Acid-base chemistry

Class 12 Chemistry, unit 4, book pages 68–78

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Acids and bases are everywhere, from stomach acid to the soil in a field. This unit measures acidity with the pH scale, links $[H X^{+}]$ and $[OH X^{-}]$ through the ionic product of water $K_{w}$ , and compares strong and weak acids and bases using $K_{a}$ , $K_{b}$ and the degree of ionisation. It ends with pH titration curves, which show how the pH changes as an acid is neutralised by a base and tell us which indicator to use.

Learning outcomes (4)

State that $pH = - lo g [H X^{+}]$ and $[H X^{+}] = 1 0^{- pH}$ .
State that a change of one pH unit represents a 10-fold change in the hydrogen ion concentration $[H X^{+}]$ .
Use the ionic product of water, $K_{w} = [H X^{+}] [OH X^{-}] = 1 0^{- 14} mo l^{2} d m^{- 6}$ at $298 K$ , to solve problems.
Sketch the pH titration curves of titrations using combinations of strong and weak acids with strong and weak alkalis.

4Acids and bases: what you already know

Acid-base chemistry describes acids and bases and how they react in water. These reactions matter everywhere: in digestion, in blood, in soil and in industry. How acidic or basic a solution is, is usually given by its pH.

Arrhenius acid: A substance that gives hydrogen ions, $H X^{+}$ , in water. Example: $HCl H X^{+} + Cl X^{-}$ .
Arrhenius base: A substance that gives hydroxide ions, $OH X^{-}$ , in water. Example: $NaOH Na X^{+} + OH X^{-}$ . A base that dissolves in water is called an alkali.
Brønsted-Lowry acid and base: An acid is a proton ( $H X^{+}$ ) donor. A base is a proton acceptor.
Conjugate acid-base pair: An acid and the species formed when it loses one proton, for example $CH X_{3} COOH$ and $CH X_{3} COO X^{-}$ , or $NH X_{4} X^{+}$ and $NH X_{3}$ .

In water, $H X^{+}$ never exists on its own. It joins a water molecule to form the hydronium ion, $H X_{3} O X^{+}$ . So $[H X^{+}]$ and $[H X_{3} O X^{+}]$ mean the same thing in this unit. Square brackets mean concentration in $mol d m^{- 3}$ (written M).

HCl (g) + H X_{2} O (l) H X_{3} O X^{+} (aq) + Cl X^{-} (aq)

An acid in water

H X^{+} (aq) + OH X^{-} (aq) H X_{2} O (l)

Neutralisation

The ionic equation for every reaction of a strong acid with a strong alkali.

4.1The pHOften examined

The pH scale was introduced in 1909 by the Danish biochemist S.P.L. Sorensen while he was studying how $H X^{+}$ ions affect the activity of enzymes. The letter p comes from the German word potenz, meaning power or exponent (here the power of 10). The term is often explained as pondus hydrogenii, Latin for the weight (amount) of hydrogen.

Concentrations of $H X^{+}$ are usually tiny numbers such as $0.0000001 M$ . The pH turns them into simple numbers between about 0 and 14.

pH: The negative logarithm to base 10 of the hydrogen ion concentration in $mol d m^{- 3}$ . Equivalently, it is the logarithm of the reciprocal of $[H X^{+}]$ .

pH = - lo g [H X^{+}] = lo g \frac{1}{[ H X ^{+} ]} or pH = - lo g [H X_{3} O X^{+}]

Definition of pH

$lo g$ means $lo g_{10}$ . pH has no unit: only the number part of the concentration is used.

To go back from pH to concentration, take the antilog:

[H X^{+}] = 1 0^{- pH} mol d m^{- 3}

Activity: the more exact definition

In 1924 Sorensen realised that pH really depends on the activity of the $H X^{+}$ ion (its effective concentration), not simply on its concentration. So the exact definition is $pH = - lo g a_{H X^{+}}$ , where $a$ is the activity. Activity depends on several things, one of which is concentration. In dilute solutions activity and concentration are almost equal, so all calculations use $[H X^{+}]$ .

One pH unit is a 10-fold change in [H+]

Derivation Why one pH unit means a 10-fold change in $[H X^{+}]$

Aim. Show that when the pH changes by 1, $[H X^{+}]$ changes by a factor of 10.

Take two solutions A and B whose pH values differ by exactly 1.
$pH_{B} = pH_{A} + 1$
Write each concentration using $[H X^{+}] = 1 0^{- pH}$ .
$[H X^{+}]_{A} = 1 0^{- pH_{A}}, [H X^{+}]_{B} = 1 0^{- (pH_{A} + 1)} = 1 0^{- pH_{A}} \times 1 0^{- 1}$
Divide the two. The factor $1 0^{- pH_{A}}$ cancels.
$\frac{[ H X ^{+} ] _{A}}{[ H X ^{+} ] _{B}} = \frac{1 0 ^{- pH_{A}}}{1 0 ^{- pH_{A}} \times 1 0 ^{- 1}} = 10$
In the same way, a difference of $n$ pH units means a factor of $1 0^{n}$ .
$\frac{[ H X ^{+} ] _{A}}{[ H X ^{+} ] _{B}} = 1 0^{(pH_{B} - pH_{A})}$

Result

Δ pH = 1 ⟺ [H X^{+}] changes 10 times

A solution of pH 3 has 10 times more $H X^{+}$ than one of pH 4, and 100 times more than one of pH 5. A small change in pH is a big change in acidity.

Each step of 1 in pH changes $[H X^{+}]$ ten times (at $2 5^{\circ} C$ ).
pH	$[H X^{+}]$ / M	$[OH X^{-}]$ / M	Nature
1	$1 0^{- 1}$	$1 0^{- 13}$	strongly acidic
4	$1 0^{- 4}$	$1 0^{- 10}$	weakly acidic
7	$1 0^{- 7}$	$1 0^{- 7}$	neutral
10	$1 0^{- 10}$	$1 0^{- 4}$	weakly basic
13	$1 0^{- 13}$	$1 0^{- 1}$	strongly basic

pH of bases and pOH

The concentration of hydroxide ions is expressed in the same way as pOH:

pOH = - lo g [OH X^{-}] [OH X^{-}] = 1 0^{- pOH}

pH and pOH are linked through the ionic product of water (section 4.3):

p K_{w} = pH + pOH at 2 5^{\circ} C : pH + pOH = 14

4.2The pH scaleOften examined

The pH scale is a numerical scale that shows how acidic or alkaline a solution is. For most solutions at $2 5^{\circ} C$ the values run from 0 to 14.

pH below 7: acidic, $[H X^{+}] > [OH X^{-}]$ . The lower the pH, the more acidic the solution.
pH equal to 7: neutral, $[H X^{+}] = [OH X^{-}] = 1 0^{- 7} M$ , for example pure water at $2 5^{\circ} C$ .
pH above 7: basic (alkaline), $[OH X^{-}] > [H X^{+}]$ . The higher the pH, the more alkaline the solution.

The ends of the scale correspond to $1 M$ solutions: at pH 0 the hydronium ion concentration is $1 M$ , and at pH 14 the hydroxide ion concentration is $1 M$ (so $[H X^{+}] = 1 0^{- 14} M$ ). Most solutions have $[H X^{+}]$ between $1 M$ and $1 0^{- 14} M$ .

Common solutions on the pH scale (after the book's Fig 4.1). Pure rain is slightly acidic (about 5.6) because it dissolves carbon dioxide from the air.

Example 4.1: pH of a strong acid

Find the pH of a

0.002 M

solution of HCl.

HCl is a strong acid, so it ionises completely and each HCl gives one $H X^{+}$ .
$HCl H X^{+} + Cl X^{-} [H X^{+}] = 0.002 = 2.0 \times 1 0^{- 3} M$
Apply the definition of pH.
$pH = - lo g (2.0 \times 1 0^{- 3}) = 2.70$

pH = 2.70

Example 4.2: [H+] from pH

If moist soil has a pH of 7.84, what is the

H X^{+}

concentration of the soil solution?

Rearrange the definition of pH.
$pH = - lo g [H X^{+}] ⟹ [H X^{+}] = 1 0^{- pH}$
Key in $- 7.84$ and press $1 0^{x}$ (the antilog).
$[H X^{+}] = 1 0^{- 7.84} = 1.45 \times 1 0^{- 8} M$

The pH is above 7, so this soil is slightly basic.

[H X^{+}] = 1.45 \times 1 0^{- 8} mol d m^{- 3}

Concept assessment 4.1, question 1

What is the pH of a solution of

2 g

pure

H X_{3} PO X_{4}

per

d m^{3}

of solution?

Find the molar mass of phosphoric acid.
$M (H X_{3} PO X_{4}) = 3 (1) + 31 + 4 (16) = 98 g mo l^{- 1}$
Convert mass concentration to molar concentration.
$C = \frac{2 g d m ^{- 3}}{98 g mo l ^{- 1}} = 0.0204 mol d m^{- 3}$
The exercise expects us to treat the acid as fully ionised, so each $H X_{3} PO X_{4}$ gives three $H X^{+}$ .
$H X_{3} PO X_{4} 3 H X^{+} + PO X_{4} X^{3 -} [H X^{+}] = 3 (0.0204) = 0.0612 M$
Find the pH.
$pH = - lo g (0.0612) = 1.21$

pH \approx 1.21 (complete ionisation assumed)

Concept assessment 4.1, question 2

Calculate the concentration of hydrogen ion (

H X^{+}

) in a solution of sulphuric acid having pH of 1.5.

The pH gives $[H X^{+}]$ directly, whatever the acid is.
$[H X^{+}] = 1 0^{- 1.5} = 3.16 \times 1 0^{- 2} M$
If the concentration of the acid itself is wanted: each $H X_{2} SO X_{4}$ gives two $H X^{+}$ .
$[H X_{2} SO X_{4}] = \frac{3.16 \times 1 0 ^{- 2}}{2} = 1.58 \times 1 0^{- 2} M$

[H X^{+}] = 3.16 \times 1 0^{- 2} mol d m^{- 3}

4.3Ionic product of water, K_wOften examined

Even the purest water conducts electricity very slightly. A tiny fraction of water molecules pass a proton to each other, forming hydronium and hydroxide ions. This is the self-ionisation (auto-ionisation) of water.

\begin{aligned}\ce{H2O(l) + H2O(l) &<=> H3O+(aq) + OH-(aq)}\\\text{or simply}\quad\ce{H2O &<=> H+ + OH-}\end{aligned}

Derivation The ionic product of water, K_w

Aim. Obtain K_w=[\ce{H3O+}][\ce{OH-}] from the equilibrium law.

Assumptions.

The solution is dilute, so water is in huge excess and its concentration is practically constant.

