Weak Acids and Bases: Ka, Kb, pKa and pKb – pH of Acidic and Alkaline Solutions (Senior 6 Chemistry) | Ganzaa

pH of Acidic and Alkaline Solutions

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Senior 6 · Chemistry

pH of Acidic and Alkaline Solutions

5 lessons

Lesson 3 of 5

Weak Acids and Bases: Ka, Kb, pKa and pKb

25 min

Same concentration, different acids

Take two beakers, one of 0.10 mol dm⁻³ hydrochloric acid and one of 0.10 mol dm⁻³ ethanoic acid (the acid in vinegar). Drop a strip of magnesium into each. The hydrochloric acid fizzes vigorously; the ethanoic acid only bubbles gently. A pH meter reads 1.00 in the first beaker and 2.88 in the second, and the ethanoic acid conducts electricity far less well. Yet titrating each with sodium hydroxide takes exactly the same volume of alkali.

So both beakers contain the same amount of acid, but in the ethanoic acid only a small fraction of the molecules has given its proton to water at any moment. In this lesson you will measure that fraction, express it as an equilibrium constant, and use the constant to calculate the pH of weak acids and weak bases.

Definition

Weak acid: an acid that is only partly ionised in water, so that un-ionised molecules and ions are in equilibrium. Example: $\mathrm{CH_3COOH + H_2O \rightleftharpoons CH_3COO^{-} + H_3O^{+}}$ .

Degree of ionisation, $\alpha$ : the fraction of the acid (or base) molecules that are ionised at equilibrium. Example: $\alpha = 0.013$ (1.3 %) for 0.10 mol dm⁻³ ethanoic acid.

Acid dissociation constant, $K_{a}$ : the equilibrium constant for $\mathrm{HA + H_2O \rightleftharpoons H_3O^{+} + A^{-}}$ , $K_a = \frac{[\mathrm{H_3O^{+}}][\mathrm{A^{-}}]}{[\mathrm{HA}]}$ , in $\mathrm{mol\,dm^{-3}}$ . p $K_{a}$ $= -\log_{10}K_a$ .

Base dissociation constant, $K_{b}$ : for $\mathrm{B + H_2O \rightleftharpoons BH^{+} + OH^{-}}$ , $K_b = \frac{[\mathrm{BH^{+}}][\mathrm{OH^{-}}]}{[\mathrm{B}]}$ . p $K_{b}$ $= -\log_{10}K_b$ .

Measuring strength: $K_{a}$ and p $K_{a}$

Water is the solvent and is left out of the expression, as in Unit 7. The larger $K_{a}$ , the further the equilibrium lies to the right and the stronger the acid. Because p $K_{a}$ has a minus sign, a smaller p $K_{a}$ means a stronger acid: each unit of p $K_{a}$ is a factor of 10 in $K_{a}$ .

Acid (25 °C)	Formula	$K_{a}$ / $\mathrm{mol\,dm^{-3}}$	p $K_{a}$
hydrofluoric acid	HF	$6.3 \times 10^{-4}$	3.20
methanoic acid	HCOOH	$1.8 \times 10^{-4}$	3.75
lactic acid	$\mathrm{CH_3CH(OH)COOH}$	$1.4 \times 10^{-4}$	3.86
benzoic acid	$\mathrm{C_6H_5COOH}$	$6.3 \times 10^{-5}$	4.20
ethanoic acid	$\mathrm{CH_3COOH}$	$1.7 \times 10^{-5}$	4.76
carbonic acid (first step)	$\mathrm{H_2CO_3}$	$4.5 \times 10^{-7}$	6.35
ammonium ion	$\mathrm{NH_4^{+}}$	$5.6 \times 10^{-10}$	9.25
phenol	$\mathrm{C_6H_5OH}$	$1.0 \times 10^{-10}$	9.99

Table 3.1: Acid dissociation constants at 25 °C, strongest at the top.

For bases, $K_{b}$ plays the same role: methylamine (p $K_{b}$ 3.36) is a stronger base than ammonia (4.75), which is far stronger than phenylamine (9.13), whose lone pair is drawn into the benzene ring (Unit 4).

The link between $K_{a}$ and $K_{b}$

For a conjugate pair such as $\mathrm{NH_4^{+}/NH_3}$ , multiply the two constants:

$K_a \times K_b = \frac{[\mathrm{NH_3}][\mathrm{H_3O^{+}}]}{[\mathrm{NH_4^{+}}]} \times \frac{[\mathrm{NH_4^{+}}][\mathrm{OH^{-}}]}{[\mathrm{NH_3}]} = [\mathrm{H_3O^{+}}][\mathrm{OH^{-}}] = K_w$

$K_a \times K_b = K_w \qquad \mathrm{p}K_a + \mathrm{p}K_b = 14.00 \text{ at 25 °C}$

So the weaker an acid, the stronger its conjugate base. Check: p $K_a(\mathrm{NH_4^{+}})$ + p $K_b(\mathrm{NH_3})$ = 9.25 + 4.75 = 14.00.

