Weak Acids and Bases: From Ka and Kb to pH

Reading Ka, Kb and their p-scales

For a weak acid HA, the ionization HA(aq) + H₂O(l) ⇌ H₃O⁺(aq) + A⁻(aq) has the constant Ka = [H₃O⁺][A⁻]/[HA]. Water is the solvent and is left out. A larger Ka means a stronger acid. For a weak base B, the reaction B(aq) + H₂O(l) ⇌ BH⁺(aq) + OH⁻(aq) has Kb = [BH⁺][OH⁻]/[B].

Because these constants span many powers of ten, they are often reported as pKa = −log Ka and pKb = −log Kb. The scale runs backward: a smaller pKa means a stronger acid. Appendix H gives Ka = 1.8 × 10⁻⁵ for acetic acid (pKa 4.74), 6.4 × 10⁻⁴ for hydrofluoric acid (pKa 3.19) and 4.9 × 10⁻¹⁰ for hydrocyanic acid (pKa 9.31). HF's Ka is about a million times larger than HCN's, even though both are called weak.

Worked example: pH of a weak acid with the small-x shortcut

Find the pH of 0.200 M acetic acid, CH₃COOH. Let x be the amount that ionizes. At equilibrium [H₃O⁺] = [CH₃COO⁻] = x and [CH₃COOH] = 0.200 − x. Water's own contribution to H₃O⁺ (10⁻⁷ M) is negligible next to the acid's.

Write 1.8 × 10⁻⁵ = x² / (0.200 − x). Because Ka is small, assume x is much less than 0.200 and drop it from the denominator. Then x² = 1.8 × 10⁻⁵ × 0.200 = 3.6 × 10⁻⁶ and x = 1.9 × 10⁻³ M. Check: 1.9 × 10⁻³ / 0.200 is just under 1 percent, well below the 5 percent limit, so the shortcut holds. pH = −log(1.9 × 10⁻³) = 2.72, reported to two decimal places because Ka has two significant figures.

Compare that with a strong acid at the same concentration, which would give pH 0.70. The weak acid has about one hundredth of the hydronium concentration.

Worked example: when the shortcut fails and you need the quadratic

Find the pH of 0.0500 M nitrous acid, HNO₂, with Ka = 4.6 × 10⁻⁴ from Appendix H. The shortcut gives x = √(4.6 × 10⁻⁴ × 0.0500) = 4.8 × 10⁻³ M, which is 9.6 percent of 0.0500. That fails the 5 percent check, so solve the full equation.

Rearrange 4.6 × 10⁻⁴ = x² / (0.0500 − x) into x² + (4.6 × 10⁻⁴)x − 2.3 × 10⁻⁵ = 0. The positive root is x = 4.57 × 10⁻³ M, so pH = 2.34 and the acid is 9.1 percent ionized. The shortcut would have given pH 2.32. A difference of 0.02 looks small, but it is larger than the 0.01 precision the data support, so the quadratic answer is the one to report.

A quick rule of thumb: the shortcut is usually safe when the starting concentration divided by Ka is above a few hundred. Here 0.0500 / 4.6 × 10⁻⁴ is only about 110, which is why it failed. Still run the 5 percent check every time.

Percent ionization rises as the acid is diluted

Percent ionization is [H₃O⁺] at equilibrium divided by the starting acid concentration, times 100. For a weak acid it is not fixed. Solving the full equation for acetic acid gives 0.94 percent at 0.200 M, 3.0 percent at 0.0200 M and 9.0 percent at 0.00200 M.

The reason is Le Chatelier's principle applied to dilution. Diluting lowers every concentration, which lowers Q below Ka, and the equilibrium shifts toward more ions. The pH still rises on dilution, from 2.72 to 3.23 to 3.74 in this series, but by only about half a unit per tenfold dilution, not the full unit you would see for a strong acid.

