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pH of weak acids and bases

Covers AS91392

Why a weak acid needs more work

For a strong acid, $[\text{H}_3\text{O}^+]$ equals the concentration of the acid, so the pH is one step away.
For a weak acid only a small fraction dissociates, so $[\text{H}_3\text{O}^+]$ is much less than the concentration and must be calculated from $K_a$ .

The method

Write the equilibrium:

$\text{HA} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{A}^-$

Note that the two ions are produced in a 1:1 ratio, so $[\text{H}_3\text{O}^+] = [\text{A}^-] = x$ .
Approximate the acid: $[\text{HA}] \approx c$ , its initial concentration.
Substitute into the $K_a$ expression:

$K_a = \frac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]} = \frac{x^2}{c}$

Solve and take the log:

$x = \sqrt{K_a \, c} \qquad \text{pH} = -\log x$

Why the approximation is allowed

Strictly, $[\text{HA}]$ at equilibrium is $c - x$ , because some of the acid has dissociated.
But EN6 of the standard states that the acids and bases you meet are those "in which the extent of reaction is small so that the equilibrium concentration of a dissolved weak acid or base can be approximated by the initial concentration".
So $c - x \approx c$ , and you may use the initial concentration throughout.
For a typical weak acid, $x$ is around 1 % of $c$ , so the error introduced is negligible.

Weak bases: derive $K_b$ first

The specification says $K_b$ values will not be provided, but may be derived. You get given the $K_a$ of the conjugate acid instead.

$K_b = \frac{K_w}{K_a}$

Then the calculation mirrors the acid case, but produces hydroxide ions:

$\text{B} + \text{H}_2\text{O} \rightleftharpoons \text{BH}^+ + \text{OH}^-$

$[\text{OH}^-] = \sqrt{K_b \, c}$

Finish by converting to pH, either via $\text{pH} = 14 - \text{pOH}$ or via $K_w$ .

The steps for a weak base

Identify the conjugate acid and look up its $K_a$ .
Derive $K_b = K_w / K_a$ .
Calculate $[\text{OH}^-] = \sqrt{K_b c}$ .
Convert: $\text{pOH} = -\log[\text{OH}^-]$ , then $\text{pH} = 14 - \text{pOH}$ .
That is four steps rather than two — leaving out the derivation or the conversion is where the marks go.

Percentage dissociation

Sometimes you are asked how much of the acid has dissociated:

$\text{percentage dissociation} = \frac{[\text{H}_3\text{O}^+]}{c} \times 100$

A typical weak acid gives around 1 %, which is what justifies the approximation above.

Tips and Tricks

Do not set $[\text{H}_3\text{O}^+]$ equal to the concentration of a weak acid. That is only true for a strong one. If your weak acid comes out with the same pH a strong acid would have, you have used the wrong method.

Square root, then log — in that order. A common slip is to take the log of $K_a c$ without the square root, which gives a pH roughly twice too large.

For a base, you are not finished when you have $[\text{OH}^-]$ . Convert to pOH and then to pH, and check the answer is above 7.

Derive $K_b$ explicitly and show it. The specification says $K_b$ will not be given, so the derivation $K_b = K_w / K_a$ is itself worth a mark.

State the approximation when you use it: "the extent of dissociation is small, so $[\text{HA}] \approx c$ ". Markers look for the assumption to be acknowledged.

Worked ExampleA weak acid and a weak base

Calculate, at 25 °C:

(a) the pH of $0.100$ mol L⁻¹ ethanoic acid, $K_a = 1.74 \times 10^{-5}$ (b) the pH of $0.150$ mol L⁻¹ ammonia, given $K_a(\text{NH}_4^+) = 5.6 \times 10^{-10}$

Part (a) — the weak acid

Step 1 — Write the equilibrium.

$\text{CH}_3\text{COOH} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{CH}_3\text{COO}^-$

Step 2 — Set up the unknowns. The two ions are formed in a 1:1 ratio, so if $[\text{H}_3\text{O}^+] = x$ then $[\text{CH}_3\text{COO}^-] = x$ too.

