\mathrm{B}_n that dissolves gives m moles of cations and n moles of anions.
\ce{A_{m}B_{n}(s) <=> m A^{n+}(aq) + n B^{m-}(aq)}\qquad[\ce{A^{n+}}]=ms,\quad[\ce{B^{m-}}]=ns
Substitute into the K_{sp} expression.
K_{sp}=(ms)^{m}(ns)^{n}=m^{m}n^{n}\,s^{m+n}
Rearrange for s.
s=\left(\dfrac{K_{sp}}{m^{m}n^{n}}\right)^{\frac{1}{m+n}}

ResultK_{sp}=m^{m}n^{n}\,s^{m+n}

For a 1:1 salt K_{sp}=s^{2}; for a 1:2 or 2:1 salt K_{sp}=4s^{3}; for a 1:3 salt K_{sp}=27s^{4}. Once s is known, each ion concentration follows from the formula of the salt.

Salt type	Examples	Ion concentrations	K_{sp} in terms of s
AB	\ce{AgCl}, \ce{AgBr}, \ce{BaSO4}	s and s	s^{2}
\mathrm{AB_2}	\ce{CaF2}, \ce{PbF2}	s and 2s	4s^{3}
\mathrm{A_2B}	\ce{Ag2CrO4}	2s and s	4s^{3}
\mathrm{AB_3}	\ce{Al(OH)3}	s and 3s	27s^{4}

Concept Assessment Exercise 3.2 (1)

K_{sp} for \ce{PbF2} is 4\times10^{-8}. Calculate the concentration of \ce{Pb^2+} and \ce{F-} ions in solution.

Write the dissociation and let the molar solubility be s. One \ce{PbF2} gives one \ce{Pb^2+} and two \ce{F-}.
\ce{PbF2(s) <=> Pb^2+(aq) + 2F-(aq)}\qquad[\ce{Pb^2+}]=s,\quad[\ce{F-}]=2s
Write K_{sp} and substitute. Double the fluoride concentration before squaring it.
K_{sp}=[\ce{Pb^2+}][\ce{F-}]^{2}=s(2s)^{2}=4s^{3}
Solve for s.
4s^{3}=4\times10^{-8}\;\Longrightarrow\;s^{3}=1\times10^{-8}\;\Longrightarrow\;s=2.15\times10^{-3}\,\mathrm{mol\,dm^{-3}}
Find each ion concentration.
[\ce{Pb^2+}]=2.15\times10^{-3}\,\mathrm{mol\,dm^{-3}},\qquad[\ce{F-}]=2s=4.3\times10^{-3}\,\mathrm{mol\,dm^{-3}}

[\ce{Pb^2+}]=2.2\times10^{-3}\,\mathrm{mol\,dm^{-3}},\quad[\ce{F-}]=4.3\times10^{-3}\,\mathrm{mol\,dm^{-3}}

Concept Assessment Exercise 3.2 (2)

K_{sp} for \ce{BaSO4} is 1\times10^{-10}. Write the equation and the equilibrium expression for the dissolving of barium sulphate. Also calculate the concentration of each ion.

Equation: one \ce{BaSO4} gives one \ce{Ba^2+} and one \ce{SO4^2-}.
\ce{BaSO4(s) <=> Ba^2+(aq) + SO4^2-(aq)}
Equilibrium expression.
K_{sp}=[\ce{Ba^2+}][\ce{SO4^2-}]
Let [\ce{Ba^2+}]=[\ce{SO4^2-}]=x and solve.
x^{2}=1\times10^{-10}\;\Longrightarrow\;x=1\times10^{-5}\,\mathrm{mol\,dm^{-3}}

[\ce{Ba^2+}]=[\ce{SO4^2-}]=1\times10^{-5}\,\mathrm{mol\,dm^{-3}}

From solubility to K_{sp}

The molar solubility of \ce{CaF2} at 25\,^\circ\mathrm{C} is 2.1\times10^{-4}\,\mathrm{mol\,dm^{-3}}. Calculate its K_{sp}.

Find the ion concentrations from the formula \ce{CaF2}.
[\ce{Ca^2+}]=2.1\times10^{-4}\,\mathrm{M},\qquad[\ce{F-}]=2\times2.1\times10^{-4}=4.2\times10^{-4}\,\mathrm{M}
Substitute into the K_{sp} expression.
K_{sp}=[\ce{Ca^2+}][\ce{F-}]^{2}=(2.1\times10^{-4})(4.2\times10^{-4})^{2}=3.7\times10^{-11}

K_{sp}(\ce{CaF2})=3.7\times10^{-11}

3.4Common ion effectOften examined

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

\begin{aligned}&\ce{HF(aq) <=>[H2O] H+(aq) + F-(aq)}\\&\ce{NaF(s) ->[H2O] Na+(aq) + F-(aq)}\end{aligned}

HF is a weak electrolyte, so it dissociates only slightly. NaF is a strong electrolyte and breaks up completely into its ions. The common ion \ce{F-} produced by NaF increases [\ce{F-}] and upsets the HF equilibrium. According to Le Chatelier's principle, the equilibrium shifts to the left to use up some of the extra \ce{F-} ions. So the dissociation of HF decreases in the presence of dissolved NaF, [\ce{H+}] falls and the concentration of un-ionized HF increases.

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. This causes a decrease in its solubility, and some of it may precipitate.

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

The term describes the behaviour of a solution in which the same ion is produced by two different compounds.

Examples from the book

(i) Precipitation of potassium perchlorate. \ce{KClO4} is moderately soluble in water. When highly soluble KCl is added to its saturated solution, the concentration of \ce{K+} ions increases. According to Le Chatelier's principle, the extra \ce{K+} ions combine with \ce{ClO4-} ions to form solid \ce{KClO4}. The ionization of \ce{KClO4} is suppressed and it precipitates out.

\begin{aligned}&\ce{KClO4(s) <=> K+(aq) + ClO4-(aq)}\\&\ce{KCl(s) -> K+(aq) + Cl-(aq)}\end{aligned}

(ii) Purification of common salt. When HCl gas is passed through a saturated solution of NaCl (brine), the concentration of \ce{Cl-} ions increases. According to Le Chatelier's principle, \ce{Cl-} ions combine with \ce{Na+} ions and pure NaCl precipitates, because the ionic product [\ce{Na+}][\ce{Cl-}] now exceeds the solubility product. Soluble impurities stay in solution.

\begin{aligned}&\ce{NaCl(s) <=> Na+(aq) + Cl-(aq)}\\&\ce{HCl(aq) -> H+(aq) + Cl-(aq)}\end{aligned}

How large is the effect

The common ion effect in numbers

(a) Compare [\ce{H+}] in 1.0\,\mathrm{M} HF alone with [\ce{H+}] in a solution that is 1.0\,\mathrm{M} in HF and also 1.0\,\mathrm{M} in NaF (K_a=7.2\times10^{-4}). (b) Compare the solubility of AgCl in pure water and in 0.10\,\mathrm{M} NaCl (K_{sp}=1.8\times10^{-10}).

(a) HF alone (Example 3.1).
[\ce{H+}]=\sqrt{K_a c}=\sqrt{7.2\times10^{-4}\times1.0}=2.7\times10^{-2}\,\mathrm{M}\quad(\mathrm{pH}=1.57)
With NaF, [\ce{F-}]\approx1.0\,\mathrm{M} (from the salt) and [\ce{HF}]\approx1.0\,\mathrm{M} (hardly ionized). Rearrange K_a for [\ce{H+}].
[\ce{H+}]=K_a\dfrac{[\ce{HF}]}{[\ce{F-}]}=7.2\times10^{-4}\times\dfrac{1.0}{1.0}=7.2\times10^{-4}\,\mathrm{M}\quad(\mathrm{pH}=3.14)
(b) AgCl in pure water.
s=\sqrt{K_{sp}}=\sqrt{1.8\times10^{-10}}=1.34\times10^{-5}\,\mathrm{mol\,dm^{-3}}
In 0.10\,\mathrm{M} NaCl, [\ce{Cl-}]\approx0.10\,\mathrm{M} (the tiny amount from AgCl is negligible), and s=[\ce{Ag+}].
s=\dfrac{K_{sp}}{[\ce{Cl-}]}=\dfrac{1.8\times10^{-10}}{0.10}=1.8\times10^{-9}\,\mathrm{mol\,dm^{-3}}

The common ion lowers [\ce{H+}] of HF about 37 times, and it lowers the solubility of AgCl about 7500 times.

Common ion effect on the solubility of AgCl

Solubility s of AgCl against the concentration c of added \ce{Cl-}, both in units of 10^{-5}\,\mathrm{mol\,dm^{-3}}. Adding even a little \ce{Cl-} (from NaCl) sharply lowers the solubility s of AgCl, because [\ce{Ag+}][\ce{Cl-}] cannot exceed K_{sp}=1.8\times10^{-10}.

Self Check Exercise 3.3

Self Check Exercise 3.3 (i)

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. It ionizes completely and gives a large amount of \ce{NH4+}, the ion common to both compounds.