Write the equilibrium constant for the self-ionisation of water.
K_c=\dfrac{[\ce{H3O+}][\ce{OH-}]}{[\ce{H2O}]^{2}}
Only about 2 molecules in every 10^{9} ionise, so [\ce{H2O}] stays at its pure-water value: 1000\,\mathrm{g} of water per \mathrm{dm^3} divided by 18\,\mathrm{g\,mol^{-1}}.
[\ce{H2O}]=\dfrac{1000}{18}=55.5\,\mathrm{mol\,dm^{-3}}\ (\text{constant})
Multiply both sides by the constant [\ce{H2O}]^2. A constant times a constant is a new constant, called K_w.
K_c[\ce{H2O}]^{2}=K_w=[\ce{H3O+}][\ce{OH-}]
In pure water at 25^\circ\mathrm{C} (298\,\mathrm{K}) measurements give [\ce{H3O+}]=[\ce{OH-}]=1.0\times10^{-7}\,\mathrm{M}.
K_w=(1.0\times10^{-7})(1.0\times10^{-7})=1.0\times10^{-14}\,\mathrm{mol^2\,dm^{-6}}

ResultK_w=[\ce{H3O+}][\ce{OH-}]=1.0\times10^{-14}\,\mathrm{mol^2\,dm^{-6}}\ \ \text{at }298\,\mathrm{K}

In every aqueous solution at 25^\circ\mathrm{C} the product [\ce{H+}][\ce{OH-}] is fixed. If an acid raises [\ce{H+}] 1000 times, [\ce{OH-}] must fall 1000 times.

Ionic product of water, K_w: The product of the molar concentrations of hydronium (hydrogen) ions and hydroxide ions in water or in any aqueous solution at a given temperature. Unit: \mathrm{mol^2\,dm^{-6}}.

Taking the negative logarithm of K_w gives \mathrm{p}K_w:

\mathrm{p}K_w=-\log K_w=-\log(1.0\times10^{-14})=14

Derivation The relation \mathrm{pH}+\mathrm{pOH}=14

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

Start from the ionic product.
[\ce{H+}][\ce{OH-}]=K_w
Take logarithms of both sides. The log of a product is the sum of the logs.
\log[\ce{H+}]+\log[\ce{OH-}]=\log K_w
Multiply every term by -1 and use the p-notation (\mathrm{p}X=-\log X).
-\log[\ce{H+}]-\log[\ce{OH-}]=-\log K_w\ \Longrightarrow\ \mathrm{pH}+\mathrm{pOH}=\mathrm{p}K_w
At 25^\circ\mathrm{C}, \mathrm{p}K_w=14.
\mathrm{pH}+\mathrm{pOH}=14

Result\mathrm{pH}+\mathrm{pOH}=\mathrm{p}K_w=14\quad(298\,\mathrm{K})

If you know pH you know pOH, and the other way round. This is the quickest route to the pH of an alkali: find pOH from [\ce{OH-}] and subtract it from 14.

K_w changes with temperature

Self-ionisation breaks bonds, so it is endothermic. By Le Chatelier's principle, heating shifts the equilibrium to the right, so K_w increases with temperature. Pure water is always neutral because [\ce{H+}]=[\ce{OH-}], but its pH is exactly 7 only at 25^\circ\mathrm{C}.

For pure water [\ce{H+}]=\sqrt{K_w}, so \mathrm{pH}=\tfrac12\mathrm{p}K_w.
Temperature	K_w / \mathrm{mol^2\,dm^{-6}}	pH of pure water
0^\circ\mathrm{C}	1.1\times10^{-15}	7.47
25^\circ\mathrm{C}	1.0\times10^{-14}	7.00
40^\circ\mathrm{C}	2.9\times10^{-14}	6.77
60^\circ\mathrm{C}	9.6\times10^{-14}	6.51

Link with K_a and K_b

The ionic product also links the strength of an acid with the strength of its conjugate base (derived in the next section but one):

\mathrm{p}K_a+\mathrm{p}K_b=\mathrm{p}K_w=14\quad(298\,\mathrm{K})

So if the \mathrm{p}K_a of an acid is known, the \mathrm{p}K_b of its conjugate base can be found. For example, ethanoic acid has \mathrm{p}K_a=4.74, so the ethanoate ion has \mathrm{p}K_b=14-4.74=9.26.

4.3Calculating pH and pOHOften examined

Every pH problem starts with two questions: is the solute an acid or a base, and is it strong or weak? Then find [\ce{H+}] or [\ce{OH-}] and convert.

Quick guide at 25^\circ\mathrm{C}; C in \mathrm{mol\,dm^{-3}}.
Solution (concentration C)	Find first	Then
Strong monoprotic acid (\ce{HCl}, \ce{HNO3})	[\ce{H+}]=C	\mathrm{pH}=-\log C
Strong diprotic acid (\ce{H2SO4}, both protons counted)	[\ce{H+}]=2C	\mathrm{pH}=-\log 2C
Strong base with one \ce{OH} (\ce{NaOH}, \ce{KOH})	[\ce{OH-}]=C	\mathrm{pOH}=-\log C, then \mathrm{pH}=14-\mathrm{pOH}
Strong base with two \ce{OH} (\ce{Ca(OH)2}, \ce{Ba(OH)2})	[\ce{OH-}]=2C	\mathrm{pOH}=-\log 2C, then \mathrm{pH}=14-\mathrm{pOH}
Weak acid, K_a given	[\ce{H+}]=\sqrt{K_aC}	\mathrm{pH}=-\log[\ce{H+}]
Weak base, K_b given	[\ce{OH-}]=\sqrt{K_bC}	\mathrm{pH}=14-\mathrm{pOH}

Example 4.3: [H3O+] in a base

If the concentration of NaOH in a solution is 2.5\times10^{-4}\,\mathrm{M}, what is the concentration of \ce{H3O+} ion at 25^\circ\mathrm{C}?

NaOH is a strong base, so it ionises completely.
[\ce{OH-}]=2.5\times10^{-4}\,\mathrm{M}
At room temperature K_w=1.0\times10^{-14}. Solve K_w=[\ce{H3O+}][\ce{OH-}] for [\ce{H3O+}].
[\ce{H3O+}]=\dfrac{K_w}{[\ce{OH-}]}=\dfrac{1.0\times10^{-14}}{2.5\times10^{-4}}=4.0\times10^{-11}\,\mathrm{M}

[\ce{H3O+}]=4.0\times10^{-11}\,\mathrm{mol\,dm^{-3}}

Example 4.4: pH of NaOH through Kw

Calculate the pH value of a 0.001\,\mathrm{mol\,dm^{-3}} solution of NaOH at 25^\circ\mathrm{C}.

Strong base, fully ionised.
[\ce{OH-}]=0.001=1.0\times10^{-3}\,\mathrm{mol\,dm^{-3}}
Use K_w=1.0\times10^{-14}\,\mathrm{mol^2\,dm^{-6}} at 298\,\mathrm{K}.
[\ce{H3O+}]=\dfrac{1.0\times10^{-14}}{1.0\times10^{-3}}=1.0\times10^{-11}\,\mathrm{mol\,dm^{-3}}
Apply the definition of pH.
\mathrm{pH}=-\log(1.0\times10^{-11})=11

\mathrm{pH}=11

Example 4.5: pH of NaOH through pOH

Calculate the pH of a 5.0\times10^{-5}\,\mathrm{M} solution of sodium hydroxide.

Find pOH from the hydroxide ion concentration.
\mathrm{pOH}=-\log[\ce{OH-}]=-\log(5.0\times10^{-5})=4.30
Use \mathrm{pH}+\mathrm{pOH}=14.
\mathrm{pH}=14-4.30=9.70

\mathrm{pH}=9.70

Concept assessment 4.2, question 1

The concentration of hydroxide ion in a given solution of slaked lime (\ce{Ca(OH)2}) is 0.001\,\mathrm{M}. Calculate the concentration of hydrogen ion in it.

The question already gives [\ce{OH-}], so do not double it. (Double only when the concentration of \ce{Ca(OH)2} itself is given.)
[\ce{OH-}]=1.0\times10^{-3}\,\mathrm{M}
Use the ionic product of water.
[\ce{H+}]=\dfrac{K_w}{[\ce{OH-}]}=\dfrac{1.0\times10^{-14}}{1.0\times10^{-3}}=1.0\times10^{-11}\,\mathrm{M}\quad(\mathrm{pH}=11)

[\ce{H+}]=1.0\times10^{-11}\,\mathrm{mol\,dm^{-3}}

Concept assessment 4.2, question 2

An aqueous solution contains 2\times10^{-3}\,\mathrm{M} of hydrogen ions (\ce{H+}). Calculate the pOH of this solution.

Find the pH first.
\mathrm{pH}=-\log(2\times10^{-3})=2.70
Subtract from 14.
\mathrm{pOH}=14-2.70=11.30

\mathrm{pOH}=11.30

Dilution, mixing and very dilute solutions

Dilution. The number of moles of \ce{H+} does not change, so C_1V_1=C_2V_2. Diluting a strong acid 10 times raises its pH by exactly 1.
Mixing an acid with a base. Find moles of \ce{H+} and of \ce{OH-}, let them neutralise each other, then divide the moles left over by the total volume.
Mixing two acids. Add the moles of \ce{H+} from each and divide by the total volume. Never add or average pH values.

Worked example: mixing an acid with a base

Find the pH when 50\,\mathrm{cm^3} of 0.10\,\mathrm{M} HCl is mixed with 30\,\mathrm{cm^3} of 0.10\,\mathrm{M} NaOH.

Moles of each ion (n=CV, with V in \mathrm{dm^3}).
\begin{aligned}n(\ce{H+})&=0.10\times0.050=5.0\times10^{-3}\,\mathrm{mol}\\n(\ce{OH-})&=0.10\times0.030=3.0\times10^{-3}\,\mathrm{mol}\end{aligned}
They react 1 : 1, so the acid is in excess.
n(\ce{H+})_{\text{left}}=5.0\times10^{-3}-3.0\times10^{-3}=2.0\times10^{-3}\,\mathrm{mol}
Divide by the total volume, 80\,\mathrm{cm^3}=0.080\,\mathrm{dm^3}.
[\ce{H+}]=\dfrac{2.0\times10^{-3}}{0.080}=0.025\,\mathrm{M}\qquad\mathrm{pH}=-\log0.025=1.60

\mathrm{pH}=1.60

Derivation pH of a very dilute strong acid

Aim. Find the correct [\ce{H+}] of a strong acid so dilute that the \ce{H+} from water cannot be ignored (for example 10^{-8}\,\mathrm{M} HCl).

Assumptions.

The acid ionises completely and supplies C\,\mathrm{mol\,dm^{-3}} of \ce{H+}.
Water supplies equal amounts, x, of \ce{H+} and \ce{OH-}.

Total [\ce{H+}] comes from the acid and from water. [\ce{OH-}] comes only from water.
[\ce{H+}]=C+x,\qquad[\ce{OH-}]=x
Put these into K_w=[\ce{H+}][\ce{OH-}].
(C+x)\,x=K_w\ \Longrightarrow\ x^{2}+Cx-K_w=0
Solve the quadratic, keep the positive root, and add C.
x=\dfrac{-C+\sqrt{C^{2}+4K_w}}{2}\ \Longrightarrow\ [\ce{H+}]=C+x=\dfrac{C+\sqrt{C^{2}+4K_w}}{2}
Example: C=1.0\times10^{-8}\,\mathrm{M} HCl.
[\ce{H+}]=\dfrac{10^{-8}+\sqrt{10^{-16}+4\times10^{-14}}}{2}=1.05\times10^{-7}\,\mathrm{M}\qquad\mathrm{pH}=6.98

Result[\ce{H+}]=\dfrac{C+\sqrt{C^{2}+4K_w}}{2}

An acid, however dilute, can never give a pH above 7. When C is much larger than 10^{-7}\,\mathrm{M} the formula reduces to [\ce{H+}]=C; when C is much smaller it gives [\ce{H+}]\approx10^{-7}\,\mathrm{M} and \mathrm{pH}\approx7.