Worked example: a weak base and its pOH

Find the pH of 0.120 M ammonia, with Kb = 1.8 × 10⁻⁵ from Appendix I. The table is the same shape as for an acid, but the product is hydroxide: [NH₄⁺] = [OH⁻] = x and [NH₃] = 0.120 − x.

Then x² = 1.8 × 10⁻⁵ × 0.120, so x = [OH⁻] = 1.5 × 10⁻³ M. The check gives 1.2 percent, so the shortcut holds. pOH = −log(1.47 × 10⁻³) = 2.83, and pH = 14.00 − 2.83 = 11.17. The most common slip here is to report 2.83 as the pH. A base must give a pH above 7.

Salts whose ions react with water

For any conjugate acid–base pair in water, Ka × Kb = Kw = 1.0 × 10⁻¹⁴ at 25 °C. The stronger the acid, the weaker its conjugate base. This lets you find a constant that the tables do not list.

Worked example 1. Find the pH of 0.100 M sodium acetate. Na⁺ does not react with water, but acetate is a weak base. Its Kb is 1.0 × 10⁻¹⁴ / 1.8 × 10⁻⁵ = 5.6 × 10⁻¹⁰. Then [OH⁻] = √(5.6 × 10⁻¹⁰ × 0.100) = 7.5 × 10⁻⁶ M, pOH = 5.13 and pH = 8.87. The solution is mildly basic.

Worked example 2. Find the pH of 0.200 M ammonium chloride. Cl⁻ is the conjugate base of a strong acid and has no measurable effect, but NH₄⁺ is a weak acid with Ka = 1.0 × 10⁻¹⁴ / 1.8 × 10⁻⁵ = 5.6 × 10⁻¹⁰. Then [H₃O⁺] = √(5.6 × 10⁻¹⁰ × 0.200) = 1.1 × 10⁻⁵ M, and pH = 4.98.

• Cation from a strong base (Na⁺, K⁺) with anion from a strong acid (Cl⁻, NO₃⁻): neutral.
• Cation from a strong base with the anion of a weak acid (CH₃COO⁻, F⁻, CN⁻): basic.
• Cation that is the conjugate acid of a weak base (NH₄⁺) with the anion of a strong acid: acidic.
• Both ions reactive: compare the cation's Ka with the anion's Kb. Whichever constant is larger wins.

Polyprotic acids lose their protons one at a time

Acids such as carbonic acid, H₂CO₃, and phosphoric acid, H₃PO₄, have one Ka for each proton, and each is much smaller than the one before. For carbonic acid, Appendix H lists Ka1 = 4.3 × 10⁻⁷ and Ka2 = 4.7 × 10⁻¹¹. Because Ka1 is thousands of times larger than Ka2, almost all the hydronium comes from the first step.

Worked example. For 0.0330 M carbonic acid, treat the first step as a single weak acid: [H₃O⁺] = √(4.3 × 10⁻⁷ × 0.0330) = 1.2 × 10⁻⁴ M, so pH = 3.92. The second step barely changes this. Because [H₃O⁺] and [HCO₃⁻] are nearly equal, they cancel in the Ka2 expression, and [CO₃²⁻] comes out approximately equal to Ka2, 4.7 × 10⁻¹¹ M.

Weak acid and base errors to watch for

Before you finish, ask whether the answer makes sense. A weak acid solution should have a higher pH than a strong acid of the same concentration, and a weak base a lower pH than a strong base.

• Treating a weak acid as strong and setting [H₃O⁺] equal to the starting concentration.
• Skipping the 5 percent check after dropping x.
• Reporting pOH as pH for a base, or forgetting to convert at the end.
• Using Ka where Kb is needed for a salt's anion. Convert with Kw first.
• Treating Cl⁻, NO₃⁻ or Na⁺ as if they change the pH.
• Adding the second ionization's hydronium to the first for a polyprotic acid when Ka2 is far smaller than Ka1.
• Reporting more decimal places in pH than there are significant figures in Ka.