\ce{NH4Cl(s) -> NH4+(aq) + Cl-(aq)}

The extra \ce{NH4+} shifts the \ce{NH4OH} equilibrium to the left (Le Chatelier's principle). The ionization of \ce{NH4OH} is suppressed, so [\ce{OH-}] decreases and the pH falls a little. The mixture of \ce{NH4OH} and \ce{NH4Cl} is a basic buffer. In qualitative analysis this low [\ce{OH-}] precipitates only the least soluble hydroxides, such as \ce{Fe(OH)3} and \ce{Al(OH)3} (group III).

Self Check Exercise 3.3 (ii)

Carbonic acid is a weak acid. It ionizes 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} ionizes completely and supplies the common ion \ce{CO3^2-}.

\ce{Na2CO3(s) -> 2Na+(aq) + CO3^2-(aq)}

The increase in [\ce{CO3^2-}] shifts the carbonic acid equilibrium to the left. The ionization of \ce{H2CO3} is suppressed, [\ce{H+}] decreases and the pH of the solution rises. (Carbonic acid really ionizes in two steps, but the conclusion is the same.)

Applications of the common ion effect

Buffer solutions: the common ion from the salt suppresses the ionization of the weak acid or weak base (section 3.2).
Purification of common salt: passing HCl gas through brine precipitates pure NaCl.
Qualitative analysis: group II sulphides are precipitated by \ce{H2S} in the presence of HCl; the \ce{H+} suppresses the ionization of \ce{H2S}, so [\ce{S^2-}] stays low and only the least soluble sulphides precipitate. Group III hydroxides are precipitated by \ce{NH4OH} in the presence of \ce{NH4Cl}.
Salting out of soap: adding NaCl to the soap solution supplies the common ion \ce{Na+}, so the soap (a sodium salt) separates out.

3.5Partition coefficient and the distribution lawOften examined

If you put two immiscible liquids, such as ether and water, in a separatory funnel and shake them, two layers appear on standing. Ether is less dense than water, so it forms the upper layer.

Suppose a solute X dissolves in both liquids. Prepare a solution of X in water of known concentration, add ether, shake well and let the layers separate. The concentration of X in the water layer is now lower, because some X has moved into the ether layer: analysis of the ether layer shows that it also contains X. The solute is distributed between the two solvents. Molecules of X keep crossing the boundary in both directions, so a dynamic equilibrium is set up. At equilibrium the concentration of X in each layer becomes constant.

Partition of a solute X between ether (upper layer) and water (lower layer). At equilibrium the ratio of its concentrations in the two layers is constant.

\ce{X}\ (\text{in water})\ \ce{<=>}\ \ce{X}\ (\text{in ether})\qquad K_{pc}=\dfrac{[\ce{X}\text{ in ether}]}{[\ce{X}\text{ in water}]}

This equilibrium constant is called the partition coefficient and is represented by the symbol K_{pc}. It has a constant value at a constant temperature.

Partition coefficient, K_{pc} (book definition): The ratio of the concentrations of a solute in two different immiscible solvents in contact with each other when equilibrium has been established at a particular temperature. It is a ratio of two concentrations, so it has no units.

Which concentration goes on top of the ratio

The definition only says 'ratio', so you must know the book's convention. In every example in this book, the concentration in the organic solvent (ether, benzene, chloroform, 'the organic layer') is written on top, and the concentration in water is written at the bottom. Long question vi writes the same thing as K_{\text{benzene/water}}.

K_{pc}=\dfrac{[\text{solute}]_{\text{organic solvent}}}{[\text{solute}]_{\text{water}}}

K_{pc}>1: the solute is more soluble in the organic (non-polar) solvent. Example: methylamine in Activity 3.1, K_{pc}=1.82.
K_{pc}<1: the solute is more soluble in water (the polar solvent). Example: ethanol between octanol and water.
K_{pc}=1: the solute is equally soluble in both solvents.

Nernst distribution law

Concentration of solute in the organic layer, c_{\text{org}}, against its concentration in water, c_{\text{aq}}. At a fixed temperature c_{\text{org}} is a fixed multiple of c_{\text{aq}}, so the graph is a straight line through the origin with slope K_{pc}. Above the dashed line the solute prefers the organic solvent; below it, water.

Nernst's distribution law

The distribution law, or Nernst distribution law, is the generalisation that governs the distribution of a solute between two immiscible solvents. It was first given by Nernst, who studied the partitioning of several solutes between various suitable pairs of solvents.

Nernst distribution law: At a constant temperature, a solute distributes itself between two immiscible solvents in contact with each other so that the ratio of its concentrations in the two solvents is constant, whatever the total amount of solute.

The law holds only when:

the temperature is constant;
the two solvents are immiscible and do not react with each other or with the solute;
the solute is in the same physical (molecular) state in both solvents, that is, it does not associate or dissociate in either of them;
the solutions are dilute and equilibrium has been reached.

3.5.1Calculating a partition coefficientOften examined

The partition coefficient for a system in which the solute is in the same physical state in the two solvents can be calculated from the equilibrium expression. We need the equilibrium concentration of the solute in each layer. When amounts and volumes are given:

K_{pc}=\dfrac{n_{\text{org}}/V_{\text{org}}}{n_{\text{aq}}/V_{\text{aq}}}=\dfrac{m_{\text{org}}/V_{\text{org}}}{m_{\text{aq}}/V_{\text{aq}}}

Here n is the number of moles and m the mass of solute in a layer, and V is the volume of that layer. Moles or grams may be used, and any volume unit, provided the same units are used for both layers.

Activity 3.1: partition coefficient of methylamine

Measure 100\,\mathrm{cm^3} of a 0.150\,\mathrm{M} solution of aqueous methylamine and add it to a separatory funnel.
Add 75\,\mathrm{cm^3} of an organic solvent to the separatory funnel.
Shake the funnel gently but thoroughly so that the solute distributes between the two solvents. Allow the system to reach equilibrium (about 5 to 10 minutes).
Let the layers settle and separate the two solvents into different beakers.
Measure the concentration of the solute in each solvent. Here a sample of the aqueous layer is titrated with HCl.

Activity 3.1: calculation

100\,\mathrm{cm^3} of a 0.150\,\mathrm{mol\,dm^{-3}} solution of aqueous methylamine (\ce{CH3NH2}) is shaken with 75.0\,\mathrm{cm^3} of an organic solvent in a separating funnel and allowed to come to equilibrium. Only 25.0\,\mathrm{cm^3} of the aqueous layer is run off and titrated against 0.225\,\mathrm{M} HCl; 7.05\,\mathrm{cm^3} of HCl is used. Calculate the partition coefficient of methylamine between the organic solvent and water.

Given

V_{\text{aq}} = 100\,\mathrm{cm^3}=0.100\,\mathrm{dm^3} volume of aqueous methylamine
c_0 = 0.150\,\mathrm{mol\,dm^{-3}} initial concentration of methylamine
V_{\text{org}} = 75.0\,\mathrm{cm^3}=0.0750\,\mathrm{dm^3} volume of organic solvent
V_{\text{titrated}} = 25.0\,\mathrm{cm^3} volume of aqueous layer titrated
V_{\ce{HCl}} = 7.05\,\mathrm{cm^3} volume of HCl used
[\ce{HCl}] = 0.225\,\mathrm{mol\,dm^{-3}} concentration of HCl

To find

K_{pc} of methylamine (organic/water)

Step 1. Write the equilibrium equation.
\ce{CH3NH2(aq)}\ \ce{<=>}\ \ce{CH3NH2}\ (\text{organic solvent})
Step 2. Write the K_{pc} expression, organic layer on top.
K_{pc}=\dfrac{[\ce{CH3NH2}]_{\text{organic}}}{[\ce{CH3NH2}]_{\text{aq}}}
Step 3. Total moles of methylamine in the original solution, n=cV.
n_{\text{total}}=0.150\times\dfrac{100}{1000}=0.0150\,\mathrm{mol}
Step 4. Moles in the 25.0\,\mathrm{cm^3} titrated. Methylamine reacts with HCl in a 1:1 ratio, \ce{CH3NH2 + HCl -> CH3NH3Cl}, so moles of methylamine equal moles of HCl.
n=0.225\times\dfrac{7.05}{1000}=1.586\times10^{-3}\,\mathrm{mol}
Scale up from 25.0\,\mathrm{cm^3} to the whole 100\,\mathrm{cm^3} aqueous layer.
n_{\text{aq}}=1.586\times10^{-3}\times\dfrac{100}{25.0}=6.345\times10^{-3}\,\mathrm{mol}
Step 5. Moles in the organic layer, by difference.
n_{\text{org}}=0.0150-0.006345=8.655\times10^{-3}\,\mathrm{mol}
Step 6. Change moles into concentrations (c=n/V).
[\ce{CH3NH2}]_{\text{aq}}=\dfrac{6.345\times10^{-3}}{0.100}=0.0634\,\mathrm{M},\qquad[\ce{CH3NH2}]_{\text{org}}=\dfrac{8.655\times10^{-3}}{0.0750}=0.115\,\mathrm{M}
Step 7. Substitute into the K_{pc} expression.
K_{pc}=\dfrac{0.115}{0.0634}=1.82

Since K_{pc} is larger than 1, methylamine is more soluble in the organic solvent than in water.