Strength of acids and bases: K_a, K_b and degree of ionisationOften examined

The strength of an acid or base is how completely it ionises in water. It is different from its concentration, which is how much of it is dissolved in each \mathrm{dm^3}.

Type	Behaviour in water	Examples
Strong acid	ionises completely (one-way arrow)	\ce{HCl}, \ce{HBr}, \ce{HI}, \ce{HNO3}, \ce{H2SO4}, \ce{HClO4}
Weak acid	ionises partly (equilibrium)	\ce{CH3COOH}, \ce{HCOOH}, \ce{H2CO3}, \ce{HCN}, \ce{H3PO4}
Strong base	ionic hydroxide, dissociates completely	\ce{NaOH}, \ce{KOH}, \ce{Ba(OH)2}
Weak base	accepts protons from water only partly	\ce{NH3}, amines such as \ce{CH3NH2}

\ce{HCl(aq) -> H+(aq) + Cl-(aq)}\qquad(\text{about }100\%\text{ ionised})

Strong acid

\ce{CH3COOH(aq) <=> H+(aq) + CH3COO-(aq)}\qquad(\text{about }1\%\text{ ionised in }0.1\,\mathrm{M})

Weak acid

Acid dissociation constant, K_a

For a weak acid \ce{HA} the equilibrium constant of its ionisation is the acid dissociation constant, K_a. Water is in large excess, so it does not appear in the expression.

\ce{HA(aq) <=> H+(aq) + A-(aq)}\qquad K_a=\dfrac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}\qquad\mathrm{p}K_a=-\log K_a

The larger K_a, the further the equilibrium lies to the right, so the more the acid ionises and the stronger it is.
The smaller \mathrm{p}K_a, the stronger the acid (the minus sign in -\log reverses the order).
K_a changes only with temperature, not with concentration. Unit: \mathrm{mol\,dm^{-3}}.

Base dissociation constant, K_b

\begin{aligned}\ce{B(aq) + H2O(l) &<=> BH+(aq) + OH-(aq)}\\K_b&=\dfrac{[\ce{BH+}][\ce{OH-}]}{[\ce{B}]}\\\mathrm{p}K_b&=-\log K_b\end{aligned}

Example: \ce{NH3 + H2O <=> NH4+ + OH-} with K_b=1.8\times10^{-5}\,\mathrm{mol\,dm^{-3}}. A larger K_b (smaller \mathrm{p}K_b) means a stronger base.

Typical data-book values at 25^\circ\mathrm{C}. Note the pairs: \ce{NH4+}/\ce{NH3} and \ce{CH3COOH}/\ce{CH3COO-} each have \mathrm{p}K_a+\mathrm{p}K_b=14.
Acid or base	K_a or K_b / \mathrm{mol\,dm^{-3}}	\mathrm{p}K_a or \mathrm{p}K_b
methanoic acid, \ce{HCOOH}	K_a=1.8\times10^{-4}	\mathrm{p}K_a=3.75
benzoic acid, \ce{C6H5COOH}	K_a=6.3\times10^{-5}	\mathrm{p}K_a=4.20
ethanoic acid, \ce{CH3COOH}	K_a=1.8\times10^{-5}	\mathrm{p}K_a=4.74
ammonium ion, \ce{NH4+}	K_a=5.6\times10^{-10}	\mathrm{p}K_a=9.25
ethylamine, \ce{C2H5NH2}	K_b=5.6\times10^{-4}	\mathrm{p}K_b=3.25
ammonia, \ce{NH3}	K_b=1.8\times10^{-5}	\mathrm{p}K_b=4.74
ethanoate ion, \ce{CH3COO-}	K_b=5.6\times10^{-10}	\mathrm{p}K_b=9.26
phenylamine, \ce{C6H5NH2}	K_b=4.3\times10^{-10}	\mathrm{p}K_b=9.37

Degree of ionisation, \alpha

Degree of ionisation (dissociation), \alpha: The fraction of the dissolved acid or base molecules that are ionised at equilibrium. Multiply by 100 for the percentage ionisation.

\alpha=\dfrac{\text{amount ionised}}{\text{total amount dissolved}}\qquad0<\alpha\le1

Strong acids and bases have \alpha\approx1 (100%). Weak ones have a small \alpha, often 1% to 5%. At the same concentration, a larger K_a (or K_b) means a larger \alpha.

Derivation Ostwald's dilution law and the pH of a weak acid

Aim. Relate K_a, the concentration C and the degree of ionisation \alpha of a weak acid, and find [\ce{H+}].

Assumptions.

C\,\mathrm{mol\,dm^{-3}} of \ce{HA} is dissolved.
The \ce{H+} from water is negligible.

Write the amounts at equilibrium. A fraction \alpha of the acid ionises.
\begin{array}{lccc} & \ce{HA} & \ce{H+} & \ce{A-}\\ \text{start} & C & 0 & 0\\ \text{equilibrium} & C(1-\alpha) & C\alpha & C\alpha\end{array}
Substitute into the expression for K_a.
K_a=\dfrac{(C\alpha)(C\alpha)}{C(1-\alpha)}=\dfrac{C\alpha^{2}}{1-\alpha}
For a weak acid \alpha is small, so 1-\alpha\approx1.
K_a\approx C\alpha^{2}\ \Longrightarrow\ \alpha=\sqrt{\dfrac{K_a}{C}}
The hydrogen ion concentration is C\alpha.
[\ce{H+}]=C\alpha=C\sqrt{\dfrac{K_a}{C}}=\sqrt{K_aC}
Take -\log of both sides to get the pH.
\mathrm{pH}=\tfrac12\left(\mathrm{p}K_a-\log C\right)

Result\alpha=\sqrt{\dfrac{K_a}{C}}\qquad[\ce{H+}]=\sqrt{K_aC}

Diluting a weak acid (smaller C) increases its degree of ionisation, even though [\ce{H+}] falls. This is Ostwald's dilution law. For a weak base, replace K_a by K_b and [\ce{H+}] by [\ce{OH-}].

Degree of ionisation of ethanoic acid

Diluting a weak acid increases the fraction that ionises (Ostwald's dilution law). A strong acid would sit at \alpha=100\% at every concentration. C: concentration in \mathrm{mol\,dm^{-3}}.

Worked example: pH and degree of ionisation of a weak acid

Calculate the pH and the degree of ionisation of 0.10\,\mathrm{M} ethanoic acid (K_a=1.8\times10^{-5}\,\mathrm{mol\,dm^{-3}}). Compare with 0.10\,\mathrm{M} HCl.

Weak acid: use [\ce{H+}]=\sqrt{K_aC}.
[\ce{H+}]=\sqrt{1.8\times10^{-5}\times0.10}=\sqrt{1.8\times10^{-6}}=1.34\times10^{-3}\,\mathrm{M}
pH from the definition.
\mathrm{pH}=-\log(1.34\times10^{-3})=2.87
Degree of ionisation.
\alpha=\dfrac{[\ce{H+}]}{C}=\dfrac{1.34\times10^{-3}}{0.10}=0.0134\ (1.3\%)

0.10\,\mathrm{M} HCl is fully ionised: \alpha=1 and \mathrm{pH}=1.00. Same concentration, but the strong acid has about 75 times more \ce{H+}.

\mathrm{pH}=2.87,\qquad\alpha=0.013\ (1.3\%)

Derivation K_a\times K_b=K_w for a conjugate acid-base pair

Aim. Show that K_aK_b=K_w, so that \mathrm{p}K_a+\mathrm{p}K_b=14.

Write K_a for the acid \ce{HA}.
\ce{HA <=> H+ + A-}\qquad K_a=\dfrac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}
Its conjugate base \ce{A-} takes a proton from water. Write K_b.
\ce{A- + H2O <=> HA + OH-}\qquad K_b=\dfrac{[\ce{HA}][\ce{OH-}]}{[\ce{A-}]}
Multiply the two expressions. [\ce{HA}] and [\ce{A-}] cancel.
K_aK_b=\dfrac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}\times\dfrac{[\ce{HA}][\ce{OH-}]}{[\ce{A-}]}=[\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(298\,\mathrm{K})

ResultK_a\times K_b=K_w\qquad\mathrm{p}K_a+\mathrm{p}K_b=14

The stronger an acid, the weaker its conjugate base. Ethanoate (\mathrm{p}K_b=9.26) is a weak base, but it is basic enough to make sodium ethanoate solution slightly alkaline. This explains where the equivalence point lies in titrations.

4.4Acid-base titrationOften examined

A titration is used to find the concentration of an unknown solution by a neutralisation reaction. A solution of known concentration, the titrant, is added from a burette into a conical flask that holds a measured volume of the solution of unknown concentration, the analyte. A few drops of an indicator change colour when the reaction is complete.

Apparatus for an acid-base titration.

Equivalence point: The point at which the amounts of acid and alkali are exactly in the ratio of the balanced equation; for \ce{HCl + NaOH}, moles of alkali added = moles of acid. Neutralisation is complete here. On a titration curve it is the middle of the vertical (steep) section.
End point: The point at which the indicator changes colour. A suitable indicator changes colour at, or very close to, the equivalence point.
pH titration curve: A graph of the pH of the solution in the flask against the volume of titrant added.

\dfrac{M_aV_a}{n_a}=\dfrac{M_bV_b}{n_b}

Titration formula

M = molarity, V = volume, n = coefficient in the balanced equation; a = acid, b = base. For 1 : 1 reactions, M_aV_a=M_bV_b.

Concept assessment 4.3, question 1

In a titration it is found that 25\,\mathrm{cm^3} of 0.1\,\mathrm{M} solution of NaOH is neutralised with 19\,\mathrm{cm^3} of HCl of unknown concentration. Calculate the concentration of the given HCl solution. \ce{HCl + NaOH -> NaCl + H2O}

The equation is 1 : 1, so n_a=n_b=1.
M_aV_a=M_bV_b
Make M_a the subject and substitute. The volume units cancel, so \mathrm{cm^3} can be used on both sides.
M_a=\dfrac{M_bV_b}{V_a}=\dfrac{0.1\times25}{19}=0.132\,\mathrm{M}

[\ce{HCl}]=0.132\,\mathrm{mol\,dm^{-3}}

Reading any titration curve

Starting pH: the pH of the solution in the flask before any titrant is added.
Gradual region: pH changes slowly while the solution in the flask is still in excess.
Vertical region: near the equivalence point one or two drops change the pH by several units.
Final region: pH levels off towards the pH of the titrant.

4.4 (a)Strong acid and strong alkali titration curveOften examined

Take 25\,\mathrm{cm^3} of hydrochloric acid in the conical flask and add sodium hydroxide from the burette. The book's Fig 4.2 uses 1.0\,\mathrm{mol\,dm^{-3}} solutions; the calculated curve below uses 0.1\,\mathrm{M}, which starts at pH 1 but has the same shape.