Degree of ionisation and dilution

If a weak acid of concentration $c$ is ionised to the extent $\alpha$ , then $[\mathrm{H_3O^{+}}] = [\mathrm{A^{-}}] = c\alpha$ and $[\mathrm{HA}] = c(1-\alpha)$ :

$K_a = \frac{c\alpha \cdot c\alpha}{c(1-\alpha)} = \frac{c\alpha^{2}}{1-\alpha} \approx c\alpha^{2} \quad \text{when } \alpha \ll 1, \text{ so } \alpha \approx \sqrt{\frac{K_a}{c}}$

This is Ostwald's dilution law, which you met in S5 with conductivity: a weak acid ionises more as it is diluted (Figure 3.1).

Figure 3.1: Percentage of ethanoic acid ionised at different concentrations ( $K_a = 1.74 \times 10^{-5}$ ).

The pH of a weak acid or weak base

When $\alpha$ is small (below about 5 %), $[\mathrm{HA}] \approx c$ , and

$[\mathrm{H_3O^{+}}] = \sqrt{K_a\,c} \qquad \text{and, for a weak base,} \qquad [\mathrm{OH^{-}}] = \sqrt{K_b\,c}$

Two assumptions are made: the $\mathrm{H_3O^{+}}$ from water is negligible, and so little acid ionises that its concentration is unchanged. Always check the second one; if $\alpha$ comes out above 5 %, solve the quadratic instead.

Worked example

Problem. Calculate the pH and the degree of ionisation of 0.100 mol dm⁻³ ethanoic acid at 25 °C ( $K_a = 1.74 \times 10^{-5}\,\mathrm{mol\,dm^{-3}}$ ).

Solution.

$[\mathrm{H_3O^{+}}] = \sqrt{K_a\,c} = \sqrt{1.74 \times 10^{-5} \times 0.100} = 1.32 \times 10^{-3}\,\mathrm{mol\,dm^{-3}}$

$\mathrm{pH} = -\log_{10}(1.32 \times 10^{-3}) = 2.88 \qquad \alpha = \frac{1.32 \times 10^{-3}}{0.100} = 0.0132$

Check: $\alpha = 1.3\,\%$ , well below 5 %, so the approximation is valid.

Answer: pH = 2.88; 1.3 % of the molecules are ionised. Hydrochloric acid of the same concentration has pH 1.00: 76 times more $\mathrm{H_3O^{+}}$ .

Worked example

Problem. A 0.0200 mol dm⁻³ solution of benzoic acid has pH 2.95 at 25 °C. Calculate $K_{a}$ and p $K_{a}$ .

Solution. $[\mathrm{H_3O^{+}}] = [\mathrm{C_6H_5COO^{-}}] = 10^{-2.95} = 1.12 \times 10^{-3}\,\mathrm{mol\,dm^{-3}}$ , and $[\mathrm{C_6H_5COOH}] = 0.0200 - 0.00112 = 0.0189\,\mathrm{mol\,dm^{-3}}$ .

$K_a = \frac{(1.12 \times 10^{-3})^2}{0.0189} = 6.7 \times 10^{-5}\,\mathrm{mol\,dm^{-3}} \qquad \mathrm{p}K_a = 4.18$

Answer: $K_a = 6.7 \times 10^{-5}\,\mathrm{mol\,dm^{-3}}$ , p $K_{a}$ = 4.18, close to the value in Table 3.1. (Ignoring the 0.00112 in the bottom line would give $6.3 \times 10^{-5}$ ; here $\alpha = 5.6\,\%$ , so the subtraction matters.)

Worked example

Problem. Calculate the pH of 0.150 mol dm⁻³ aqueous ammonia at 25 °C ( $K_b = 1.8 \times 10^{-5}\,\mathrm{mol\,dm^{-3}}$ ), and the $K_{a}$ of its conjugate acid.

Solution.

$[\mathrm{OH^{-}}] = \sqrt{K_b\,c} = \sqrt{1.8 \times 10^{-5} \times 0.150} = 1.64 \times 10^{-3}\,\mathrm{mol\,dm^{-3}}$

$\mathrm{pOH} = 2.78 \qquad \mathrm{pH} = 14.00 - 2.78 = 11.22$

$K_a(\mathrm{NH_4^{+}}) = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.6 \times 10^{-10}\,\mathrm{mol\,dm^{-3}}$

Answer: pH = 11.22 ( $\alpha = 1.1\,\%$ , so the approximation holds); $K_a(\mathrm{NH_4^{+}}) = 5.6 \times 10^{-10}\,\mathrm{mol\,dm^{-3}}$ .

Warning

Common mistakes with weak acids

Treating a weak acid as strong: 0.100 mol dm⁻³ ethanoic acid does not have pH 1.00.
Using $\sqrt{K_a c}$ without checking $\alpha$ . For 0.00100 mol dm⁻³ ethanoic acid it gives $\alpha = 13\,\%$ , and the exact (quadratic) value is 12 %.
Reading p $K_{a}$ the wrong way round: a larger p $K_{a}$ means a weaker acid.
Confusing "weak" with "dilute": a 5 mol dm⁻³ ethanoic acid solution is concentrated but still weak.