K_{pc}\approx1.82\ \ (\text{book: }1.83)

Concept Assessment Exercise 3.3

When 2 grams of a solute is shaken with a mixture of 150\,\mathrm{cm^3} of water and 20\,\mathrm{cm^3} of chloroform, 1.5 grams of the solute is found in the chloroform layer after shaking. Calculate the partition coefficient.

Given

m_{\text{total}} = 2.0\,\mathrm{g}
m_{\text{chloroform}} = 1.5\,\mathrm{g}
V_{\text{water}} = 150\,\mathrm{cm^3}
V_{\text{chloroform}} = 20\,\mathrm{cm^3}

To find

K_{pc}=[\text{solute}]_{\text{chloroform}}/[\text{solute}]_{\text{water}}

Mass left in the water layer.
m_{\text{water}}=2.0-1.5=0.5\,\mathrm{g}
Concentration in each layer (mass per volume; the same units in both layers).
[\text{solute}]_{\text{chloroform}}=\dfrac{1.5}{20}=0.075\,\mathrm{g\,cm^{-3}},\qquad[\text{solute}]_{\text{water}}=\dfrac{0.5}{150}=3.33\times10^{-3}\,\mathrm{g\,cm^{-3}}
Divide, organic layer on top.
K_{pc}=\dfrac{0.075}{3.33\times10^{-3}}=22.5

The solute is 22.5 times more concentrated in chloroform than in water, so it is far more soluble in chloroform. Note that chloroform is the lower layer in the funnel, yet it is still written on top of the ratio.

K_{pc}=22.5

Derivation Amount of solute left in water after one extraction

Aim. Find the amount x of solute that ends up in the water layer when a total of n mol is shared between V_{\text{aq}} of water and V_{\text{org}} of an organic solvent.

If x mol is in the water layer, the rest is in the organic layer.
n_{\text{aq}}=x,\qquad n_{\text{org}}=n-x
Write K_{pc} with concentrations n/V.
K_{pc}=\dfrac{(n-x)/V_{\text{org}}}{x/V_{\text{aq}}}
Cross-multiply and collect the x terms.
n-x=K_{pc}\,x\,\dfrac{V_{\text{org}}}{V_{\text{aq}}}\;\Longrightarrow\;x\left(1+K_{pc}\dfrac{V_{\text{org}}}{V_{\text{aq}}}\right)=n

Resultx=\dfrac{n}{1+K_{pc}\,V_{\text{org}}/V_{\text{aq}}}

The larger K_{pc} and the larger the volume of organic solvent, the less solute stays in the water. With n=0.20\,\mathrm{mol}, K_{pc}=4 and equal volumes, x=0.20/5=0.04\,\mathrm{mol} (long question vi).

3.5.2Factors affecting the value of the partition coefficient

The partition coefficient shows how a solute distributes itself between two immiscible phases, usually a non-polar solvent (such as octanol, ether or benzene) and a polar solvent (water). The guiding rule is like dissolves like: a solute goes preferentially into the solvent whose polarity matches its own. The factors that affect the numerical value of K_{pc} are:

1. The polarity of the solute. A polar solute has a greater affinity for a polar phase such as water, because it interacts with polar solvent molecules through dipole–dipole forces or hydrogen bonding. For example, ethanol is polar because of its hydroxyl group. Between octanol and water, ethanol is more soluble in water than in octanol, so its partition coefficient is in favour of the polar phase (K_{\text{octanol/water}}\approx0.5, less than 1).

ethanol (polar)

1-octanol (mostly non-polar)

benzene (non-polar)

2. The polarity of the solvent. The nature of the solvents plays a decisive role. Non-polar solvents promote the partitioning of non-polar solutes, while polar solvents promote the partitioning of polar solutes. For example, the partition coefficient of non-polar benzene is higher in octanol than in water, because benzene interacts more with non-polar octanol (K_{\text{octanol/water}}\approx135).

3. Temperature. Temperature changes the solubility and the kinetic energy of the solute in both solvents, so K_{pc} is constant only at a fixed temperature. In general, an increase in temperature can increase the solubility of solutes in polar solvents. For example, the partitioning of a polar solute such as aspirin between octanol and water gives different values at different temperatures; at higher temperatures aspirin is more soluble in water than in octanol.

4. Molecular structure and size. Large solute molecules can interact with solvents differently from smaller molecules, and the functional groups present also affect the distribution. The book compares two molecules of similar structure but different size, ethyl acetate and butyl acetate, partitioned between octanol and water.

ethyl acetate

butyl acetate

Summary: factors affecting the partition coefficient
Factor	Effect on K_{pc}=[\text{organic}]/[\text{water}]	Book example
Polar solute	Smaller K_{pc} (the solute prefers water)	Ethanol between octanol and water
Non-polar solute	Larger K_{pc} (the solute prefers the organic solvent)	Benzene between octanol and water
Polarity of the solvents	A non-polar organic layer takes non-polar solutes; water takes polar solutes	Octanol and water
Temperature	Changes the solubility in both solvents, so K_{pc} changes	Aspirin
Molecular size and structure	A larger non-polar part raises K_{pc}; polar groups (\ce{-OH}, \ce{-COOH}, \ce{-NH2}) lower it	Ethyl acetate and butyl acetate

Formula sheet

Equilibrium constant (law of mass action)

K_c=\dfrac{[\mathrm{C}]^{c}[\mathrm{D}]^{d}}{[\mathrm{A}]^{a}[\mathrm{B}]^{b}}\quad\text{for }\ce{aA + bB <=> cC + dD}

square brackets: equilibrium molar concentrations; pure solids, pure liquids and the solvent are left out

Unit: depends on the reaction

Acid dissociation constant

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

\ce{HX}: weak acid, \ce{X-}: its conjugate base; the greater K_a, the stronger the acid

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

\mathrm{p}K_a

\mathrm{p}K_a=-\log K_a

lower \mathrm{p}K_a means a stronger acid

no unit

Weak acid: degree of ionization and [H+]

\alpha=\sqrt{\dfrac{K_a}{c}},\qquad[\ce{H+}]=\sqrt{K_a\,c}

c: initial concentration of the acid; valid when \alpha is small (below about 0.05)

[\ce{H+}] in \mathrm{mol\,dm^{-3}}; \alpha has no unit

Base dissociation constant

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

\ce{B}: weak base, \ce{BH+}: its conjugate acid; the greater K_b, the stronger the base

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

\mathrm{p}K_b

\mathrm{p}K_b=-\log K_b

lower \mathrm{p}K_b means a stronger base

no unit

Weak base: [OH-]

[\ce{OH-}]=\sqrt{K_b\,c},\qquad\alpha=\sqrt{\dfrac{K_b}{c}}

c: initial concentration of the base

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

Conjugate acid–base pair

K_a\times K_b=K_w=1.0\times10^{-14},\qquad\mathrm{p}K_a+\mathrm{p}K_b=14

K_a of the acid and K_b of its conjugate base, at 25\,^\circ\mathrm{C}

no unit for the p-form

pH and pOH

\mathrm{pH}=-\log[\ce{H+}],\quad\mathrm{pOH}=-\log[\ce{OH-}],\quad\mathrm{pH}+\mathrm{pOH}=14

at 25\,^\circ\mathrm{C}; treated fully in Unit 4

no unit

Henderson equation (acidic buffer)

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

[salt]: concentration (or moles) of the conjugate base from the salt; [acid]: concentration (or moles) of the weak acid

no unit

Henderson equation (basic buffer)

\mathrm{pOH}=\mathrm{p}K_b+\log\dfrac{[\text{salt}]}{[\text{base}]},\qquad\mathrm{pH}=14-\mathrm{pOH}

[salt]: concentration of the conjugate acid (e.g. \ce{NH4+}); [base]: concentration of the weak base

no unit

Solubility product

K_{sp}=[\ce{A^{n+}}]^{m}[\ce{B^{m-}}]^{n}\quad\text{for }\ce{A_{m}B_{n}(s) <=> m A^{n+} + n B^{m-}}

equilibrium ion concentrations in a saturated solution; the solid is left out

(\mathrm{mol\,dm^{-3}})^{m+n} (often written without a unit)

Ksp and molar solubility s

K_{sp}=m^{m}n^{n}s^{m+n}:\quad s^{2}\ (\mathrm{AB}),\ 4s^{3}\ (\mathrm{AB_2},\ \mathrm{A_2B}),\ 27s^{4}\ (\mathrm{AB_3})

s: moles of salt dissolved per \mathrm{dm^3} of saturated solution

Unit: s in \mathrm{mol\,dm^{-3}}

Precipitation condition

Q>K_{sp}\ \Rightarrow\ \text{precipitate};\quad Q=K_{sp}\ \Rightarrow\ \text{saturated};\quad Q<K_{sp}\ \Rightarrow\ \text{no precipitate}