\ce{HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l)}

Start: only \ce{H+} ions from the strong acid are present, so the pH is low (about 1 to 2).
Before the equivalence point: \ce{OH-} ions added from the burette react with \ce{H+} to form water, so pH rises slightly. The change is small until the volume added gets close to the equivalence point (after 24.5\,\mathrm{cm^3} the pH is still only about 3).
Near the equivalence point: the last \ce{H+} ions are used up and the pH surges upwards very steeply, from about 3 to about 11, for a fraction of a \mathrm{cm^3}.
At the equivalence point: all \ce{H+} ions have been neutralised. The flask holds only \ce{NaCl} and water, and neither \ce{Na+} nor \ce{Cl-} reacts with water, so the pH is 7. 25\,\mathrm{cm^3} of NaOH neutralises 25\,\mathrm{cm^3} of HCl of the same concentration.
After the equivalence point: there is now an excess of \ce{OH-} ions, so the pH levels off at a high value (about 12.5 here; 13 to 14 with 1.0\,\mathrm{M} NaOH).

Strong acid titrated with strong base

25\,\mathrm{cm^3} of 0.1\,\mathrm{M} \ce{HCl} in the flask, 0.1\,\mathrm{M} \ce{NaOH} from the burette (calculated curve). The book's Fig 4.2 uses 1.0\,\mathrm{M} HCl: same shape, starting at a lower pH. V: volume of titrant added in \mathrm{cm^3}.

Reverse titration: acid added to alkali

If HCl is run from the burette into NaOH in the flask (book Fig 4.3), the curve has the same shape turned upside down. The initial and final pH swap places: the curve starts high (about 13), falls slowly, drops steeply at the equivalence point, which is still pH 7, and levels off at a low pH.

Strong base titrated with strong acid

25\,\mathrm{cm^3} of 0.1\,\mathrm{M} \ce{NaOH} in the flask, 0.1\,\mathrm{M} \ce{HCl} from the burette (book Fig 4.3). The curve is the strong acid curve turned upside down. V: volume of titrant added in \mathrm{cm^3}.

4.4 (b)Strong acid and weak alkali titration curveOften examined

Now 25\,\mathrm{cm^3} of hydrochloric acid in the flask is titrated with aqueous ammonia, a weak alkali, from the burette.

\ce{HCl(aq) + NH3(aq) -> NH4Cl(aq)}

Start: only \ce{H+} ions are present (same acid as before), so the initial pH is about 1 to 2.
Before the equivalence point: added \ce{NH3} removes \ce{H+} ions (\ce{NH3 + H+ -> NH4+}), so pH rises slightly. The change is small until the equivalence point is near.
At the equivalence point: the flask holds ammonium chloride. The \ce{NH4+} ion is the conjugate acid of a weak base, so it gives some \ce{H3O+} to water. The solution is therefore acidic: pH about 5.3 (the book says about 5.5).
The steep section is short and lies on the acid side, from about pH 3 to 7.5.
After the equivalence point: more \ce{NH3} is added. It is a weak alkali and forms a buffer with \ce{NH4+}, so the pH rises above 7 but levels off at about 9 to 10, well below the 12 to 13 of a strong alkali.

\ce{NH4+(aq) + H2O(l) <=> NH3(aq) + H3O+(aq)}

Why the equivalence point is acidic

Strong acid titrated with weak base

25\,\mathrm{cm^3} of 0.1\,\mathrm{M} \ce{HCl} in the flask, 0.1\,\mathrm{M} \ce{NH3} from the burette (book Fig 4.4). The steep part is short and lies on the acid side. V: volume of titrant added in \mathrm{cm^3}.

Reverse titration: strong acid added to weak alkali

If HCl is added from the burette to ammonia in the flask, the curve is the same shape upside down, with the initial and final pH the other way around. It starts at about pH 11, falls slowly through a buffer region (pH 9.25 half-way to the equivalence point), drops steeply from about 7.5 to 3, passes the equivalence point at about pH 5.3, and levels off near pH 1.5. This is the curve asked for in short question vi and in the 2026 paper.

Weak base titrated with strong acid

25\,\mathrm{cm^3} of 0.1\,\mathrm{M} \ce{NH3} in the flask, 0.1\,\mathrm{M} \ce{HCl} from the burette. Half-way to the equivalence point, \mathrm{pH}=\mathrm{p}K_a(\ce{NH4+})=9.25. V: volume of titrant added in \mathrm{cm^3}.

Concept assessment 4.3, question 2

Draw the pH titration curve when 20\,\mathrm{cm^3} of HCl from the burette is added to 20\,\mathrm{cm^3} of aqueous solution of ammonia present in the conical flask.

The question gives no concentrations, so assume both solutions are 0.1\,\mathrm{M}. Then 20\,\mathrm{cm^3} of acid exactly neutralises the ammonia: the equivalence point is at 20\,\mathrm{cm^3}.

Starting pH: ammonia is a weak base (K_b=1.8\times10^{-5}).
\begin{aligned}{}[\ce{OH-}]&=\sqrt{K_bC}=\sqrt{1.8\times10^{-5}\times0.1}=1.34\times10^{-3}\,\mathrm{M}\\\mathrm{pH}&=14-2.87=11.1\end{aligned}
Equivalence point (20\,\mathrm{cm^3}): the flask holds 0.002\,\mathrm{mol} of \ce{NH4Cl} in 40\,\mathrm{cm^3}, that is 0.05\,\mathrm{M}. \ce{NH4+} is a weak acid with K_a=K_w/K_b=5.6\times10^{-10}.
[\ce{H+}]=\sqrt{5.6\times10^{-10}\times0.05}=5.3\times10^{-6}\,\mathrm{M}\qquad\mathrm{pH}=5.3
After 20\,\mathrm{cm^3}, excess HCl makes the pH fall steeply and then level off near 1.5.

HCl added to aqueous ammonia

20\,\mathrm{cm^3} of 0.1\,\mathrm{M} \ce{NH3} in the flask, 0.1\,\mathrm{M} HCl from the burette (concentrations assumed). The curve has been continued to 40\,\mathrm{cm^3} to show its full shape. V: volume of titrant added in \mathrm{cm^3}.

\text{equivalence point at }20\,\mathrm{cm^3},\ \mathrm{pH}\approx5.3

4.4 (c)Weak acid and strong alkali titration curveOften examined

Here 25\,\mathrm{cm^3} of ethanoic acid, \ce{CH3COOH}, in the flask is titrated with sodium hydroxide from the burette.

\ce{CH3COOH(aq) + NaOH(aq) -> CH3COONa(aq) + H2O(l)}

ethanoic acid (weak acid)

ethanoate ion (its conjugate base)

Start: ethanoic acid is weak and only about 1% ionised, so the initial pH is higher than for HCl: about 2 to 3 (2.9 for 0.1\,\mathrm{M}).
Before the equivalence point: \ce{OH-} reacts with the acid, and a mixture of \ce{CH3COOH} and \ce{CH3COO-} forms. This mixture is a buffer, so the pH rises only slowly. Half-way to the equivalence point (12.5\,\mathrm{cm^3}), [\ce{CH3COOH}]=[\ce{CH3COO-}] and \mathrm{pH}=\mathrm{p}K_a=4.74.
At the equivalence point: all the acid has reacted and the flask holds sodium ethanoate. The ethanoate ion is a weak base: it takes \ce{H+} from water and releases \ce{OH-}. So the solution is slightly basic: pH about 8.7 (the book says about 9).
The steep section is short and lies on the alkaline side, from about pH 7 to 11.
After the equivalence point: excess \ce{OH-} from the strong alkali; the pH levels off at about 12.5 to 13 (13 to 14 with 1\,\mathrm{M} NaOH).

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

Why the equivalence point is basic

Weak acid titrated with strong base

25\,\mathrm{cm^3} of 0.1\,\mathrm{M} ethanoic acid in the flask, 0.1\,\mathrm{M} \ce{NaOH} from the burette (book Fig 4.5). At half-neutralisation (12.5\,\mathrm{cm^3}) \mathrm{pH}=\mathrm{p}K_a. V: volume of titrant added in \mathrm{cm^3}.

Derivation pH at the equivalence point of a weak acid and strong base titration

Aim. Find the pH of the salt solution formed at the equivalence point.

Assumptions.

The salt \ce{NaA} is fully dissociated; only a small fraction of \ce{A-} reacts with water.

The anion of the salt reacts with water (salt hydrolysis). Its K_b comes from K_aK_b=K_w.
\ce{A- + H2O <=> HA + OH-}\qquad K_b=\dfrac{K_w}{K_a}
\ce{A-} is a weak base, so use the weak-base result [\ce{OH-}]=\sqrt{K_bC_s}, where C_s is the salt concentration at equivalence.
[\ce{OH-}]=\sqrt{\dfrac{K_wC_s}{K_a}}
Find pOH, then pH.
\mathrm{pOH}=-\log[\ce{OH-}]\qquad\mathrm{pH}=14-\mathrm{pOH}
Combining the logarithms gives a one-line formula.
\mathrm{pH}=7+\tfrac12\mathrm{p}K_a+\tfrac12\log C_s

Result\mathrm{pH}_{\text{eq}}=7+\tfrac12\mathrm{p}K_a+\tfrac12\log C_s

For ethanoic acid (\mathrm{p}K_a=4.74) with C_s=0.05\,\mathrm{M}: \mathrm{pH}=7+2.37-0.65=8.72. The equivalence point is above 7 because the anion of a weak acid is basic.

4.4 (d)Weak acid and weak alkali titration curveOften examined

Finally, ethanoic acid in the flask is titrated with aqueous ammonia.

\ce{CH3COOH(aq) + NH3(aq) -> CH3COONH4(aq)}

Start: only the \ce{H+} ions from the partial ionisation of ethanoic acid are present: initial pH about 2 to 3.
Before the equivalence point: the pH rises gradually (ethanoic acid and ethanoate form a buffer).
Around the equivalence point: there is no vertical region. The curve only bends through a 'point of inflexion' at the equivalence point; the pH changes by only about 1 unit per \mathrm{cm^3}.
At the equivalence point: here pH is about 7, because K_a of ethanoic acid and K_b of ammonia happen to be equal (1.8\times10^{-5}).
After the equivalence point: excess ammonia forms a buffer with \ce{NH4+} and the pH levels off at about 9 to 10.

Weak acid titrated with weak base

25\,\mathrm{cm^3} of 0.1\,\mathrm{M} ethanoic acid in the flask, 0.1\,\mathrm{M} \ce{NH3} from the burette (book Fig 4.6). There is no long vertical section, only a point of inflexion. V: volume of titrant added in \mathrm{cm^3}.

Because the pH changes slowly everywhere, no indicator gives a sharp colour change. The curve 'does not provide much other information', as the book says. Such titrations are followed with a pH meter (or by measuring conductivity) instead.

4.4Comparing the four titration curvesOften examined

Values for 0.1\,\mathrm{M} solutions with 25\,\mathrm{cm^3} of acid in the flask.
Titration	Starting pH	Vertical region	pH at equivalence	Final pH	Indicator
strong acid + strong base (\ce{HCl} + \ce{NaOH})	1	about 3 to 11	7	12.5 to 13	methyl orange or phenolphthalein
strong acid + weak base (\ce{HCl} + \ce{NH3})	1	about 3 to 7.5	about 5.3	about 9 to 10	methyl orange (or methyl red)
weak acid + strong base (\ce{CH3COOH} + \ce{NaOH})	about 2.9	about 7 to 11	about 8.7	12.5 to 13	phenolphthalein
weak acid + weak base (\ce{CH3COOH} + \ce{NH3})	about 2.9	none	about 7	about 9.3	none suitable (use a pH meter)

The four titration curves compared

Each curve: 25\,\mathrm{cm^3} of 0.1\,\mathrm{M} acid in the flask, 0.1\,\mathrm{M} base from the burette. All reach equivalence at 25\,\mathrm{cm^3}, but the starting pH, the length of the steep section and the pH at equivalence differ. V: volume of titrant added in \mathrm{cm^3}.

