What if the 5% rule fails?

You must expand the equilibrium expression into standard quadratic form, $ax^2 + bx + c = 0$, and use the quadratic formula to solve for $x$.

What if the weak acid is polyprotic?

For most polyprotic acids, $K_{a1}$ is significantly larger than $K_{a2}$. The first dissociation step produces almost all the hydronium ions, so you generally only calculate using $K_{a1}$.

What if the acid concentration is very low (e.g., $< 10^{-6}$ M)?

The autoionization of water becomes significant. You can no longer assume the initial $[H_3O^+]$ is zero, and must set up mass and charge balance equations to solve the system.

How to calculate pH of a weak acid

Q: What if the acid concentration is very low (e.g., $< 10^{-6}$ M)?

The autoionization of water becomes significant. You can no longer assume the initial $[H_3O^+]$ is zero, and must set up mass and charge balance equations to solve the system.

The pH of a weak acid is calculated by determining the equilibrium concentration of hydronium ions $[H_3O^+]$ using the acid dissociation constant $K_a$ and an ICE (Initial, Change, Equilibrium) table.

This method applies to aqueous solutions of monoprotic weak acids where the dissociation is incomplete, and the initial acid concentration is significantly larger than the ions provided by the autoionization of water.

The setup

Write the balanced chemical equation for the dissociation of the weak acid $HA$ in water: $HA(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + A^-(aq)$ . Identify the initial concentration of the acid $[HA]_0$ and the given $K_a$ value for the acid.

The steps

Set up an ICE table with columns for $HA$ , $H_3O^+$ , and $A^-$ .
Enter initial concentrations: $[HA]_0$ for the acid, and $0$ for both products (neglecting the $1.0 imes 10^{-7}$ M from water).
Define the change in concentration as $-x$ for $HA$ , and $+x$ for the products.
Write the equilibrium concentrations as $[HA]_0 - x$ , $x$ , and $x$ .
Substitute these into the equilibrium expression: $K_a = \frac{[H_3O^+][A^-]}{[HA]} = \frac{x^2}{[HA]_0 - x}$ .
Solve for $x$ . If $[HA]_0 / K_a > 100$ , assume $x$ is negligible compared to $[HA]_0$ to simplify to $K_a \approx \frac{x^2}{[HA]_0}$ .
Calculate pH using $ext{pH} = -\log_{10}(x)$ .

Checking the result

Verify the small $x$ approximation by checking the percent ionization: $\frac{x}{[HA]_0} imes 100\%$ . If this value is less than 5%, the approximation is valid. If it is 5% or greater, you must solve the exact quadratic equation $x^2 + K_a x - K_a [HA]_0 = 0$ .

Common errors

A frequent error is forgetting to take the negative base-10 logarithm of $x$ to find the pH. Another common mistake is applying the small $x$ approximation without checking the 5% rule, leading to inaccurate concentrations for relatively strong weak acids or very dilute solutions.

Worked example

Calculate the pH of a 0.10 M solution of acetic acid ( $CH_3COOH$ ), given $K_a = 1.8 imes 10^{-5}$ .

$CH_3COOH \rightleftharpoons H^+ + CH_3COO^-$

Initial: $0.10$ M, $0$ , $0$ Change: $-x$ , $+x$ , $+x$ Equilibrium: $0.10 - x$ , $x$ , $x$

$K_a = \frac{x^2}{0.10 - x} = 1.8 imes 10^{-5}$

Assume $x \ll 0.10$ , so $0.10 - x \approx 0.10$ .

$\frac{x^2}{0.10} = 1.8 imes 10^{-5}$ $x^2 = 1.8 imes 10^{-6}$ $x = \sqrt{1.8 imes 10^{-6}} = 1.34 imes 10^{-3} ext{ M}$

Check approximation: $\frac{1.34 imes 10^{-3}}{0.10} imes 100 = 1.34\%$ (valid).

$ext{pH} = -\log_{10}(1.34 imes 10^{-3})$ $ext{pH} = 2.87$

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References: OpenStax Chemistry 2e, Chapter 14: Acid-Base Equilibria · Zumdahl Chemistry, Chapter 7: Acids and Bases