Q: ionic product, the same expression as K_{sp} with the actual concentrations

Unit: same as K_{sp}

Solubility with a common ion (1:1 salt)

s=\dfrac{K_{sp}}{[\text{common ion}]}

valid when the common ion concentration is much larger than s

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

Partition coefficient (book convention)

K_{pc}=\dfrac{[\text{solute}]_{\text{organic}}}{[\text{solute}]_{\text{water}}}=\dfrac{[\ce{X}\text{ in ether}]}{[\ce{X}\text{ in water}]}

equilibrium concentrations in the two immiscible layers at a fixed temperature; K_{pc}>1: solute prefers the organic solvent

no units

Partition coefficient from amounts

K_{pc}=\dfrac{n_{\text{org}}/V_{\text{org}}}{n_{\text{aq}}/V_{\text{aq}}}

n: moles (or grams) of solute in a layer; V: volume of that layer, same units for both layers

no units

Amount left in water after one extraction

x=\dfrac{n}{1+K_{pc}\,V_{\text{org}}/V_{\text{aq}}}

n: total amount of solute; x: amount in the water layer at equilibrium

same unit as n

Key terms

Dynamic equilibrium: The state of a reversible reaction in which the forward and reverse reactions go on at equal rates, so the concentrations stay constant.
Le Chatelier's principle: If a system at equilibrium is disturbed by a change in concentration, pressure or temperature, the equilibrium shifts in the direction that tends to cancel the change.
Equilibrium constant: The ratio of the product of equilibrium concentrations of products to that of reactants, each raised to the power of its coefficient; it changes only with temperature.
Bronsted acid and base: An acid is a proton (\ce{H+}) donor; a base is a proton acceptor.
Conjugate acid–base pair: Two species that differ by one proton, such as \ce{CH3COOH}/\ce{CH3COO-} or \ce{NH4+}/\ce{NH3}.
Acid strength: The degree to which a Bronsted acid gives off protons.
Base strength: The ability of a base to accept a proton from a solvent.
Strong acid: An acid that ionizes almost completely in water, for example HCl; it has a very large K_a and a low or negative \mathrm{p}K_a.
Weak acid: An acid that ionizes only partly in water, for example \ce{CH3COOH}; it has a small K_a.
Strong base: A base that ionizes completely in water, for example NaOH or KOH.
Weak base: A base that reacts with water only partly, for example \ce{NH3} or an amine; it has a small K_b.
Extent (degree) of ionization: The fraction of the dissolved acid or base that is ionized at equilibrium, \alpha; multiplied by 100 it gives the percentage ionization.
Acid dissociation constant, K_a: The equilibrium constant for the ionization of an acid in water, K_a=[\ce{H3O+}][\ce{X-}]/[\ce{HX}]; a measure of the extent of ionization. The greater K_a, the stronger the acid.
\mathrm{p}K_a: The negative logarithm of K_a; the lower the \mathrm{p}K_a, the stronger the acid.
Base dissociation constant, K_b: The equilibrium constant for the reaction of a base with water, K_b=[\ce{BH+}][\ce{OH-}]/[\ce{B}]. The greater K_b, the stronger the base.
\mathrm{p}K_b: The negative logarithm of K_b; the lower the \mathrm{p}K_b, the stronger the base.
Buffer solution: A solution whose pH does not change significantly when a small amount of acid or base is added to it.
Acidic buffer: A buffer of a weak acid and its salt with a strong base, pH below 7, e.g. \ce{CH3COOH + CH3COONa}.
Basic buffer: A buffer of a weak base and its salt with a strong acid, pH above 7, e.g. \ce{NH4OH + NH4Cl}.
Buffer action: The removal of added \ce{H+} by the conjugate base and of added \ce{OH-} by the weak acid of a buffer, which keeps the pH nearly constant.
Buffer capacity: The amount of acid or base a buffer can absorb before its pH changes appreciably; it is greatest when the acid and salt are concentrated and present in similar amounts.
Henderson equation: \mathrm{pH}=\mathrm{p}K_a+\log([\text{salt}]/[\text{acid}]) for an acidic buffer; \mathrm{pOH}=\mathrm{p}K_b+\log([\text{salt}]/[\text{base}]) for a basic buffer.
Bicarbonate buffer system: The \ce{H2CO3}/\ce{HCO3-} buffer of blood plasma that keeps blood pH at 7.35 to 7.45.
Carbonic anhydrase: The enzyme in red blood cells that catalyses \ce{CO2 + H2O <=> H2CO3} and the related steps of the bicarbonate buffer.
Saturated solution: A solution in dynamic equilibrium with undissolved solute, holding the maximum amount of dissolved solute at that temperature.
Molar solubility: The number of moles of a salt that dissolve in 1\,\mathrm{dm^3} of solvent to give a saturated solution.
Solubility product constant, K_{sp}: The product of the molar concentrations of the ions in a saturated solution of a slightly soluble salt, each raised to a power equal to its coefficient in the balanced equation.
Ionic product: The product of the actual ion concentrations written in the same form as K_{sp}; a precipitate forms when it exceeds K_{sp}.
Common ion: An ion produced by both of two electrolytes present in the same solution, such as \ce{F-} in a solution of HF and NaF.
Common ion effect: The suppression of the degree of ionization or solubility of an electrolyte by the addition of a highly soluble electrolyte containing a common ion.
Immiscible liquids: Liquids that do not mix and form two separate layers, such as ether and water.
Separatory funnel: A glass funnel with a stopcock, used to shake two immiscible liquids together and then run off the lower layer.
Partition coefficient, K_{pc}: The ratio of the concentrations of a solute in two immiscible solvents in contact with each other at equilibrium at a particular temperature; it has no units. The book writes the organic layer on top.
Nernst distribution law: At constant temperature a solute distributes itself between two immiscible solvents so that the ratio of its concentrations in them is constant.

Multiple choice questionsp. 65–66

i1 mark

Which phenomenon describes a shift in equilibrium due to the addition of an ion already involved in the equilibrium?

ALe Chatelier's Principle
BCommon Ion EffectAnswer
CNernst's Law
DAvogadro's Principle

Why

Adding an ion that is already part of an equilibrium (a common ion) pushes the equilibrium back and suppresses ionization or solubility. This particular phenomenon is called the common ion effect.

Le Chatelier's principle is the general rule that explains the shift, but it is not the name of this phenomenon. Nernst's law is about the distribution of a solute between two solvents, and Avogadro's principle is about gas volumes.

ii1 mark

What property is used to distinguish between strong and weak acids?

AMolar mass
BExtent of ionizationAnswer
CDensity
DMelting point

Why

A strong acid ionizes almost completely in water, while a weak acid ionizes only partly. The extent of ionization (measured by K_a) is therefore what separates them; molar mass, density and melting point say nothing about proton donation.

iii1 mark

Which factor is considered when distinguishing between strong and weak bases?

AColour
BOdor
CExtent of ionizationAnswer
DSolubility

Why

A strong base such as NaOH ionizes completely, while a weak base such as \ce{NH3} reacts with water only partly. The extent of ionization, expressed by K_b, decides the strength. A base can be very soluble and still weak (ammonia).

iv1 mark

What defines a buffer solution?

AHigh concentration of ions
BPresence of a weak acid and its conjugate base (or a weak base and its conjugate acid)Answer
CLow pH
DAbsence of ions

Why

A buffer resists changes in pH because it contains a weak acid and its conjugate base (or a weak base and its conjugate acid) in large amounts: one member removes added \ce{H+} and the other removes added \ce{OH-}. Basic buffers have a pH above 7, so a low pH is not what defines a buffer.

v1 mark

How can a buffer solution be made?

AMixing strong acids and bases
BMixing a weak acid and its conjugate base (or a weak base and its conjugate acid)Answer
CDiluting a strong acid
DAdding water to a strong base

Why

For example, \ce{CH3COOH} mixed with \ce{CH3COONa} gives an acidic buffer, and \ce{NH4OH} mixed with \ce{NH4Cl} gives a basic buffer. A strong acid or base, diluted or neutralised, contains no reserve to absorb added acid or base.

vi1 mark

What role does \ce{HCO3-} play in controlling pH in blood?

ANeutralization of acids
BBuffering against changes in pHAnswer
CCatalysing reactions
DEnhancing oxygen transport

Why

\ce{HCO3-} together with \ce{H2CO3} forms the bicarbonate buffer, which resists pH change in both directions. Option A describes only half of its job (removing \ce{H+}); catalysis is done by the enzyme carbonic anhydrase, not by \ce{HCO3-}.

vii1 mark

How is the concentration of a slightly soluble salt calculated?

AUsing the solubility product constant (K_{sp})Answer
BTitration with a strong base
CMeasuring density
DConductivity measurement

Why

The ion concentrations in a saturated solution follow from K_{sp}; for a 1:1 salt such as AgCl, [\ce{Ag+}]=[\ce{Cl-}]=\sqrt{K_{sp}}. This is the method taught in section 3.3.2.

viii1 mark

Which term is used to describe the strength of an acid in terms of its ionization in water?