Because the extent of dissociation is small, the acid concentration at equilibrium is approximately its initial value:

$[\text{CH}_3\text{COOH}] \approx 0.100 \text{ mol L}^{-1}$

Step 3 — Substitute into $K_a$ .

$K_a = \frac{x^2}{c} \quad \Rightarrow \quad 1.74 \times 10^{-5} = \frac{x^2}{0.100}$

$x^2 = 1.74 \times 10^{-6}$

$x = \sqrt{1.74 \times 10^{-6}} = 1.319 \times 10^{-3} \text{ mol L}^{-1}$

Step 4 — Take the negative log.

$\text{pH} = -\log(1.319 \times 10^{-3}) = 2.879$

Part (b) — the weak base

Step 1 — Derive $K_b$ . Ammonia is a weak base, and its conjugate acid is $\text{NH}_4^+$ . The specification states $K_b$ will not be provided, so derive it:

$K_b = \frac{K_w}{K_a} = \frac{1 \times 10^{-14}}{5.6 \times 10^{-10}} = 1.786 \times 10^{-5}$

Step 2 — Write the equilibrium.

$\text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^-$

Step 3 — Find the hydroxide concentration. Again the two ions are formed 1:1, and the extent of reaction is small so $[\text{NH}_3] \approx 0.150$ .

$[\text{OH}^-] = \sqrt{K_b \, c} = \sqrt{1.786 \times 10^{-5} \times 0.150}$

$= \sqrt{2.679 \times 10^{-6}} = 1.637 \times 10^{-3} \text{ mol L}^{-1}$

Step 4 — Convert to pH. This is the step that must not be forgotten.

$\text{pOH} = -\log(1.637 \times 10^{-3}) = 2.786$

$\text{pH} = 14 - 2.786 = 11.214$

pH of weak acids and bases

Covers AS91392

Why a weak acid needs more work

For a strong acid, $[\text{H}_3\text{O}^+]$ equals the concentration of the acid, so the pH is one step away.
For a weak acid only a small fraction dissociates, so $[\text{H}_3\text{O}^+]$ is much less than the concentration and must be calculated from $K_a$ .

The method

Write the equilibrium:

$\text{HA} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{A}^-$

Note that the two ions are produced in a 1:1 ratio, so $[\text{H}_3\text{O}^+] = [\text{A}^-] = x$ .
Approximate the acid: $[\text{HA}] \approx c$ , its initial concentration.
Substitute into the $K_a$ expression:

$K_a = \frac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]} = \frac{x^2}{c}$

Solve and take the log:

$x = \sqrt{K_a \, c} \qquad \text{pH} = -\log x$

Why the approximation is allowed

Strictly, $[\text{HA}]$ at equilibrium is $c - x$ , because some of the acid has dissociated.
But EN6 of the standard states that the acids and bases you meet are those "in which the extent of reaction is small so that the equilibrium concentration of a dissolved weak acid or base can be approximated by the initial concentration".
So $c - x \approx c$ , and you may use the initial concentration throughout.
For a typical weak acid, $x$ is around 1 % of $c$ , so the error introduced is negligible.

Weak bases: derive $K_b$ first

The specification says $K_b$ values will not be provided, but may be derived. You get given the $K_a$ of the conjugate acid instead.

$K_b = \frac{K_w}{K_a}$

Then the calculation mirrors the acid case, but produces hydroxide ions:

$\text{B} + \text{H}_2\text{O} \rightleftharpoons \text{BH}^+ + \text{OH}^-$

$[\text{OH}^-] = \sqrt{K_b \, c}$

Finish by converting to pH, either via $\text{pH} = 14 - \text{pOH}$ or via $K_w$ .

The steps for a weak base

Identify the conjugate acid and look up its $K_a$ .
Derive $K_b = K_w / K_a$ .
Calculate $[\text{OH}^-] = \sqrt{K_b c}$ .
Convert: $\text{pOH} = -\log[\text{OH}^-]$ , then $\text{pH} = 14 - \text{pOH}$ .
That is four steps rather than two — leaving out the derivation or the conversion is where the marks go.