How to sketch any titration curve in the exam

Draw the axes: pH from 0 to 14 (vertical) against volume of titrant added in \mathrm{cm^3} (horizontal).
Calculate the equivalence volume from \dfrac{M_aV_a}{n_a}=\dfrac{M_bV_b}{n_b} and mark it on the volume axis.
Mark the starting pH: the pH of the solution in the flask (low for a strong acid, 2 to 3 for a weak acid, about 11 for ammonia, 13 for 0.1\,\mathrm{M} NaOH).
Mark the pH at the equivalence point: 7 for strong + strong, below 7 for strong acid + weak base, above 7 for weak acid + strong base.
Draw the steep section through the equivalence point (long for strong + strong, short for strong + weak, none for weak + weak), then level off towards the pH of the titrant.
Label the initial pH, the equivalence point (volume and pH) and the final pH.

4.4Acid-base indicators and how to choose one

An acid-base indicator is a weak organic acid, written \ce{HIn}, whose un-ionised form and ionised form (\ce{In-}) have different colours.

\ce{HIn(aq) <=> H+(aq) + In^-(aq)}

In acid (high [\ce{H+}]) the equilibrium lies to the left and the colour of \ce{HIn} is seen. In alkali, \ce{OH-} removes \ce{H+}, the equilibrium shifts right and the colour of \ce{In-} is seen.

The colour changes over a range of about 2 pH units, centred on the indicator's own \mathrm{p}K_{In}, where [\ce{HIn}]=[\ce{In-}].

\text{colour-change range}\approx\mathrm{p}K_{In}\pm1

Indicator	pH range of colour change	Colour in acid	Colour in alkali
methyl orange	3.1 to 4.4	red	yellow
methyl red	4.4 to 6.2	red	yellow
bromothymol blue	6.0 to 7.6	yellow	blue
phenolphthalein	8.3 to 10.0	colourless	pink

Rule for choosing an indicator: its whole colour-change range must lie inside the steep (vertical) part of the titration curve. Then a single drop of titrant changes the colour completely, and the end point matches the equivalence point.

Choosing an indicator

Shaded bands: colour-change ranges. Both bands cross the steep part of the strong acid curve, so either indicator works there. For ethanoic acid only the phenolphthalein band lies in the steep part; methyl orange would change colour far too early. V: volume of titrant added in \mathrm{cm^3}.

Strong acid + strong base: methyl orange or phenolphthalein (both ranges lie inside pH 3 to 11).
Strong acid + weak base: methyl orange (or methyl red). Not phenolphthalein.
Weak acid + strong base: phenolphthalein. Not methyl orange.
Weak acid + weak base: no indicator is suitable; use a pH meter.

Key points

The pH of a solution is the logarithm to base 10 of the reciprocal of the hydrogen ion concentration: \mathrm{pH}=-\log[\ce{H+}].
pH is used to compare the acidity and alkalinity of solutions. One pH unit is a 10-fold change in [\ce{H+}].
The pH scale runs from 0 to 14 for most solutions. As pH increases, basic character increases, and the other way round.
The ionic product of water is K_w=[\ce{H+}][\ce{OH-}]=1\times10^{-14}\,\mathrm{mol^2\,dm^{-6}} at 25^\circ\mathrm{C}; so \mathrm{pH}+\mathrm{pOH}=14. K_w increases with temperature.
Strong acids and bases ionise completely; weak ones only partly. K_a, K_b and the degree of ionisation \alpha measure strength; K_aK_b=K_w for a conjugate pair.
Titration is a technique used in neutralisation reactions between acids and alkalis to find the concentration of an unknown solution.
At the equivalence point neutralisation is complete (\ce{H+ + OH- -> H2O}); the indicator changes colour at the end point, which should coincide with it.
Equivalence point pH: 7 (strong + strong), below 7 (strong acid + weak base), above 7 (weak acid + strong base). Weak + weak has no vertical region.

Formula sheet

\mathrm{pH}=-\log[\ce{H+}]=\log\dfrac{1}{[\ce{H+}]}

[\ce{H+}] (or [\ce{H3O+}]): hydrogen ion concentration in \mathrm{mol\,dm^{-3}}

no unit

Hydrogen ion concentration from pH

[\ce{H+}]=10^{-\mathrm{pH}}

antilog of -\mathrm{pH}

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

pOH

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

[\ce{OH-}]: hydroxide ion concentration

no unit

Ionic product of water

K_w=[\ce{H3O+}][\ce{OH-}]=1.0\times10^{-14}\ (298\,\mathrm{K})

increases with temperature

Unit: \mathrm{mol^2\,dm^{-6}}

pH and pOH

\mathrm{pH}+\mathrm{pOH}=\mathrm{p}K_w=14\ (298\,\mathrm{K})

\mathrm{p}K_w=-\log K_w

no unit

Acid dissociation constant

K_a=\dfrac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}\qquad\mathrm{p}K_a=-\log K_a

for \ce{HA <=> H+ + A-}; larger K_a (smaller \mathrm{p}K_a) means a stronger acid

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

Base dissociation constant

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

for \ce{B + H2O <=> BH+ + OH-}; larger K_b means a stronger base

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

Conjugate acid-base pair

K_a\times K_b=K_w\qquad\mathrm{p}K_a+\mathrm{p}K_b=14

K_a of the acid, K_b of its conjugate base

Unit: \mathrm{mol^2\,dm^{-6}} for the product

Degree of ionisation (Ostwald's dilution law)

K_a=\dfrac{C\alpha^2}{1-\alpha}\approx C\alpha^2\qquad\alpha=\sqrt{\dfrac{K_a}{C}}

C: concentration of weak acid, \alpha: fraction ionised

\alpha has no unit

Weak acid

[\ce{H+}]=\sqrt{K_aC}\qquad\mathrm{pH}=\tfrac12(\mathrm{p}K_a-\log C)

valid when \alpha is small

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

Weak base

[\ce{OH-}]=\sqrt{K_bC}

then \mathrm{pH}=14-\mathrm{pOH}

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

Very dilute strong acid

[\ce{H+}]=\dfrac{C+\sqrt{C^2+4K_w}}{2}

use when C is below about 10^{-6}\,\mathrm{M}

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

Dilution

C_1V_1=C_2V_2

moles of solute unchanged on dilution

any consistent units

Titration formula

\dfrac{M_aV_a}{n_a}=\dfrac{M_bV_b}{n_b}

M: molarity, V: volume, n: coefficient in the balanced equation (a acid, b base)

volumes in the same unit on both sides

pH at equivalence (weak acid, strong base)

\mathrm{pH}=7+\tfrac12\mathrm{p}K_a+\tfrac12\log C_s

C_s: concentration of the salt at the equivalence point

no unit

Half-neutralisation of a weak acid

\mathrm{pH}=\mathrm{p}K_a

when [\ce{HA}]=[\ce{A-}], half-way to the equivalence point

no unit

Key terms

pH: The negative logarithm to base 10 of the hydrogen ion concentration in \mathrm{mol\,dm^{-3}}: \mathrm{pH}=-\log[\ce{H+}].
pOH: The negative logarithm to base 10 of the hydroxide ion concentration: \mathrm{pOH}=-\log[\ce{OH-}].
pH scale: A numerical scale, usually 0 to 14, showing how acidic (below 7), neutral (7) or alkaline (above 7) a solution is at 25^\circ\mathrm{C}.
Activity: The effective concentration of an ion; in dilute solutions it is almost equal to the concentration.
Self-ionisation of water: The reaction \ce{H2O + H2O <=> H3O+ + OH-} that occurs to a tiny extent in pure water.
Ionic product of water, K_w: K_w=[\ce{H3O+}][\ce{OH-}]; 1.0\times10^{-14}\,\mathrm{mol^2\,dm^{-6}} at 298\,\mathrm{K}. It increases with temperature.
\mathrm{p}K_w: -\log K_w; equal to 14 at 298\,\mathrm{K}, so \mathrm{pH}+\mathrm{pOH}=14.
Hydronium ion: \ce{H3O+}, the form in which a hydrogen ion exists in water.
Strong acid (base): An acid (base) that ionises completely in water.
Weak acid (base): An acid (base) that ionises only partly in water, so an equilibrium is set up.
Acid dissociation constant, K_a: The equilibrium constant K_a=[\ce{H+}][\ce{A-}]/[\ce{HA}] for the ionisation of a weak acid; a larger value means a stronger acid.
Base dissociation constant, K_b: The equilibrium constant K_b=[\ce{BH+}][\ce{OH-}]/[\ce{B}] for a weak base; a larger value means a stronger base.
Degree of ionisation, \alpha: The fraction of the dissolved acid or base that is ionised at equilibrium.
Ostwald's dilution law: For a weak electrolyte K_a=C\alpha^2/(1-\alpha): the degree of ionisation increases on dilution.
Conjugate acid-base pair: Two species that differ by one proton, such as \ce{CH3COOH} and \ce{CH3COO-}; K_aK_b=K_w.
Salt hydrolysis: Reaction of the ion of a weak acid or weak base with water, which makes a salt solution basic (e.g. \ce{CH3COO-}) or acidic (e.g. \ce{NH4+}).
Titration: A technique that uses a neutralisation reaction to find the concentration of an unknown solution.
Titrant: The solution of known concentration added from the burette.
Analyte: The solution of unknown concentration placed in the conical flask.
Equivalence point: The point at which the acid and base have reacted in exactly the ratio of the balanced equation; the middle of the vertical part of the curve.
End point: The point at which the indicator changes colour; it should coincide with the equivalence point.
Indicator: A weak organic acid \ce{HIn} whose ionised and un-ionised forms have different colours; it changes colour over about 2 pH units.
pH titration curve: A graph of pH against the volume of titrant added.
Buffer region: The flat part of a weak acid (or weak base) titration curve where both the weak acid and its conjugate base are present.
Point of inflexion: The point where a curve changes the direction of its bend; in a weak acid and weak base titration it marks the equivalence point.
Neutralisation: The reaction of an acid with a base to form a salt and water: \ce{H+ + OH- -> H2O}.

Multiple choice questionsp. 76–77

i1 mark

The value of the ionic product of water

Adepends on volume of water
Bdepends on temperatureAnswer
Cchanges by adding acid or alkali
Dalways remains constant

Why

K_w is an equilibrium constant, and equilibrium constants change only with temperature. Self-ionisation of water is endothermic, so K_w rises on heating (1.0\times10^{-14} at 25^\circ\mathrm{C}, about 9.6\times10^{-14} at 60^\circ\mathrm{C}). Adding acid or alkali changes [\ce{H+}] and [\ce{OH-}], but their product stays the same, so option C is wrong.

ii1 mark

A base when dissolved in water yields a solution with a hydroxyl ion concentration of 0.05\,\mathrm{mol\,dm^{-3}}. The solution is

AbasicAnswer
Bacidic
Cneutral
Deither b or c

Why

[\ce{OH-}]=0.05\,\mathrm{M} is far greater than 10^{-7}\,\mathrm{M}. \mathrm{pOH}=-\log0.05=1.30, so \mathrm{pH}=14-1.30=12.7: the solution is basic.

iii1 mark

pH scale was introduced by

AArrhenius
BSorensenAnswer
CLewis
DLowry

Why

The Danish biochemist S.P.L. Sorensen introduced the pH scale in 1909 while studying the effect of \ce{H+} ions on enzymes. Arrhenius, Lewis and Lowry are known for definitions of acids and bases, not for the pH scale.