AAcid Concentration
BAcid Dissociation Constant (K_a)Answer
CAcid Molarity
DAcid Equilibrium Constant

Why

K_a measures the extent to which an acid is ionized at equilibrium: the greater K_a, the stronger the acid. Concentration and molarity describe how much acid is dissolved, not how far it ionizes.

ix1 mark

Which type of solvent would favour the partitioning of a polar solute?

ANon-polar solvent
BHydrophobic solvent
CPolar solventAnswer
DAprotic solvent

Why

Like dissolves like: a polar solute interacts with a polar solvent through dipole–dipole forces or hydrogen bonds, so it partitions in favour of the polar phase (for example ethanol prefers water to octanol). Hydrophobic solvents are non-polar.

x1 mark

What is the partition coefficient defined as?

ARatio of solute concentration in one solvent to the otherAnswer
BRatio of solute concentration in a single solvent
CRatio of solute mass in one solvent to the other
DRatio of solute mass in a single solvent

Why

K_{pc} is the ratio of the concentrations of a solute in two immiscible solvents at equilibrium at a particular temperature. A mass ratio (option C) equals it only when the two volumes are equal.

Short answer questionsp. 66

i3 marks

Explain the common ion effect with a suitable example.

Solution

Common ion effect: the suppression of the degree of ionization (or the solubility) of an electrolyte by adding a highly soluble electrolyte that contains a common ion.

Example. Hydrofluoric acid is a weak electrolyte and ionizes only slightly. Sodium fluoride is a strong electrolyte and ionizes completely. Both give \ce{F-}, the common ion.

\begin{aligned}&\ce{HF(aq) <=> H+(aq) + F-(aq)}\\&\ce{NaF(s) -> Na+(aq) + F-(aq)}\end{aligned}

Adding NaF raises [\ce{F-}]. By Le Chatelier's principle the HF equilibrium shifts to the left to use up the extra \ce{F-}. The ionization of HF is suppressed: [\ce{H+}] falls and [\ce{HF}] rises. In the same way, passing HCl gas into saturated brine raises [\ce{Cl-}] and pure NaCl precipitates.

ii3 marks

Differentiate between strong and weak acids using the extent of ionization and K_a.

Solution

For an acid \ce{HX} in water, \ce{HX + H2O <=> H3O+ + X-} and K_a=\dfrac{[\ce{H3O+}][\ce{X-}]}{[\ce{HX}]}. The greater the K_a, the more the acid is ionized and the stronger it is.

	Strong acid	Weak acid
Extent of ionization	Almost complete (about 100%)	Partial (a few per cent or less)
Equation	\ce{HCl + H2O -> H3O+ + Cl-}	\ce{CH3COOH + H2O <=> H3O+ + CH3COO-}
K_a	Very large, e.g. 1.0\times10^{6} for HCl	Small, e.g. 1.8\times10^{-5} for \ce{CH3COOH}
\mathrm{p}K_a	Low or negative (-6.0)	Positive (+4.7)
[\ce{H+}] in 1.0\,\mathrm{M} solution	1.0\,\mathrm{M}	Only 2.7\times10^{-2}\,\mathrm{M} for HF

So at the same concentration a strong acid gives far more \ce{H+} ions than a weak acid, and it has the larger K_a and the lower \mathrm{p}K_a.

iii3 marks

Differentiate between strong and weak bases using the extent of ionization and K_b.

Solution

For a base B in water, \ce{B + H2O <=> BH+ + OH-} and K_b=\dfrac{[\ce{BH+}][\ce{OH-}]}{[\ce{B}]}. K_b is large when the degree of ionization of the base is high, so the larger the K_b (the lower the \mathrm{p}K_b), the stronger the base.

	Strong base	Weak base
Extent of ionization	Complete (100%)	Partial (a few per cent)
Equation	\ce{NaOH -> Na+ + OH-}	\ce{NH3 + H2O <=> NH4+ + OH-}
K_b	Very large	Small, e.g. 1.7\times10^{-5} for \ce{NH3}
\mathrm{p}K_b	Very low	Higher, e.g. 4.76 for \ce{NH3}, 9.37 for aniline
Examples	\ce{NaOH}, \ce{KOH}	\ce{NH3}, \ce{CH3NH2}, pyridine, aniline

So a 0.1\,\mathrm{M} NaOH solution contains 0.1\,\mathrm{M} \ce{OH-} (pH 13), but 0.1\,\mathrm{M} ammonia contains only about 1.3\times10^{-3}\,\mathrm{M} \ce{OH-} (pH about 11.1).

iv3 marks

Define a buffer solution and provide an example of how it can be made.

Solution

Buffer solution: a solution whose pH does not change significantly when a small amount of acid or base is added to it.

How it is made. A buffer is made by mixing a weak acid with its salt of a strong base (acidic buffer, pH below 7), or a weak base with its salt of a strong acid (basic buffer, pH above 7).

Example. Dissolve 1 mol of acetic acid and 1 mol of sodium acetate in water and make the volume up to 1\,\mathrm{dm^3}. This acidic buffer has \mathrm{pH}=\mathrm{p}K_a=4.74.

\ce{CH3COOH + CH3COONa}\ \ (\text{acidic buffer}),\qquad\ce{NH4OH + NH4Cl}\ \ (\text{basic buffer, pH }9.26)

The same acetate buffer can also be made by adding 0.5 mol NaOH to 1 mol acetic acid, which converts half of the acid into its salt.

v3 marks

How does a buffer solution control pH? Include chemical equations in your explanation.

Solution

A buffer contains large amounts of a weak acid and its conjugate base, for example \ce{CH3COOH} and \ce{CH3COO-} (from fully ionized \ce{CH3COONa}). Together they remove added \ce{H+} or \ce{OH-} ions, so [\ce{H+}] stays almost constant.

Added acid is removed by the acetate ions, which turn it into weakly ionized acetic acid:

\ce{CH3COO-(aq) + H+(aq) -> CH3COOH(aq)}

Added base is removed by the acetic acid, which turns it into water:

\ce{CH3COOH(aq) + OH-(aq) -> CH3COO-(aq) + H2O(l)}

In both cases the ratio [\ce{CH3COO-}]/[\ce{CH3COOH}] changes only slightly. Since \mathrm{pH}=\mathrm{p}K_a+\log\dfrac{[\text{salt}]}{[\text{acid}]}, the pH also changes only slightly.

vi3 marks

Describe the uses of buffer solutions in various applications.

Solution

Buffers keep a stable pH wherever a sudden pH change would cause harm. Important uses are:

Blood plasma: the \ce{H2CO3}/\ce{HCO3-} buffer keeps blood pH between 7.35 and 7.45.
Household cleaning products: buffers in detergents keep a stable pH, so they clean well without harming surfaces or skin.
Personal care products: buffers keep the pH of shampoos and soaps steady and prevent skin irritation.
Swimming pools: buffers in water treatment keep the pH stable, preventing corrosion of equipment and keeping the water comfortable for swimmers.
Food and beverages: buffers control acidity, preserve flavours, keep the taste consistent and prevent spoilage.
Photography: developer solutions are buffered so that film or prints develop in a controlled way.

vii2026 Annual I, Set D, Q.2 (ii) OR3 marks

Explain how \ce{HCO3-} plays a role in controlling pH in the blood.

Solution

Blood pH is kept at 7.35 to 7.45 mainly by the bicarbonate buffer system, \ce{H2CO3}/\ce{HCO3-}. Carbon dioxide dissolves in blood to form carbonic acid, which ionizes to give bicarbonate ions; the enzyme carbonic anhydrase in red blood cells speeds up these steps.

\ce{CO2(g) + H2O(l) <=> H2CO3(aq) <=> H+(aq) + HCO3-(aq)}

Excess acid: bicarbonate acts as a base and removes the extra \ce{H+}. The \ce{H2CO3} formed breaks down to \ce{CO2}, which the lungs breathe out.

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

Excess base: carbonic acid releases \ce{H+}, which neutralises the added base.

\ce{OH-(aq) + H2CO3(aq) <=> HCO3-(aq) + H2O(l)}

The lungs (which control \ce{CO2}) and the kidneys (which control \ce{HCO3-}) support the buffer and keep blood pH in the normal range.

viii3 marks

Calculate the concentration of a slightly soluble salt given its solubility product constant (K_{sp}).

Solution

Let s be the molar solubility of the salt (mol dissolved per \mathrm{dm^3} of saturated solution). Write the dissociation equation, write K_{sp}, express each ion concentration in terms of s from the formula, and solve for s.

Example 1 (1:1 salt). AgCl, K_{sp}=1.8\times10^{-10}:

\begin{aligned}\ce{AgCl(s) &<=> Ag+(aq) + Cl-(aq)}\\K_{sp}&=[\ce{Ag+}][\ce{Cl-}]=s^{2}\\\Longrightarrow\ s&=\sqrt{1.8\times10^{-10}}=1.34\times10^{-5}\,\mathrm{mol\,dm^{-3}}\end{aligned}

Example 2 (1:2 salt). \ce{PbF2}, K_{sp}=4\times10^{-8}; here [\ce{Pb^2+}]=s and [\ce{F-}]=2s:

\begin{aligned}K_{sp}&=s(2s)^{2}=4s^{3}=4\times10^{-8}\\\Longrightarrow\ s&=2.15\times10^{-3}\,\mathrm{mol\,dm^{-3}}\\{}[\ce{F-}]&=4.3\times10^{-3}\,\mathrm{mol\,dm^{-3}}\end{aligned}

The molar solubility s is the concentration of the dissolved salt; multiplying by the molar mass gives the solubility in \mathrm{g\,dm^{-3}}.

