pH of a Weak Acid

For a weak acid, $[\mathrm{H_3O^+}]$ is not the acid's concentration; only a small part ionizes. Its Ka and an ICE table give the answer, and the small-x approximation usually makes it quick.

Key idea

For a weak acid at concentration $C$: $K_a = \dfrac{x^2}{C - x}$ with $x = [\mathrm{H_3O^+}]$. If $x$ is under 5% of $C$, then $x \approx \sqrt{K_aC}$. Then $\text{pH} = -\log x$.

Why it works

Each HA that ionizes makes one $\mathrm{H_3O^+}$ and one $\mathrm{A^-}$, so both equal $x$, and $C - x$ acid is left. The water's own $\mathrm{H_3O^+}$ ($10^{-7}\ \text{M}$ at most) is tiny next to $x$ in any ordinary weak acid solution, so it is ignored. Because $K_a$ is small, $x$ is usually much smaller than $C$, which lets you drop it from $C - x$.

Same concentration, different strength: HCl is fully ionized, while only a small fraction of acetic acid molecules give up their protons at any moment.

Worked examples

Worked example: Acetic acid

Find the pH of $0.10\ \text{M}$ acetic acid, $K_a = 1.8 \times 10^{-5}$.

ICE: $[\mathrm{H_3O^+}] = [\mathrm{CH_3COO^-}] = x$, $[\mathrm{CH_3COOH}] = 0.10 - x$.
Approximate: $x = \sqrt{(1.8 \times 10^{-5})(0.10)} = 1.3 \times 10^{-3}\ \text{M}$. Check: $1.3 \times 10^{-3}/0.10 = 1.3\%$, under 5%.
$\text{pH} = -\log(1.34 \times 10^{-3}) = 2.87$. Compare: $0.10\ \text{M}$ HCl has pH 1.00.

Answer: $\text{pH} = 2.87$.

Worked example: When the shortcut fails

Find the pH of $0.20\ \text{M}$ HF, $K_a = 6.8 \times 10^{-4}$.

Approximate: $x = \sqrt{(6.8 \times 10^{-4})(0.20)} = 0.0117$. Check: $0.0117/0.20 = 5.8\%$. Over the limit.
Solve exactly: $x^2 + (6.8 \times 10^{-4})x - 1.36 \times 10^{-4} = 0$, so $b^2 - 4ac = 5.44 \times 10^{-4}$ (square root $0.02333$) and $x = \dfrac{-0.00068 + 0.02333}{2} = 0.0113\ \text{M}$.
$\text{pH} = -\log(0.0113) = 1.95$.

Answer: $\text{pH} = 1.95$.

Worked example: Ka from a measured pH

A $0.050\ \text{M}$ solution of a weak acid has pH 3.00. Find $K_a$.

$x = 10^{-3.00} = 1.0 \times 10^{-3}\ \text{M}$, so $[\mathrm{A^-}] = 1.0 \times 10^{-3}$ and $[\mathrm{HA}] = 0.050 - 0.0010 = 0.049\ \text{M}$.
$K_a = \dfrac{(1.0 \times 10^{-3})^2}{0.049} = 2.0 \times 10^{-5}$.

Answer: $K_a = 2.0 \times 10^{-5}$.

Two assumptions hide inside the shortcut, and both can be checked. The 5% test checks that $x$ is small next to $C$. And we ignored the $\mathrm{H_3O^+}$ that water itself provides; that is safe as long as $x$ is much larger than $10^{-7}\ \text{M}$, which is true unless the acid is extremely weak or extremely dilute. The percent ionization lesson shows what happens as an acid is diluted.

Common mistake

Treating a weak acid as fully ionized, so that $0.10\ \text{M}$ acetic acid would have pH 1.00. A weak acid gives far less $\mathrm{H_3O^+}$ than its concentration. Check: a weak acid's pH must be higher than that of a strong acid at the same concentration.

pH of a Weak Acid

Worked examples

Practice

Builds on

Leads to

Sources and further reading