Tip

For any weak-acid problem, write the three-line ICE table even if you then use the shortcut: $\mathrm{HA}$ : $c \to c - x$ ; $\mathrm{H_3O^{+}}$ : $0 \to x$ ; $\mathrm{A^{-}}$ : $0 \to x$ . It shows the examiner which assumption you made, and it makes the quadratic easy when the assumption fails.

Chemistry Student: Acid Dissociation Constant, Ka and pKa (A-Level Chemistry). Watch how the Ka expression is used to find the hydrogen-ion concentration of a weak acid, and how Ka and pKa are converted into each other.

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Practice

Write expressions for $K_{a}$ of methanoic acid and $K_{b}$ of methylamine, $\mathrm{CH_3NH_2}$ .
Use Table 3.1 to arrange phenol, ethanoic acid, methanoic acid and carbonic acid in order of increasing acid strength.
Calculate the pH of 0.0500 mol dm⁻³ methanoic acid ( $K_a = 1.8 \times 10^{-4}$ ). Check whether the approximation is valid.
A 0.100 mol dm⁻³ solution of lactic acid has pH 2.43. Calculate $K_{a}$ and p $K_{a}$ .
Calculate the pH of 0.0200 mol dm⁻³ methylamine ( $K_b = 4.4 \times 10^{-4}$ ). Why must the quadratic be used here?
Calculate $K_{b}$ and p $K_{b}$ of the ethanoate ion, $\mathrm{CH_3COO^{-}}$ .
A student argues: "0.10 mol dm⁻³ HCl and 0.10 mol dm⁻³ ethanoic acid need the same volume of NaOH to neutralise them, so they are equally strong acids." Analyse this argument.

Show answersHide answers

$K_a = \frac{[\mathrm{H_3O^{+}}][\mathrm{HCOO^{-}}]}{[\mathrm{HCOOH}]}$ ; $K_b = \frac{[\mathrm{CH_3NH_3^{+}}][\mathrm{OH^{-}}]}{[\mathrm{CH_3NH_2}]}$ .
Phenol (9.99) < carbonic acid (6.35) < ethanoic acid (4.76) < methanoic acid (3.75).
$\sqrt{1.8 \times 10^{-4} \times 0.0500} = 3.0 \times 10^{-3}$ , pH 2.52; $\alpha = 6\,\%$ , slightly above 5 %, so the approximation is borderline. The quadratic gives $2.91 \times 10^{-3}$ , pH 2.54.
$[\mathrm{H_3O^{+}}] = 10^{-2.43} = 3.72 \times 10^{-3}$ ; $K_a = \frac{(3.72 \times 10^{-3})^2}{0.100 - 0.00372} = 1.4 \times 10^{-4}\,\mathrm{mol\,dm^{-3}}$ ; p $K_{a}$ = 3.84.
The shortcut gives $[\mathrm{OH^{-}}] = 2.97 \times 10^{-3}$ , $\alpha = 15\,\%$ : too large. Solving $x^2 + 4.4 \times 10^{-4}x - 8.8 \times 10^{-6} = 0$ gives $x = 2.75 \times 10^{-3}$ ; pOH = 2.56, pH = 11.44.
$K_b = \frac{1.0 \times 10^{-14}}{1.74 \times 10^{-5}} = 5.7 \times 10^{-10}\,\mathrm{mol\,dm^{-3}}$ ; p $K_{b}$ = 9.24.
The equal titres show the two solutions contain the same amount of acid, because as NaOH removes $\mathrm{H_3O^{+}}$ , the ethanoic acid equilibrium keeps shifting right until all of it has reacted. Strength is the extent of ionisation before any base is added: HCl is fully ionised (pH 1.00), ethanoic acid about 1.3 % (pH 2.88). The argument confuses amount with strength.

Sources and credits

Dissociation constants in Table 3.1 and the text: values as tabulated in the CRC Handbook of Chemistry and Physics ("Dissociation constants of organic acids and bases" and "of inorganic acids and bases"), rounded by Ganzaa.
Image: HCl & magnesium.jpg by Dr.T, via Wikimedia Commons, licensed under CC BY-SA 3.0.
Video: Acid Dissociation Constant, Ka and pKa (A-Level Chemistry), Chemistry Student, embedded from YouTube.
The text, graph, table, worked examples and Practice items were written and drawn by Ganzaa.
This lesson adapts CC BY-SA material and is shared under CC BY-SA 4.0.

1Same concentration, different acids

2Measuring strength: Ka K_a Ka​ and pKa K_a Ka​

3The link between Ka K_a Ka​ and Kb K_b Kb​

4Degree of ionisation and dilution

5The pH of a weak acid or weak base

Same concentration, different acids

Measuring strength: $K_{a}$ and p $K_{a}$

The link between $K_{a}$ and $K_{b}$

Degree of ionisation and dilution

The pH of a weak acid or weak base