Long answer questionsp. 67

i7 marks

Discuss the principles behind the common ion effect, providing two examples to illustrate its application.

Solution

Definition

The common ion effect is the suppression of the degree of ionization or the solubility of an electrolyte by the addition of a highly soluble electrolyte that contains a common ion, that is, an ion already present in the equilibrium.

Principle

A weak electrolyte (or a slightly soluble salt) is in dynamic equilibrium with its ions. A strong electrolyte that shares one of these ions ionizes completely, so it raises the concentration of that ion sharply. According to Le Chatelier's principle, the equilibrium shifts in the direction that uses up the extra ion, that is, back towards the un-ionized molecules or the undissolved solid. The value of K_a or K_{sp} does not change, because the temperature is constant; only the position of the equilibrium changes. As a result the weak electrolyte ionizes less, or the slightly soluble salt dissolves less and may precipitate.

In terms of the solubility product: the added common ion makes the ionic product Q greater than K_{sp}, so the solid precipitates until Q=K_{sp} again.

Example 1: suppression of the ionization of hydrofluoric acid

\begin{aligned}&\ce{HF(aq) <=> H+(aq) + F-(aq)}&&\text{(weak, slightly ionized)}\\&\ce{NaF(s) -> Na+(aq) + F-(aq)}&&\text{(strong, completely ionized)}\end{aligned}

NaF supplies a large amount of the common ion \ce{F-}. The HF equilibrium shifts to the left, the dissociation of HF decreases and [\ce{H+}] falls. For 1.0\,\mathrm{M} HF, [\ce{H+}] drops from 2.7\times10^{-2}\,\mathrm{M} to 7.2\times10^{-4}\,\mathrm{M} when the solution is also made 1.0\,\mathrm{M} in NaF. The same effect in \ce{CH3COOH}/\ce{CH3COONa} is the basis of buffer solutions.

Example 2: purification of sodium chloride

\begin{aligned}&\ce{NaCl(s) <=> Na+(aq) + Cl-(aq)}\\&\ce{HCl(g) -> H+(aq) + Cl-(aq)}\end{aligned}

When HCl gas is passed into a saturated solution of NaCl (brine), [\ce{Cl-}] increases. The ionic product [\ce{Na+}][\ce{Cl-}] becomes greater than K_{sp}, so \ce{Na+} and \ce{Cl-} combine and pure NaCl crystallises out, while soluble impurities stay in solution. In the same way, adding highly soluble KCl to a saturated solution of \ce{KClO4} precipitates \ce{KClO4} because of the common \ce{K+} ion.

Other applications

Buffers: the common ion from the salt keeps the weak acid (or base) almost un-ionized, so large reserves of both acid and conjugate base remain.
Qualitative analysis: group II sulphides are precipitated by \ce{H2S} in the presence of HCl, and group III hydroxides by \ce{NH4OH} in the presence of \ce{NH4Cl}; the common ion keeps [\ce{S^2-}] or [\ce{OH-}] low so that only the least soluble compounds precipitate.
Salting out of soap: NaCl supplies the common \ce{Na+} ion, which makes the soap separate from solution.

ii7 marks

Compare and contrast strong and weak acids based on their extent of ionization and the value of K_a.

Solution

Acid strength

In the Bronsted–Lowry sense an acid is a proton donor. Acid strength is the degree to which an acid gives off protons to water. It is judged from the extent of ionization and from the acid dissociation constant K_a.

Extent of ionization

A strong acid ionizes almost completely, so its solution contains practically no un-ionized molecules. A weak acid ionizes only partly; most of its molecules stay un-ionized and an equilibrium is set up.

\begin{aligned}&\ce{HCl(aq) + H2O(l) -> H3O+(aq) + Cl-(aq)}&&(\alpha\approx1)\\&\ce{CH3COOH(aq) + H2O(l) <=> H3O+(aq) + CH3COO-(aq)}&&(\alpha\ll1)\end{aligned}

The degree of ionization \alpha is the fraction of the acid that has ionized. For a weak acid of concentration c, \alpha=\sqrt{K_a/c}, so it is small and increases on dilution.

Acid dissociation constant, K_a

For \ce{HX + H2O <=> H3O+ + X-}, the equilibrium constant is K=\dfrac{[\ce{H3O+}][\ce{X-}]}{[\ce{HX}][\ce{H2O}]}. Water is in large excess, so [\ce{H2O}] is constant and K[\ce{H2O}]=K_a:

K_a=\dfrac{[\ce{H3O+}][\ce{X-}]}{[\ce{HX}]},\qquad\mathrm{p}K_a=-\log K_a

K_a measures the extent of ionization at equilibrium and depends only on temperature. The greater the K_a (the lower the \mathrm{p}K_a), the stronger the acid.

Comparison

Feature	Strong acid	Weak acid
Extent of ionization	About 100%	Small, a few per cent or less
Nature of reaction	Goes to completion (\ce{->})	Reversible equilibrium (\ce{<=>}), lies to the left
K_a	Very large, e.g. HCl 1.0\times10^{6}, \ce{HClO4} 1.0\times10^{10}	Small, e.g. HF 7.2\times10^{-4}, \ce{CH3COOH} 1.8\times10^{-5}
\mathrm{p}K_a	Low or negative (-6 to -10)	Positive (3 to 10)
[\ce{H+}] for concentration c	[\ce{H+}]=c	[\ce{H+}]\approx\sqrt{K_ac}, much less than c
Electrical conductivity	High	Low
Conjugate base	Extremely weak (\ce{Cl-})	Weak but significant (\ce{CH3COO-})

Numerical illustration

In 1.0\,\mathrm{M} solutions, HCl gives [\ce{H+}]=1.0\,\mathrm{M} (pH 0). HF (K_a=7.2\times10^{-4}) gives only [\ce{H+}]=\sqrt{7.2\times10^{-4}\times1.0}=2.7\times10^{-2}\,\mathrm{M} (pH 1.57), because only 2.7% of it ionizes.

Similarities

Both are proton donors and form \ce{H3O+} in water; both turn blue litmus red.
Both neutralise bases to form a salt and water; the same amount of base is needed to neutralise equal moles of either acid.
Both have conjugate bases, and for both, K_a changes only with temperature.

iii7 marks

Elaborate on the factors that distinguish strong and weak bases, incorporating the concept of ionization and the base dissociation constant (K_b).

Solution

Strength of a base

The strength of a base is its ability to accept a proton from a solvent, usually water. Two things distinguish strong and weak bases: the extent of ionization in water and the value of the base dissociation constant K_b.

Strong bases: complete ionization

Hydroxides of alkali metals such as NaOH and KOH ionize completely in aqueous solution. Every formula unit releases an \ce{OH-} ion, which is a Bronsted base because it accepts a proton:

\ce{NaOH(aq) -> Na+(aq) + OH-(aq)}\qquad\ce{OH- + H+ -> H2O}

Weak bases: partial ionization

Ammonia and amines take protons from water only partly. An equilibrium is set up that lies mostly to the left, so only a small fraction of the base is ionized:

\ce{B + H2O(l) <=> BH+(aq) + OH-(aq)}\qquad\ce{NH3 + H2O <=> NH4+ + OH-}

Base dissociation constant, K_b

Applying the law of mass action and taking [\ce{H2O}] as constant (water is in excess) gives:

K_b=\dfrac{[\ce{BH+}][\ce{OH-}]}{[\ce{B}]},\qquad\mathrm{p}K_b=-\log K_b

K_b is large if the degree of ionization of the base is high, that is, if the base is strong, and small for a weak base. The larger the K_b (the lower the \mathrm{p}K_b), the stronger the base. For a weak base of concentration c, \alpha=\sqrt{K_b/c} and [\ce{OH-}]=\sqrt{K_bc}.

Comparing bases with Table 3.2

Using K_b values: diethylamine (9.6\times10^{-4}) > ethylamine (5.6\times10^{-4}) > methylamine (4.5\times10^{-4}) > ammonia (1.7\times10^{-5}) > pyridine (5.6\times10^{-9}) > aniline (4.3\times10^{-10}). So ammonia is stronger than pyridine and aniline but weaker than methylamine and ethylamine, and diethylamine is the strongest base in the table.