Note. The book labels the options a, b, d, d; the third option (Lewis) should be c.

iv1 mark

pH of solution is defined by expression

A\log[\ce{H+}]
B\log\dfrac{1}{[\ce{H+}]}Answer
C\dfrac{1}{\log[\ce{H+}]}
D\dfrac{1}{-\log[\ce{H+}]}

Why

\mathrm{pH}=-\log[\ce{H+}]=\log\dfrac{1}{[\ce{H+}]}, because -\log x=\log(1/x). Option A has the wrong sign: it would give negative pH values for ordinary solutions such as 10^{-3}\,\mathrm{M} acid.

v1 mark

The pH of a 10^{-3}\,\mathrm{M} HCl solution at 25^\circ\mathrm{C} if it is diluted 1000 times, will be

A3
Bzero
C5.98Answer
D6.02

Why

After 1000-fold dilution C=10^{-6}\,\mathrm{M}, which alone would give pH 6. At such a low concentration water adds a little \ce{H+} as well: [\ce{H+}]=\dfrac{C+\sqrt{C^2+4K_w}}{2}=1.01\times10^{-6}\,\mathrm{M}, so the pH is just below 6 (5.996).

Extra \ce{H+} can only lower the pH, so 6.02 is impossible; 5.98 is the intended answer.

vi1 mark

How many \mathrm{dm^3} of water must be added to 1\,\mathrm{dm^3} of an aqueous solution of HCl with a pH of 1 to create an aqueous solution with pH of 2?

A0.1\,\mathrm{dm^3}
B0.9\,\mathrm{dm^3}
C2.0\,\mathrm{dm^3}
D9.0\,\mathrm{dm^3}Answer

Why

Going from pH 1 to pH 2 means [\ce{H+}] must fall 10 times, from 0.1 to 0.01\,\mathrm{M}. By C_1V_1=C_2V_2: 0.1\times1=0.01\times V_2, so V_2=10\,\mathrm{dm^3}. Water to add =10-1=9.0\,\mathrm{dm^3}. Option C wrongly assumes that doubling the volume doubles the pH.

vii1 mark

What is the approximate pH of a 1\times10^{-3}\,\mathrm{M} NaOH solution?

A3
B11Answer
C7
D1\times10^{-11}

Why

NaOH is a strong base: [\ce{OH-}]=10^{-3}\,\mathrm{M}, so \mathrm{pOH}=3 and \mathrm{pH}=14-3=11. Option A is the pOH and option D is [\ce{H+}], not the pH.

viii1 mark

Calculate the pOH of a solution at 25^\circ\mathrm{C} that contains 1\times10^{-10}\,\mathrm{M} of hydronium ions, i.e. \ce{H3O+}.

A4.0Answer
B9.0
C1.0
D7.0

Why

\mathrm{pH}=-\log(1\times10^{-10})=10, so \mathrm{pOH}=14-10=4.0.

ix1 mark

The pH value of a 10\,\mathrm{M} solution of HCl is

Aless than 0Answer
Bequal to 0
Cequal to 1
Dequal to 2

Why

[\ce{H+}]=10\,\mathrm{M}, so \mathrm{pH}=-\log10=-1, which is less than 0. Very concentrated acids lie below 0 on the pH scale.

x1 mark

Which of the following has the highest pH?

A\tfrac{M}{4} KOH
B\tfrac{M}{4} NaOH
C\tfrac{M}{4} \ce{NH4OH}
D\tfrac{M}{4} \ce{Ca(OH)2}Answer

Why

\ce{Ca(OH)2} gives two \ce{OH-} per formula unit, so [\ce{OH-}]=2\times0.25=0.5\,\mathrm{M} (pH 13.7). KOH and NaOH give 0.25\,\mathrm{M} (pH 13.4), and weak \ce{NH4OH} gives much less. (In practice \ce{Ca(OH)2} is only slightly soluble; the question assumes it all dissolves.)

xi1 mark

Which of the following statements are correct?

(i) K_w=[\ce{H+}][\ce{OH-}]=10^{-14}\,\mathrm{mol^2\,dm^{-6}} at 298\,\mathrm{K}
(ii) At 298\,\mathrm{K}, [\ce{H+}]=[\ce{OH-}]=10^{-7}
(iii) K_w does not depend upon temperature
(iv) Molarity of pure water =55.55\,\mathrm{M}

A(i), (ii) and (iii)
B(i), (ii) and (iv)Answer
C(i) and (iv)
D(ii) and (iii)

Why

(i) is the value of K_w at 298\,\mathrm{K}. (ii) is true for pure water at 298\,\mathrm{K}. (iv) is true: 1\,\mathrm{dm^3} of water has mass 1000\,\mathrm{g}, and 1000\div18=55.55\,\mathrm{mol}. (iii) is false because K_w increases with temperature.

Short answer questionsp. 78

Questions i to v are pH numericals: decide strong or weak, find the ion concentration, then convert. Question vi is a titration sketch.

i3 marks

Calculate \ce{H+} ion concentration of a solution prepared by dissolving 4\,\mathrm{g} of NaOH (atomic weight of Na = 23 amu) in 1000\,\mathrm{cm^3} of solution.

Solution

Given

m = 4\,\mathrm{g} mass of NaOH
V = 1000\,\mathrm{cm^3}=1.000\,\mathrm{dm^3} volume of solution
M(\ce{NaOH}) = 23+16+1=40\,\mathrm{g\,mol^{-1}} molar mass

To find

[\ce{H+}]

Moles of NaOH.
n=\dfrac{m}{M}=\dfrac{4}{40}=0.1\,\mathrm{mol}
NaOH is a strong base and ionises completely, so [\ce{OH-}] equals its molarity.
[\ce{OH-}]=\dfrac{0.1\,\mathrm{mol}}{1.000\,\mathrm{dm^3}}=0.1\,\mathrm{mol\,dm^{-3}}
Use the ionic product of water at 25^\circ\mathrm{C}.
[\ce{H+}]=\dfrac{K_w}{[\ce{OH-}]}=\dfrac{1.0\times10^{-14}}{0.1}=1.0\times10^{-13}\,\mathrm{mol\,dm^{-3}}
Check: this corresponds to \mathrm{pH}=13, a strongly alkaline solution, as expected.

[\ce{H+}]=1.0\times10^{-13}\,\mathrm{mol\,dm^{-3}}\ \ (\mathrm{pH}=13)

ii3 marks

Calculate the pH of 0.005 molar solution of \ce{H2SO4}.

Solution

Given

C = 0.005\,\mathrm{mol\,dm^{-3}} concentration of \ce{H2SO4}

To find

Sulphuric acid is a strong diprotic acid. Assuming both protons ionise, each molecule gives two \ce{H+}.
\ce{H2SO4 -> 2H+ + SO4^2-}\qquad[\ce{H+}]=2\times0.005=0.01\,\mathrm{mol\,dm^{-3}}
Apply the definition of pH.
\mathrm{pH}=-\log(0.01)=-\log(10^{-2})=2

\mathrm{pH}=2.0

Note. The book prints 'Calulate'; corrected to 'Calculate'.

iii3 marks

Calculate the pH of the following compounds: (i) 10^{-4}\,\mathrm{M} KOH (ii) 10^{-10}\,\mathrm{M} HCl (iii) 10^{-10}\,\mathrm{M} KOH (iv) 10^{-4}\,\mathrm{M} HCl

Solution

(i) 10^{-4}\,\mathrm{M} KOH

Strong base: [\ce{OH-}]=10^{-4}\,\mathrm{M}.
\mathrm{pOH}=-\log10^{-4}=4\qquad\mathrm{pH}=14-4=10

\mathrm{pH}=10

(ii) 10^{-10}\,\mathrm{M} HCl

The acid supplies only 10^{-10}\,\mathrm{M} of \ce{H+}, which is 1000 times less than the 10^{-7}\,\mathrm{M} from water itself. So water's \ce{H+} must be included.
[\ce{H+}]=\dfrac{C+\sqrt{C^2+4K_w}}{2}=\dfrac{10^{-10}+\sqrt{10^{-20}+4\times10^{-14}}}{2}=1.0005\times10^{-7}\,\mathrm{M}
Find the pH.
\mathrm{pH}=-\log(1.0005\times10^{-7})=6.9998\approx7.00

\mathrm{pH}\approx7.00\ \ (\text{just below }7)

(iii) 10^{-10}\,\mathrm{M} KOH

In the same way, water's own \ce{OH-} (10^{-7}\,\mathrm{M}) dominates.
[\ce{OH-}]=\dfrac{10^{-10}+\sqrt{10^{-20}+4\times10^{-14}}}{2}=1.0005\times10^{-7}\,\mathrm{M}
Find pOH and pH.
\mathrm{pOH}=6.9998\qquad\mathrm{pH}=14-6.9998=7.0002\approx7.00

\mathrm{pH}\approx7.00\ \ (\text{just above }7)

(iv) 10^{-4}\,\mathrm{M} HCl

Strong acid: [\ce{H+}]=10^{-4}\,\mathrm{M}.
\mathrm{pH}=-\log10^{-4}=4

\mathrm{pH}=4

iv3 marks

100\,\mathrm{cm^3} of 0.04\,\mathrm{M} HCl aqueous solution is mixed with 100\,\mathrm{cm^3} of 0.02\,\mathrm{M} NaOH solution. Calculate the pH of the resulting solution.

Solution

Given

V_a,\ C_a = 100\,\mathrm{cm^3}=0.100\,\mathrm{dm^3},\ 0.04\,\mathrm{M} HCl
V_b,\ C_b = 100\,\mathrm{cm^3}=0.100\,\mathrm{dm^3},\ 0.02\,\mathrm{M} NaOH

To find

pH of the mixture

Moles of \ce{H+} and \ce{OH-} (n=CV).
\begin{aligned}n(\ce{H+})&=0.04\times0.100=4.0\times10^{-3}\,\mathrm{mol}\\n(\ce{OH-})&=0.02\times0.100=2.0\times10^{-3}\,\mathrm{mol}\end{aligned}
They react 1 : 1 (\ce{H+ + OH- -> H2O}). The acid is in excess.
n(\ce{H+})_{\text{left}}=4.0\times10^{-3}-2.0\times10^{-3}=2.0\times10^{-3}\,\mathrm{mol}
Divide by the total volume, 200\,\mathrm{cm^3}=0.200\,\mathrm{dm^3}.
[\ce{H+}]=\dfrac{2.0\times10^{-3}}{0.200}=0.010\,\mathrm{M}
Find the pH.
\mathrm{pH}=-\log0.010=2.0

\mathrm{pH}=2.0

v3 marks

Equal volumes of three acid solutions of pH 3, 4 and 5 are mixed in a vessel. What will be the \ce{H+} ion concentration in the mixture?

