Topic 14 · page 5 of 18

pH of a weak acid

Two assumptions turn the Ka expression into a square root, so you can find the pH of a weak acid without solving a quadratic.

14 min read14.9

Read first: The acid dissociation constant

Two assumptions 14.9

For a strong acid, [H⁺] is just the concentration. A weak acid is harder. You are given KaK_\text{a} and the concentration you made up, but the equilibrium expression holds three unknown concentrations:

Ka=[HX+][AX−][HA] K_\text{a} = \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}

You cannot solve one equation for three unknowns without help. The help is two sensible assumptions.

Assumption 1 lets you replace [A⁻] with [H⁺]. Assumption 2 lets you replace [HA] with the known concentration, c.

Ka=[HX+]2c⇒[HX+]=Ka×c K_\text{a} = \frac{[\ce{H+}]^2}{c} \qquad\Rightarrow\qquad [\ce{H+}] = \sqrt{K_\text{a} \times c}

No quadratic equation is needed. The square root is the method.

MethodpH of a weak acid

  1. Write the equation and the expression for KaK_\text{a}.
  2. State the two assumptions.
  3. Calculate [H⁺] = √(KaK_\text{a} × c).
  4. Calculate pH = −lg [H⁺].
  5. (Optional check) [H⁺] ÷ c should be small, a few per cent at most.

Worked examples 14.9

Worked exampleEthanoic acid

Calculate the pH of 0.100 mol dm⁻³ ethanoic acid. KaK_\text{a} = 1.74 × 10⁻⁵ mol dm⁻³.

  1. Assume [H⁺] = [CH₃COO⁻] and [CH₃COOH] ≈ 0.100 mol dm⁻³.
  2. [H⁺] = √(1.74 × 10⁻⁵ × 0.100) = √(1.74 × 10⁻⁶) = 1.32 × 10⁻³ mol dm⁻³
  3. pH = −lg (1.32 × 10⁻³) = 2.88
  4. Check: 1.32 × 10⁻³ ÷ 0.100 = 0.013, so 1.3% is dissociated and assumption 2 holds.

This is the 2.88 from the start of the topic. It is also why hydrochloric acid of the same concentration, pH 1.00, is so different.

When the question gives pKa\text{p}K_\text{a} instead, convert it first.

Worked exampleFrom pKa

Calculate the pH of 0.050 mol dm⁻³ benzoic acid, pKa\text{p}K_\text{a} = 4.20.

  1. Ka=10−4.20K_\text{a} = 10^{-4.20} = 6.31 × 10⁻⁵ mol dm⁻³
  2. [H⁺] = √(6.31 × 10⁻⁵ × 0.050) = √(3.155 × 10⁻⁶) = 1.78 × 10⁻³ mol dm⁻³
  3. pH = −lg (1.78 × 10⁻³) = 2.75

A shortcut with logs 14.9

Take logs of [H⁺]² = KaK_\text{a} × c and change signs:

pH=12(pKa−lg⁡c) \text{pH} = \tfrac{1}{2}\left(\text{p}K_\text{a} - \lg c\right)

For benzoic acid: pH = ½ (4.20 + 1.30) = 2.75, the same answer in one step. Use it to check your first method, and remember it only works under the same two assumptions.

When the assumptions stop working 14.9

Assumption 2 is good when the acid is concentrated and KaK_\text{a} is small. It gets worse when the solution is very dilute, because a larger fraction of the acid is dissociated. The table compares the shortcut with the full equilibrium solution for ethanoic acid. The full values are model values from solving the quadratic, shown only so you can see the trend.

Concentration / mol dm⁻³ pH by the two assumptions pH from the full solution Dissociated
0.100 2.88 2.88 1.3%
0.0100 3.38 3.39 4.1%
0.00100 3.88 3.91 12%
0.000100 4.38 4.47 34%

Check yourself

  1. State the two assumptions made when calculating the pH of a weak acid from its KaK_\text{a}.
  2. Calculate the pH of 0.0100 mol dm⁻³ methanoic acid. KaK_\text{a} = 1.6 × 10⁻⁴ mol dm⁻³.
  3. Calculate the pH of 0.250 mol dm⁻³ propanoic acid. KaK_\text{a} = 1.35 × 10⁻⁵ mol dm⁻³.
  4. Use pKa\text{p}K_\text{a} = 4.76 to calculate the pH of 0.250 mol dm⁻³ ethanoic acid.
  5. A solution of ethanoic acid has pH 3.00. Calculate its concentration. KaK_\text{a} = 1.74 × 10⁻⁵ mol dm⁻³.
  6. Show that diluting a weak acid ten times raises its pH by 0.5, and explain why a strong acid rises by 1.0 instead.

Answers cover these until you have tried the questions

  1. (1) [H⁺] = [A⁻], because the acid is the only significant source of H⁺ (water's H⁺ is negligible). (2) [HA] at equilibrium ≈ the initial [HA], because only a small fraction of the weak acid dissociates.
  2. [H⁺] = √(1.6 × 10⁻⁴ × 0.0100) = √(1.6 × 10⁻⁶) = 1.26 × 10⁻³ mol dm⁻³. pH = −lg (1.26 × 10⁻³) = 2.90
  3. [H⁺] = √(1.35 × 10⁻⁵ × 0.250) = √(3.375 × 10⁻⁶) = 1.84 × 10⁻³ mol dm⁻³. pH = 2.74
  4. pH = ½ (pKa\text{p}K_\text{a} − lg c) = ½ (4.76 − lg 0.250) = ½ (4.76 + 0.602) = 2.68
  5. [H⁺] = 10−3.0010^{-3.00} = 1.00 × 10⁻³ mol dm⁻³. From [H⁺]² = KaK_\text{a} × c: c = (1.00 × 10⁻³)² ÷ 1.74 × 10⁻⁵ = 1.00 × 10⁻⁶ ÷ 1.74 × 10⁻⁵ = 0.0575 mol dm⁻³.
  6. pH = ½ (pKa\text{p}K_\text{a} − lg c). Dilution by ten makes c ten times smaller, so −lg c rises by 1, and the pH rises by ½ × 1 = 0.5. For a strong acid [H⁺] = c, so pH = −lg c and a tenfold dilution raises the pH by the full 1.0. The weak acid rises by less because more of it dissociates as it is diluted, which partly replaces the H⁺ lost.