Comparison

Feature	Strong base	Weak base
Extent of ionization	Complete (100%)	Partial (a few per cent)
Equation	One-way (\ce{->})	Equilibrium (\ce{<=>})
K_b and \mathrm{p}K_b	Very large K_b, very low \mathrm{p}K_b	Small K_b (10^{-4} to 10^{-10}), \mathrm{p}K_b 3 to 10
[\ce{OH-}] in 0.1\,\mathrm{M} solution	0.1\,\mathrm{M} (pH 13)	About 1.3\times10^{-3}\,\mathrm{M} for \ce{NH3} (pH 11.1)
Conjugate acid	No acid strength (\ce{Na+}, \ce{K+})	Weak acid (\ce{NH4+}); K_aK_b=K_w
Examples	NaOH, KOH	\ce{NH3}, \ce{CH3NH2}, \ce{C2H5NH2}, pyridine, aniline

In short, a strong base is fully ionized and has a very large K_b; a weak base is only partly ionized, has a small K_b, and its strength can be ranked by comparing K_b or \mathrm{p}K_b values at the same temperature.

iv7 marks

Provide a comprehensive explanation of buffer solutions, including how they are prepared, how they control pH, and their uses in different scenarios.

Solution

Definition

A buffer solution is a solution whose pH does not change significantly when a small amount of acid or base is added to it (or when it is diluted).

Types and preparation

Acidic buffer (pH below 7): a weak acid mixed with its salt with a strong base, e.g. \ce{CH3COOH + CH3COONa}.
Basic buffer (pH above 7): a weak base mixed with its salt with a strong acid, e.g. \ce{NH4OH + NH4Cl}.
A buffer can also be prepared by partly neutralising a weak acid with a strong base, e.g. 1 mol \ce{CH3COOH} + 0.5 mol NaOH gives equal amounts of acid and salt.

How a buffer controls pH

In the acetate buffer, \ce{CH3COONa} ionizes completely and its \ce{CH3COO-} ions suppress the ionization of \ce{CH3COOH} (common ion effect). So the solution holds large reserves of both \ce{CH3COOH} and \ce{CH3COO-}.

\begin{aligned}&\ce{CH3COOH + H2O <=> CH3COO- + H3O+}\\&\ce{CH3COONa -> CH3COO- + Na+}\end{aligned}

Added acid is removed by acetate ions, and added base is removed by acetic acid:

\begin{aligned}&\ce{CH3COO-(aq) + H+(aq) -> CH3COOH(aq)}\\&\ce{CH3COOH(aq) + OH-(aq) -> CH3COO-(aq) + H2O(l)}\end{aligned}

In a basic buffer, \ce{NH4OH} removes added \ce{H+} (\ce{NH4OH + H+ -> NH4+ + H2O}) and \ce{NH4+} removes added \ce{OH-} (\ce{NH4+ + OH- -> NH4OH}). In every case the added ion becomes a weak electrolyte or water, so [\ce{H+}] hardly changes.

pH of a buffer

\mathrm{pH}=\mathrm{p}K_a+\log\dfrac{[\text{salt}]}{[\text{acid}]},\qquad\mathrm{pOH}=\mathrm{p}K_b+\log\dfrac{[\text{salt}]}{[\text{base}]}

For 1.0 mol each of \ce{CH3COOH} and \ce{CH3COONa} in 1\,\mathrm{dm^3} (K_a=1.8\times10^{-5}), pH =4.745. After adding 0.01 mol HCl the ratio becomes 0.99/1.01 and the pH is 4.736, a change of only 0.009, whereas the same acid would change the pH of pure water from 7 to 2.

Uses of buffers

Blood: the \ce{H2CO3}/\ce{HCO3-} buffer keeps blood pH at 7.35 to 7.45 (\ce{HCO3- + H+ <=> H2CO3}; \ce{H2CO3 + OH- <=> HCO3- + H2O}).
Household cleaning products and detergents: stable pH for effective, safe cleaning.
Shampoos and soaps: prevent skin irritation.
Swimming pools: prevent corrosion of equipment and keep water comfortable.
Food and beverages: control acidity, preserve flavour and prevent spoilage.
Photography: developer solutions give controlled, consistent development.

v4 marks

Calculate the pH of an \ce{NH4OH} and \ce{NH4Cl} buffer solution containing one mole of each. K_b of \ce{NH4OH} is 1.8\times10^{-5}.

Solution

Given

n(\ce{NH4OH}) = 1\,\mathrm{mol} weak base
n(\ce{NH4Cl}) = 1\,\mathrm{mol} salt, supplies \ce{NH4+}
K_b = 1.8\times10^{-5}

To find

pH of the buffer

\ce{NH4Cl} ionizes completely, so [\ce{NH4+}] equals the salt concentration. \ce{NH4OH} is weak and hardly ionized, so its concentration stays at the starting value. Both are in the same volume, so moles can be used.
\ce{NH4OH <=> NH4+ + OH-}\qquad K_b=\dfrac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH4OH}]}
Rearrange for [\ce{OH-}] and substitute.
[\ce{OH-}]=K_b\,\dfrac{[\ce{NH4OH}]}{[\ce{NH4+}]}=1.8\times10^{-5}\times\dfrac{1}{1}=1.8\times10^{-5}\,\mathrm{M}
Find pOH.
\mathrm{pOH}=-\log(1.8\times10^{-5})=4.74
Convert to pH (at 25\,^\circ\mathrm{C}, \mathrm{pH}+\mathrm{pOH}=14).
\mathrm{pH}=14-4.74=9.26

\mathrm{pH}=9.26

Book answer. 9.26

vi4 marks

A solute has a partition coefficient K_{\text{benzene/water}}=4. 100\,\mathrm{cm^3} of water is shaken with 100\,\mathrm{cm^3} of benzene containing 0.20\,\mathrm{mol} of the solute. Find the amount of solute that remains in water at equilibrium.

Solution

Given

K_{pc} = \dfrac{[\text{solute}]_{\text{benzene}}}{[\text{solute}]_{\text{water}}}=4
V_{\text{water}} = 100\,\mathrm{cm^3}=0.100\,\mathrm{dm^3}
V_{\text{benzene}} = 100\,\mathrm{cm^3}=0.100\,\mathrm{dm^3}
n_{\text{total}} = 0.20\,\mathrm{mol} all in benzene at the start

To find

amount of solute in the water layer at equilibrium

Let x mol of solute move into the water. Then 0.20-x mol stays in the benzene.
n_{\text{water}}=x,\qquad n_{\text{benzene}}=0.20-x
Write K_{pc} with concentrations (c=n/V), benzene on top as the subscript says.
4=\dfrac{(0.20-x)/0.100}{x/0.100}
The volumes are equal and cancel.
4=\dfrac{0.20-x}{x}\;\Longrightarrow\;4x=0.20-x
Solve for x.
5x=0.20\;\Longrightarrow\;x=0.040\,\mathrm{mol}

So 0.040\,\mathrm{mol} of solute is in the water and 0.16\,\mathrm{mol} remains in the benzene. Check: \dfrac{0.16/0.100}{0.040/0.100}=\dfrac{1.6}{0.40}=4.

n_{\text{water}}=0.04\,\mathrm{mol}

Book answer. 0.04 mol

FBISE 2026 questions

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

2026 Annual I, Set B, MCQ 21 mark

If a solute is more soluble in polar solvent than in non-polar solvent, the partition coefficient will:

ABe greater than 1
BBe less than 1Answer
CBe equal to 1
DBecome zero

Why

The textbook defines K_{pc} with the concentration in the organic (non-polar) solvent on top and the concentration in water (the polar solvent) at the bottom, K_{pc}=[\text{solute}]_{\text{organic}}/[\text{solute}]_{\text{water}}. A solute that prefers the polar solvent has the larger concentration in the denominator, so K_{pc}<1.

It cannot be zero, because some solute always enters each layer; it would equal 1 only if the solute were equally soluble in both. (If the ratio were taken the other way, polar over non-polar, the value would be greater than 1, so the answer depends on the book's convention.)

Note. Answer based on the book's convention (organic layer on top of the ratio), as used in Activity 3.1 and long question vi.

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

The partition coefficient of a solute between two solvents is 2.5. If concentration in aqueous layer is 0.4\,\mathrm{mol\,dm^{-3}}, calculate concentration in organic layer.

Solution

Given

K_{pc} = 2.5 organic layer over aqueous layer
[\text{solute}]_{\text{aq}} = 0.4\,\mathrm{mol\,dm^{-3}}

To find

[\text{solute}]_{\text{organic}}

Write the partition coefficient as in the textbook, organic layer on top (1 mark).
K_{pc}=\dfrac{[\text{solute}]_{\text{organic}}}{[\text{solute}]_{\text{aq}}}
Rearrange for the organic concentration and substitute (1 mark).
[\text{solute}]_{\text{organic}}=K_{pc}\times[\text{solute}]_{\text{aq}}=2.5\times0.4
Evaluate, with the unit (1 mark).
[\text{solute}]_{\text{organic}}=1.0\,\mathrm{mol\,dm^{-3}}

The organic layer holds 2.5 times the concentration of the aqueous layer, as expected for K_{pc}>1: the solute is more soluble in the organic solvent.