Solution

Given

\mathrm{pH} = 3,\ 4,\ 5 equal volumes V of each

To find

[\ce{H+}] in the mixture

Convert each pH into [\ce{H+}]. pH values cannot be averaged, but moles can be added.
[\ce{H+}]_1=10^{-3},\quad[\ce{H+}]_2=10^{-4},\quad[\ce{H+}]_3=10^{-5}\,\mathrm{M}
Total moles of \ce{H+} in the three portions of volume V each.
n=(1.0\times10^{-3}+1.0\times10^{-4}+1.0\times10^{-5})V=1.11\times10^{-3}\,V\ \mathrm{mol}
The total volume is 3V.
[\ce{H+}]=\dfrac{1.11\times10^{-3}\,V}{3V}=3.7\times10^{-4}\,\mathrm{mol\,dm^{-3}}
The pH of the mixture (not asked, but a useful check).
\mathrm{pH}=-\log(3.7\times10^{-4})=3.43

[\ce{H+}]=3.7\times10^{-4}\,\mathrm{mol\,dm^{-3}}\ \ (\mathrm{pH}=3.43)

vi2026 Annual I, Set D, Q.2 (iii) OR3 marks

A 20.0\,\mathrm{cm^3} sample of 0.200\,\mathrm{mol\,dm^{-3}} \ce{NH3(aq)} was titrated with 0.100\,\mathrm{mol\,dm^{-3}} HCl. On the following axes, sketch how the pH changes during this titration. Mark clearly where the end point occurs.

Solution

Ammonia (weak base) is in the flask and HCl (strong acid) is added, so the curve starts high, falls, and has its steep drop on the acid side.

End point volume. Moles of \ce{NH3}, then the volume of HCl that neutralises it (1 : 1).
\begin{aligned}n(\ce{NH3})&=0.200\times0.0200=4.00\times10^{-3}\,\mathrm{mol}\\V(\ce{HCl})&=\dfrac{4.00\times10^{-3}}{0.100}=0.0400\,\mathrm{dm^3}=40.0\,\mathrm{cm^3}\end{aligned}
Starting pH. 0.200\,\mathrm{M} ammonia is a weak base.
[\ce{OH-}]=\sqrt{1.8\times10^{-5}\times0.200}=1.9\times10^{-3}\,\mathrm{M}\qquad\mathrm{pH}\approx11.3
Buffer region. \ce{NH3} and \ce{NH4+} are both present and the pH falls slowly; at 20\,\mathrm{cm^3} (half-way) \mathrm{pH}=\mathrm{p}K_a(\ce{NH4+})=9.25.
End point. At 40.0\,\mathrm{cm^3} the flask holds \ce{NH4Cl}, an acidic salt, so the steep drop (from about 7.5 to 3) is centred near pH 5.2.
After the end point. Excess HCl: at 60\,\mathrm{cm^3}, [\ce{H+}]=\dfrac{0.100\times0.020}{0.080}=0.025\,\mathrm{M}, so the pH levels off near 1.6.

\ce{NH3} (flask) titrated with HCl

20.0\,\mathrm{cm^3} of 0.200\,\mathrm{M} \ce{NH3} with 0.100\,\mathrm{M} HCl. End point at 40.0\,\mathrm{cm^3}, in the middle of the steep drop, at about pH 5.2. V: volume of titrant added in \mathrm{cm^3}.

Long answer questionsp. 78

i8 marks

Explain the concept of pH and its significance in acid-base chemistry. How does the pH scale relate to the concentration of hydrogen ions in a solution?

Solution

Concept of pH

The concentration of hydrogen ions in most solutions is a very small number, such as 0.0000001\,\mathrm{mol\,dm^{-3}}. In 1909 the Danish biochemist S.P.L. Sorensen introduced pH to express it as a simple number. The p stands for potenz (power), so pH means the power of the hydrogen ion concentration.

Definition: pH is the negative logarithm to base 10 of the hydrogen ion concentration in \mathrm{mol\,dm^{-3}}, or the logarithm of its reciprocal.

\mathrm{pH}=-\log[\ce{H+}]=\log\dfrac{1}{[\ce{H+}]}\qquad[\ce{H+}]=10^{-\mathrm{pH}}

Strictly, pH depends on the activity (effective concentration) of \ce{H+}, but in dilute solutions activity equals concentration. In the same way, \mathrm{pOH}=-\log[\ce{OH-}].

The pH scale

Water ionises slightly: \ce{H2O <=> H+ + OH-}, with K_w=[\ce{H+}][\ce{OH-}]=1.0\times10^{-14} at 25^\circ\mathrm{C}. Taking -\log of both sides gives \mathrm{pH}+\mathrm{pOH}=14, so the scale normally runs from 0 to 14.

\mathrm{pH}<7: acidic, [\ce{H+}]>10^{-7}\,\mathrm{M} (for example lemon juice, pH 2).
\mathrm{pH}=7: neutral, [\ce{H+}]=[\ce{OH-}]=10^{-7}\,\mathrm{M} (pure water at 25^\circ\mathrm{C}).
\mathrm{pH}>7: basic, [\ce{H+}]<10^{-7}\,\mathrm{M} (for example household ammonia, pH about 11.5).

Relation between pH and [H+]

Inverse relation: the higher [\ce{H+}], the lower the pH.
Logarithmic relation: a change of 1 pH unit is a 10-fold change in [\ce{H+}]; 2 units is a 100-fold change. A solution of pH 3 is 10 times more acidic than one of pH 4.
Example: [\ce{H+}]=1\times10^{-3}\,\mathrm{M} gives pH 3; diluting it 10 times to 1\times10^{-4}\,\mathrm{M} gives pH 4.

[\ce{H+}] / M	10^{0}	10^{-3}	10^{-7}	10^{-11}	10^{-14}
pH	0	3	7	11	14

[\ce{H+}] / M	pH
10^{0}	0
10^{-3}	3
10^{-7}	7
10^{-11}	11
10^{-14}	14

Significance of pH

Biology and medicine: blood must stay between pH 7.35 and 7.45; stomach juice (pH 1 to 2) helps digestion; each enzyme works best at a particular pH (Sorensen's original study).
Agriculture: crops grow best in soil of suitable pH; acidic soils are treated with lime (\ce{Ca(OH)2} or \ce{CaCO3}).
Industry and environment: pH is controlled in food processing, dyeing, water treatment and swimming pools; rain below pH 5.6 is acid rain.
Analytical chemistry: pH titration curves show how pH changes during neutralisation and help choose a suitable indicator.

ii8 marks

Compare and contrast the titration curves for a strong acid with a strong base, a weak acid with a strong base, a strong acid with a weak base, and a weak acid with a weak base.

Solution

Introduction

A pH titration curve is a graph of the pH in the conical flask against the volume of titrant added. For each case below, 25\,\mathrm{cm^3} of 0.1\,\mathrm{M} acid is in the flask and 0.1\,\mathrm{M} base is added from the burette, so every equivalence point is at 25\,\mathrm{cm^3}.

The four titration curves compared

1. Strong acid with strong base (\ce{HCl} and \ce{NaOH})

Starts at a low pH (about 1). The pH rises slowly, then jumps steeply from about 3 to 11 near 25\,\mathrm{cm^3}. The equivalence point is at pH 7 because \ce{NaCl} does not react with water. The curve levels off at about 12.5 to 13.

2. Weak acid with strong base (\ce{CH3COOH} and \ce{NaOH})

Starts higher (about 2.9) because the weak acid is only partly ionised. A buffer (\ce{CH3COOH}/\ce{CH3COO-}) makes the early part flat; at half-neutralisation \mathrm{pH}=\mathrm{p}K_a=4.74. The steep part is shorter (about 7 to 11) and the equivalence point is above 7 (about 8.7), because the ethanoate ion is basic: \ce{CH3COO- + H2O <=> CH3COOH + OH-}. The final pH is the same as for curve 1.

3. Strong acid with weak base (\ce{HCl} and \ce{NH3})

Starts at about pH 1, like curve 1. The steep part is short (about 3 to 7.5) and the equivalence point is below 7 (about 5.3), because the ammonium ion is acidic: \ce{NH4+ + H2O <=> NH3 + H3O+}. After equivalence the pH levels off at only about 9 to 10, because excess ammonia is a weak base.

4. Weak acid with weak base (\ce{CH3COOH} and \ce{NH3})

Starts at about 2.9 and ends at about 9.3. There is no vertical region, only a point of inflexion at the equivalence point (pH about 7 here, since K_a=K_b). The curve gives little information and no indicator can be used.

Comparison

Feature	Strong acid + strong base	Weak acid + strong base	Strong acid + weak base	Weak acid + weak base
Starting pH	about 1	about 2.9	about 1	about 2.9
Steep region	long, pH 3 to 11	short, pH 7 to 11	short, pH 3 to 7.5	none
pH at equivalence	7	about 8.7 (above 7)	about 5.3 (below 7)	about 7
Final pH	12.5 to 13	12.5 to 13	9 to 10	about 9.3
Indicator	methyl orange or phenolphthalein	phenolphthalein	methyl orange	none (use a pH meter)

Similarities and differences

Similar: every curve rises as base is added; with equal concentrations every equivalence point is at the same volume (25\,\mathrm{cm^3}), because the volume depends only on moles, not on strength.
Different: the starting pH, the length and position of the steep section, the pH at equivalence and the final pH, all of which depend on the strength of the acid and the base.
If the acid is added to the base instead, each curve has the same shape turned upside down, with the initial and final pH swapped.

iii8 marks

A 50\,\mathrm{cm^3} solution of 0.1\,\mathrm{M} acetic acid is titrated with 0.1\,\mathrm{M} sodium hydroxide. The K_a of acetic acid at 25^\circ\mathrm{C} is 1.8\times10^{-5}\,\mathrm{mol\,dm^{-3}}. Calculate the pH of the solution at the equivalence point and explain the shape of the titration curve for this reaction.

Solution

Part 1: pH at the equivalence point

Given

V_a,\ C_a = 50\,\mathrm{cm^3}=0.050\,\mathrm{dm^3},\ 0.1\,\mathrm{M} acetic (ethanoic) acid
C_b = 0.1\,\mathrm{M} NaOH
K_a = 1.8\times10^{-5}\,\mathrm{mol\,dm^{-3}}
K_w = 1.0\times10^{-14}\,\mathrm{mol^2\,dm^{-6}}

To find

pH at the equivalence point

Volume of NaOH needed (1 : 1 reaction).
\begin{aligned}\ce{CH3COOH + NaOH &-> CH3COONa + H2O}\\V_b&=\dfrac{C_aV_a}{C_b}=\dfrac{0.1\times50}{0.1}=50\,\mathrm{cm^3}\end{aligned}
At equivalence all the acid has become sodium ethanoate. Moles and concentration in the total volume 50+50=100\,\mathrm{cm^3}.
n=0.1\times0.050=5.0\times10^{-3}\,\mathrm{mol}\qquad[\ce{CH3COO-}]=\dfrac{5.0\times10^{-3}}{0.100}=0.050\,\mathrm{M}
The ethanoate ion is a weak base. Find its K_b from K_aK_b=K_w.
\begin{aligned}\ce{CH3COO- + H2O &<=> CH3COOH + OH-}\\K_b&=\dfrac{1.0\times10^{-14}}{1.8\times10^{-5}}=5.56\times10^{-10}\end{aligned}
Weak base: [\ce{OH-}]=\sqrt{K_bC}.
[\ce{OH-}]=\sqrt{5.56\times10^{-10}\times0.050}=\sqrt{2.78\times10^{-11}}=5.27\times10^{-6}\,\mathrm{M}
Convert to pH.
\mathrm{pOH}=-\log(5.27\times10^{-6})=5.28\qquad\mathrm{pH}=14-5.28=8.72

\mathrm{pH}_{\text{eq}}=8.72

Part 2: shape of the titration curve

Ethanoic acid titrated with sodium hydroxide

50\,\mathrm{cm^3} of 0.1\,\mathrm{M} ethanoic acid with 0.1\,\mathrm{M} NaOH (calculated). Equivalence at 50\,\mathrm{cm^3}, pH 8.72. V: volume of titrant added in \mathrm{cm^3}.