Percentage dissociation

Sometimes you are asked how much of the acid has dissociated:

$\text{percentage dissociation} = \frac{[\text{H}_3\text{O}^+]}{c} \times 100$

A typical weak acid gives around 1 %, which is what justifies the approximation above.

Tips and Tricks

Square root, then log — in that order. A common slip is to take the log of $K_a c$ without the square root, which gives a pH roughly twice too large.

For a base, you are not finished when you have $[\text{OH}^-]$ . Convert to pOH and then to pH, and check the answer is above 7.

Derive $K_b$ explicitly and show it. The specification says $K_b$ will not be given, so the derivation $K_b = K_w / K_a$ is itself worth a mark.

State the approximation when you use it: "the extent of dissociation is small, so $[\text{HA}] \approx c$ ". Markers look for the assumption to be acknowledged.

Worked ExampleA weak acid and a weak base

Calculate, at 25 °C:

(a) the pH of $0.100$ mol L⁻¹ ethanoic acid, $K_a = 1.74 \times 10^{-5}$ (b) the pH of $0.150$ mol L⁻¹ ammonia, given $K_a(\text{NH}_4^+) = 5.6 \times 10^{-10}$

Part (a) — the weak acid

Step 1 — Write the equilibrium.

$\text{CH}_3\text{COOH} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{CH}_3\text{COO}^-$

Step 2 — Set up the unknowns. The two ions are formed in a 1:1 ratio, so if $[\text{H}_3\text{O}^+] = x$ then $[\text{CH}_3\text{COO}^-] = x$ too.

Because the extent of dissociation is small, the acid concentration at equilibrium is approximately its initial value:

$[\text{CH}_3\text{COOH}] \approx 0.100 \text{ mol L}^{-1}$

Step 3 — Substitute into $K_a$ .

$K_a = \frac{x^2}{c} \quad \Rightarrow \quad 1.74 \times 10^{-5} = \frac{x^2}{0.100}$

$x^2 = 1.74 \times 10^{-6}$

$x = \sqrt{1.74 \times 10^{-6}} = 1.319 \times 10^{-3} \text{ mol L}^{-1}$

Step 4 — Take the negative log.

$\text{pH} = -\log(1.319 \times 10^{-3}) = 2.879$

Part (b) — the weak base

Step 1 — Derive $K_b$ . Ammonia is a weak base, and its conjugate acid is $\text{NH}_4^+$ . The specification states $K_b$ will not be provided, so derive it:

$K_b = \frac{K_w}{K_a} = \frac{1 \times 10^{-14}}{5.6 \times 10^{-10}} = 1.786 \times 10^{-5}$

Step 2 — Write the equilibrium.

$\text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^-$

Step 3 — Find the hydroxide concentration. Again the two ions are formed 1:1, and the extent of reaction is small so $[\text{NH}_3] \approx 0.150$ .

$[\text{OH}^-] = \sqrt{K_b \, c} = \sqrt{1.786 \times 10^{-5} \times 0.150}$

$= \sqrt{2.679 \times 10^{-6}} = 1.637 \times 10^{-3} \text{ mol L}^{-1}$

Step 4 — Convert to pH. This is the step that must not be forgotten.

$\text{pOH} = -\log(1.637 \times 10^{-3}) = 2.786$

$\text{pH} = 14 - 2.786 = 11.214$

pH of weak acids and bases

Why a weak acid needs more work

The method

Why the approximation is allowed

Weak bases: derive KbK_bKb​ first

The steps for a weak base

Percentage dissociation

Part (a) — the weak acid

Part (b) — the weak base

pH of weak acids and bases

Why a weak acid needs more work

The method

Why the approximation is allowed

Weak bases: derive KbK_bKb​ first

The steps for a weak base

Percentage dissociation

Part (a) — the weak acid

Part (b) — the weak base

Weak bases: derive $K_b$ first

Weak bases: derive $K_b$ first