[\text{solute}]_{\text{organic}}=1.0\,\mathrm{mol\,dm^{-3}}

2026 Annual I, Set B, Q.2 (iii) OR1+2 = 3 marks

Calculate the pH of an acetic acid / sodium acetate buffer containing 1mol of each component. Ka for acetic acid =1.8\times10^{-5}

Solution

Given

n(\ce{CH3COOH}) = 1\,\mathrm{mol}
n(\ce{CH3COONa}) = 1\,\mathrm{mol}
K_a = 1.8\times10^{-5}

To find

pH of the buffer

Formula (1 mark). For an acidic buffer, \ce{CH3COONa} supplies \ce{CH3COO-} and the weak acid stays almost un-ionized, so the Henderson equation applies.
\mathrm{pH}=\mathrm{p}K_a+\log\dfrac{[\text{salt}]}{[\text{acid}]}
Substitution (1 mark). First find \mathrm{p}K_a; the salt-to-acid ratio is 1/1.
\mathrm{p}K_a=-\log(1.8\times10^{-5})=4.74\qquad\mathrm{pH}=4.74+\log\dfrac{1}{1}
Answer (1 mark). \log1=0.
\mathrm{pH}=4.74

Direct method, as in the book: K_a=\dfrac{[\ce{CH3COO-}][\ce{H+}]}{[\ce{CH3COOH}]} gives 1.8\times10^{-5}=\dfrac{(1)[\ce{H+}]}{1}, so [\ce{H+}]=1.8\times10^{-5}\,\mathrm{M} and \mathrm{pH}=-\log(1.8\times10^{-5})=4.74. When salt and acid are equal, \mathrm{pH}=\mathrm{p}K_a.

\mathrm{pH}=4.74

Note. This is part (a) of Example 3.2 in the book (section 3.2.1).

2026 Annual I, Set D, Q.2 (i) OR3 marks

A solute distributes between water and ether at equilibrium, its concentration in ether is 4\,\mathrm{g/L} and in water is 2\,\mathrm{g/L}. Find the partition coefficient (K_{pc}).

Solution

Given

[\text{solute}]_{\text{ether}} = 4\,\mathrm{g\,L^{-1}}
[\text{solute}]_{\text{water}} = 2\,\mathrm{g\,L^{-1}}

To find

K_{pc}

The equilibrium is \ce{X}\,(\text{in water})\ce{<=>}\ce{X}\,(\text{in ether}). Write K_{pc} as in the textbook, ether (organic) layer on top (1 mark).
K_{pc}=\dfrac{[\ce{X}\text{ in ether}]}{[\ce{X}\text{ in water}]}
Substitute. Both concentrations are in the same unit, so the units cancel (1 mark).
K_{pc}=\dfrac{4\,\mathrm{g\,L^{-1}}}{2\,\mathrm{g\,L^{-1}}}
Evaluate (1 mark).
K_{pc}=2

K_{pc} has no units because it is a ratio of two concentrations. Since K_{pc}>1, the solute is twice as concentrated in ether: it is more soluble in ether than in water.

K_{pc}=2\ \ (\text{no units})

2026 Annual I, Set D, Q.2 (ii) OR1.5+1.5 = 3 marks

How does bicarbonate buffer system (\ce{H2CO3}/\ce{HCO3-}) in blood maintain a constant pH with the help of chemical reactions?

Solution

Blood pH is kept at 7.35 to 7.45 by the bicarbonate buffer. Carbon dioxide dissolves in blood to form carbonic acid, which ionizes to bicarbonate; carbonic anhydrase in red blood cells catalyses these steps:

\ce{CO2(g) + H2O(l) <=> H2CO3(aq) <=> H+(aq) + HCO3-(aq)}

Removal of excess acid (1.5 marks)

When acid (\ce{H+}) enters the blood, bicarbonate acts as a base and combines with it to form carbonic acid, a weak acid. The carbonic acid breaks down to \ce{CO2} and water, and the lungs breathe out the \ce{CO2}. So [\ce{H+}] does not rise.

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

Removal of excess base (1.5 marks)

When a base (\ce{OH-}) enters the blood, carbonic acid releases \ce{H+}, which neutralises it to water and forms more bicarbonate. So [\ce{H+}] does not fall.

\ce{OH-(aq) + H2CO3(aq) <=> HCO3-(aq) + H2O(l)}

The lungs (control of \ce{CO2}) and kidneys (control of \ce{HCO3-}) support these reactions, so blood pH stays nearly constant.

Exam focus

Define a buffer and explain the buffer action of \ce{CH3COOH}/\ce{CH3COONa} with equations. (long)
The core SLO of the unit; asked as short question v and long question iv.
Go to answer
Calculate the pH of an acidic or a basic buffer (Henderson equation). (numerical)
2026 Set B Q.2 (iii) OR asked the acetate buffer (pH 4.74); long question v asks the ammonium buffer (pH 9.26).
Go to answer
Explain how the \ce{H2CO3}/\ce{HCO3-} buffer keeps blood pH constant, with reactions. (short)
2026 Set D Q.2 (ii) OR and book short question vii; both directions (acid and base) carry marks.
Go to answer
Distinguish strong and weak acids using extent of ionization and K_a; derive K_a and calculate [\ce{H+}] of a weak acid. (long)
Short question ii, long question ii and Example 3.1 type numericals.
Go to answer
Identify the stronger base from K_b values and justify it by extent of ionization. (short)
Asked in 2026 Set D Q.2 (iii) (the paper map files it under Unit 4); also short question iii and long question iii.
Go to answer
Write K_{sp} expressions and calculate ion concentrations from K_{sp} for AB and \mathrm{AB_2} salts. (numerical)
SLO 'calculate concentration of slightly soluble salts'; short question viii and Concept Assessment Exercises 3.1 and 3.2.
Go to answer
Explain the common ion effect with two examples and equations. (long)
Short question i, long question i and MCQ i.
Go to answer
State the definition of K_{pc} and the book's convention (organic layer on top). (short)
2026 Set B MCQ 2 tested that a solute preferring the polar solvent has K_{pc}<1; MCQ x asks the definition.
Go to answer
Partition numericals: find K_{pc}, a missing concentration, or the amount left in water. (numerical)
Two 2026 questions (Set B Q.2 (i) OR, Set D Q.2 (i) OR) plus long question vi and Activity 3.1.
Go to answer
Explain the factors that affect the value of a partition coefficient. (short)
A direct SLO in terms of polarity of solute and solvent; MCQ ix tests it.
Go to answer

Chemical equilibrium: the ideas this unit uses

Le Chatelier's principle

The constants of this unit

pH in one line

3.1Strength of acids and bases

3.1.1Strong and weak acidsOften examined

The acid dissociation constant, Ka​

Derivation Expression for Ka​

pKa​

Extent of ionization and Ka​ together

Derivation [HX+] and degree of ionization of a weak acid

Strong and weak acids compared

3.1.2Strong and weak basesOften examined

Derivation Expression for K_b

Link between K_a and K_b

Derivation K_a\times K_b=K_w for a conjugate pair

Strong and weak bases compared

3.2Buffer solutions and buffer actionOften examined

Types of buffer solutions

Buffer action of an acidic buffer

Buffer action of a basic buffer

3.2.1Calculating the pH of a buffer (Henderson equation)Often examined

Derivation Henderson equation for an acidic buffer

Derivation Henderson equation for a basic buffer

(a) pH of the buffer

(b) After adding 0.01 mol HCl

3.2.2Applications of buffers and the bicarbonate buffer in bloodOften examined

1. The buffer system in blood plasma

2 to 6. Other uses of buffers

3.3, 3.3.1Solubility product, KspOften examined

Derivation The solubility product expression

Precipitation reactions: ionic product and K_{sp}

3.3.2Calculating concentrations of ions from KspOften examined

Derivation Molar solubility of any salt \mathrm{A}_m\mathrm{B}_n

3.4Common ion effectOften examined

Examples from the book

How large is the effect

Self Check Exercise 3.3

Applications of the common ion effect

3.5Partition coefficient and the distribution lawOften examined

Which concentration goes on top of the ratio

Nernst's distribution law

3.5.1Calculating a partition coefficientOften examined

Activity 3.1: partition coefficient of methylamine

Derivation Amount of solute left in water after one extraction

3.5.2Factors affecting the value of the partition coefficient

Formula sheet

Key terms

Multiple choice questionsp. 65–66

Short answer questionsp. 66

Long answer questionsp. 67

Definition

Principle

Example 1: suppression of the ionization of hydrofluoric acid

Example 2: purification of sodium chloride

Other applications

Acid strength

Extent of ionization

Acid dissociation constant, K_a

Comparison

Numerical illustration

Similarities

Strength of a base

Strong bases: complete ionization

Weak bases: partial ionization

Base dissociation constant, K_b

Comparing bases with Table 3.2

Comparison

Definition

Types and preparation

How a buffer controls pH

pH of a buffer

Uses of buffers

FBISE 2026 questions

Removal of excess acid (1.5 marks)

Removal of excess base (1.5 marks)

Exam focus

The acid dissociation constant, $K_{a}$

Derivation Expression for $K_{a}$

$p K_{a}$

Extent of ionization and $K_{a}$ together

Derivation $[H X^{+}]$ and degree of ionization of a weak acid