Initial pH (about 2.9): acetic acid is weak. [\ce{H+}]=\sqrt{K_aC}=\sqrt{1.8\times10^{-5}\times0.1}=1.34\times10^{-3}\,\mathrm{M}, so the curve starts at pH 2.87, not at pH 1 as for HCl.
Small initial rise, then a flat buffer region: the first NaOH forms ethanoate ions; the \ce{CH3COOH}/\ce{CH3COO-} mixture is a buffer, so the pH changes slowly. At 25\,\mathrm{cm^3} (half-neutralisation) [\ce{CH3COOH}]=[\ce{CH3COO-}] and \mathrm{pH}=\mathrm{p}K_a=4.74.
**Steep rise near 50\,\mathrm{cm^3}:** the last acid is used up and the pH jumps from about 7 to 11. This steep part is shorter than for a strong acid and lies on the alkaline side.
Equivalence point above 7 (pH 8.72): because the ethanoate ion hydrolyses and releases \ce{OH-}, as calculated above.
After equivalence: excess strong base; the pH levels off at about 12 to 13.

\mathrm{pH}\ \text{at equivalence}=8.72

FBISE 2026 questions

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

MCQ 32026 Annual I, Set B, MCQ 31 mark

Which titration curve shows equivalence point above pH 7?

AStrong acid vs strong base
BWeak acid vs strong baseAnswer
CStrong acid vs weak base
DWeak acid vs weak base

Why

In a weak acid and strong base titration the salt formed (for example \ce{CH3COONa}) contains the anion of a weak acid. It reacts with water and releases \ce{OH-} (\ce{CH3COO- + H2O <=> CH3COOH + OH-}), so the equivalence point is at about pH 8.7 to 9.

Strong acid + strong base gives pH 7, strong acid + weak base gives about 5.3, and weak acid + weak base depends on K_a and K_b (about 7 for ethanoic acid and ammonia).

Q.2 (ii) OR2026 Annual I, Set B, Q.2 (ii) OR1.5+1.5 = 3 marks

Draw the titration curve for HCl titrated with \ce{NH3}. Label: initial pH, equivalence point, and final pH. (\ce{NH3} in burette)

Solution

HCl (strong acid) is in the flask and \ce{NH3} (weak base) is added from the burette. Take 25\,\mathrm{cm^3} of 0.1\,\mathrm{M} HCl and 0.1\,\mathrm{M} \ce{NH3}, so the equivalence point is at 25\,\mathrm{cm^3}.

Curve (1.5 marks)

Strong acid titrated with weak base

Labels (1.5 marks)

Initial pH about 1: only the strong acid is present, [\ce{H+}]=0.1\,\mathrm{M}.
**Equivalence point at 25\,\mathrm{cm^3}, pH about 5.3 (below 7):** the flask contains \ce{NH4Cl}, and the \ce{NH4+} ion is acidic: \ce{NH4+ + H2O <=> NH3 + H3O+}. The steep section is short, from about pH 3 to 7.5.
Final pH about 9 to 10: excess ammonia is a weak base (it forms an \ce{NH3}/\ce{NH4+} buffer), so the pH never reaches the 12 to 13 of a strong alkali.

MCQ 42026 Annual I, Set D, MCQ 41 mark

Which of the following statements is correct?

AStrong bases have high degree of ionizationAnswer
BWeak bases completely ionize
CStrong bases have low degree of ionization
DWeak bases have high K_b

Why

A strong base such as NaOH is completely ionised in water, so its degree of ionisation is high (\alpha\approx1). Weak bases ionise only partly, which rules out B, and they have low K_b values, which rules out D. Option C contradicts the meaning of 'strong'.

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

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

Solution

Stronger base (1 mark)

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

Justification (2 marks)

\ce{B + H2O <=> BH+ + OH-}\qquad K_b=\dfrac{[\ce{BH+}][\ce{OH-}]}{[\ce{B}]}

K_b measures the extent of ionisation of a base. A larger K_b means the equilibrium lies further to the right: more base molecules accept protons from water and more \ce{OH-} ions are formed. K_b of B is about 550 times that of A (\mathrm{p}K_b = 2.00 for B and 4.74 for A).

For 0.1\,\mathrm{M} solutions, A ionises only slightly, so \alpha=\sqrt{K_b/C}.
\alpha_A=\sqrt{\dfrac{1.8\times10^{-5}}{0.1}}=0.013\ (1.3\%)
B ionises much more; solving K_b=\dfrac{C\alpha^2}{1-\alpha} exactly:
0.1\alpha^2+0.01\alpha-0.01=0\ \Longrightarrow\ \alpha_B=0.27\ (27\%)

So B is about 20 times more ionised than A and gives a much higher [\ce{OH-}] (pH 12.4 against 11.1 for 0.1\,\mathrm{M} solutions). Greater ionisation means a stronger base.

Q.2 (iii) OR2026 Annual I, Set D, Q.2 (iii) OR2+1 = 3 marks

Sketch the titration curve of a weak base with a strong acid and indicate the position of the equivalence point.

Solution

Take 25\,\mathrm{cm^3} of 0.1\,\mathrm{M} ammonia (weak base) in the flask and add 0.1\,\mathrm{M} HCl (strong acid) from the burette.

Sketch (2 marks)

Weak base titrated with strong acid

The curve starts at about pH 11 (weak base, only partly ionised).
It falls slowly through a buffer region (\ce{NH3}/\ce{NH4+}); half-way, \mathrm{pH}=9.25.
It then drops steeply from about 7.5 to 3 and levels off near pH 1.5 in excess acid.

Equivalence point (1 mark)

The equivalence point is at 25\,\mathrm{cm^3} of acid, in the middle of the steep drop, at about pH 5.3 (below 7). At this point the flask holds \ce{NH4Cl}, whose \ce{NH4+} ion is acidic: \ce{NH4+ + H2O <=> NH3 + H3O+}. Methyl orange is a suitable indicator.

Exam focus

Define pH and show that a change of one pH unit is a 10-fold change in [\ce{H+}]. (short)
A stated SLO; the definition and the 10-fold idea appear in MCQs and short questions every year.
Go to answer
Derive K_w=[\ce{H3O+}][\ce{OH-}] and \mathrm{pH}+\mathrm{pOH}=14; explain why K_w depends on temperature. (short)
Underpins every pH numerical; MCQs i, viii and xi test it.
Go to answer
Calculate the pH of strong acids and bases, including \ce{H2SO4} and \ce{Ca(OH)2}, and of mixtures of acid and base. (numerical)
Short questions i, ii and iv are this type; each is a quick 3 marks.
Go to answer
pH of very dilute solutions (10^{-8} or 10^{-10}\,\mathrm{M} acid or base). (numerical)
A favourite trap (short question iii, MCQ v): the answer stays close to 7.
Go to answer
Compare two bases (or acids) from their K_b (K_a) values using the degree of ionisation. (short)
Asked in 2026 (Set D Q.2 (iii) and MCQ 4).
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Draw the titration curve for HCl titrated with \ce{NH3}, labelling the initial pH, equivalence point and final pH. (short)
Asked in 2026 (Set B Q.2 (ii) OR); the same idea as Concept assessment 4.3 Q2.
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Sketch the titration curve of a weak base with a strong acid and mark the end point. (short)
Book short question vi; asked in 2026 (Set D Q.2 (iii) OR).
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Compare and contrast the four titration curves (strong and weak acids with strong and weak bases). (long)
The main SLO of the unit; a likely Section C question.
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Calculate the pH at the equivalence point of a weak acid and strong base titration and explain the curve. (long)
Book long question iii; combines K_a, K_b=K_w/K_a and the curve shape.
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Choose a suitable indicator for a given titration and give the reason. (short)
Links the curve shape to practical work; often asked with the curves.
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Which titration has its equivalence point above pH 7? (mcq)
Asked in 2026 (Set B MCQ 3).
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4Acids and bases: what you already know

4.1The pHOften examined

Activity: the more exact definition

One pH unit is a 10-fold change in [H+]

Derivation Why one pH unit means a 10-fold change in [HX+]

pH of bases and pOH

4.2The pH scaleOften examined

4.3Ionic product of water, K_wOften examined

Derivation The ionic product of water, K_w

Derivation The relation \mathrm{pH}+\mathrm{pOH}=14

K_w changes with temperature

Link with K_a and K_b

4.3Calculating pH and pOHOften examined

Dilution, mixing and very dilute solutions

Derivation pH of a very dilute strong acid

Strength of acids and bases: K_a, K_b and degree of ionisationOften examined

Acid dissociation constant, K_a

Base dissociation constant, K_b

Degree of ionisation, \alpha

Derivation Ostwald's dilution law and the pH of a weak acid

Derivation K_a\times K_b=K_w for a conjugate acid-base pair

4.4Acid-base titrationOften examined

Reading any titration curve

4.4 (a)Strong acid and strong alkali titration curveOften examined

Reverse titration: acid added to alkali

4.4 (b)Strong acid and weak alkali titration curveOften examined

Reverse titration: strong acid added to weak alkali

4.4 (c)Weak acid and strong alkali titration curveOften examined

Derivation pH at the equivalence point of a weak acid and strong base titration

4.4 (d)Weak acid and weak alkali titration curveOften examined

4.4Comparing the four titration curvesOften examined

4.4Acid-base indicators and how to choose one

Key points

Formula sheet

Key terms

Multiple choice questionsp. 76–77

Short answer questionsp. 78

Long answer questionsp. 78

Concept of pH

The pH scale

Relation between pH and [H+]

Significance of pH

Introduction

1. Strong acid with strong base (\ce{HCl} and \ce{NaOH})

2. Weak acid with strong base (\ce{CH3COOH} and \ce{NaOH})

3. Strong acid with weak base (\ce{HCl} and \ce{NH3})

4. Weak acid with weak base (\ce{CH3COOH} and \ce{NH3})

Comparison

Similarities and differences

Part 1: pH at the equivalence point

Part 2: shape of the titration curve

FBISE 2026 questions

Curve (1.5 marks)

Labels (1.5 marks)

Stronger base (1 mark)

Justification (2 marks)

Sketch (2 marks)

Equivalence point (1 mark)

Exam focus

Derivation Why one pH unit means a 10-fold change in $[H X^{+}]$