Topic 14 · page 4 of 18

The acid dissociation constant

Ka is the equilibrium constant for a weak acid splitting into ions. A larger Ka means a stronger acid, and pKa turns it into a small number.

12 min read14.814.12

Read first: Strong and weak acids

Writing Ka 14.8

A weak acid HA sets up an equilibrium with water, with the acid and both ions all present:

HA(aq)⇌HX+(aq)+AX−(aq) \ce{HA(aq) <=> H+(aq) + A-(aq)}

This is an ordinary equilibrium, so it has an equilibrium constant. You write it exactly as you wrote KcK_\text{c} in the last topic: concentrations of the products on the top, concentrations of the reactants on the bottom.

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

The units are mol dm⁻³: (mol dm⁻³ × mol dm⁻³) ÷ mol dm⁻³ leaves mol dm⁻³. The water is left out of the expression, because it is the solvent and its concentration is effectively constant.

Worked exampleEthanoic acid

Write the expression for KaK_\text{a} for ethanoic acid. At 298 K its value is 1.74 × 10⁻⁵ mol dm⁻³.

  1. Equation: CHX3COOH(aq)⇌HX+(aq)+CHX3COOX−(aq)\ce{CH3COOH(aq) <=> H+(aq) + CH3COO-(aq)}
  2. Ka=[CHX3COOX−][HX+][CHX3COOH]K_\text{a} = \dfrac{[\ce{CH3COO-}][\ce{H+}]}{[\ce{CH3COOH}]} = 1.74 × 10⁻⁵ mol dm⁻³

The pattern is the same for any weak acid, including a cation. The ammonium ion is a weak acid because it can give away a proton:

NHX4X+(aq)⇌HX+(aq)+NHX3(aq)Ka=[HX+][NHX3][NHX4X+] \ce{NH4+(aq) <=> H+(aq) + NH3(aq)} \qquad K_\text{a} = \frac{[\ce{H+}][\ce{NH3}]}{[\ce{NH4+}]}

pKa 14.12

KaK_\text{a} values for weak acids are tiny and awkward, such as 0.000 017 4. The same trick as pH makes them readable.

Like pH, pKa\text{p}K_\text{a} has no units, and a smaller pKa\text{p}K_\text{a} means a stronger acid.

Acid KaK_\text{a} / mol dm⁻³ pKa\text{p}K_\text{a}
Methanoic acid 1.6 × 10⁻⁴ 3.80
Benzoic acid 6.3 × 10⁻⁵ 4.20
Ethanoic acid 1.74 × 10⁻⁵ 4.76
Ammonium ion, NH₄⁺ 5.6 × 10⁻¹⁰ 9.25

Methanoic acid has the largest KaK_\text{a} and the smallest pKa\text{p}K_\text{a}, so it is the strongest of the four.

Worked exampleConverting both ways

(a) Calculate pKa\text{p}K_\text{a} for an acid with KaK_\text{a} = 1.74 × 10⁻⁵ mol dm⁻³. (b) Calculate KaK_\text{a} for an acid with pKa\text{p}K_\text{a} = 3.80.

  1. (a) pKa=−lg⁡(1.74×10−5)\text{p}K_\text{a} = -\lg(1.74\times10^{-5}) = 4.76
  2. (b) Ka=10−3.80K_\text{a} = 10^{-3.80} = 1.58 × 10⁻⁴, so 1.6 × 10⁻⁴ mol dm⁻³

What stays the same when you dilute 14.8

Add water to a solution of a weak acid. The concentrations of HA, H⁺ and A⁻ all fall, so the fraction on the right of the expression briefly drops below KaK_\text{a}. The equilibrium shifts to the right to restore the value. More acid dissociates, so the degree of dissociation rises, but KaK_\text{a} itself has not changed. Page 8 uses this to tell a weak acid from a strong one.

Check yourself

  1. Write the expression for KaK_\text{a} for propanoic acid, CH₃CH₂COOH, and give its units.
  2. Write the expression for KaK_\text{a} for the methylammonium ion, CH₃NH₃⁺, which dissociates into CH₃NH₂ and H⁺.
  3. Acid A has KaK_\text{a} = 3.2 × 10⁻⁴ mol dm⁻³ and acid B has KaK_\text{a} = 4.5 × 10⁻⁶ mol dm⁻³. Calculate pKa\text{p}K_\text{a} for each and say which is the stronger acid.
  4. A weak acid has pKa\text{p}K_\text{a} = 2.87. Calculate KaK_\text{a}.
  5. Explain why KaK_\text{a} for ethanoic acid does not change when the solution is diluted, even though the acid becomes more dissociated.
  6. Why does the expression for KaK_\text{a} not contain [H₂O]?

Answers cover these until you have tried the questions

  1. CHX3CHX2COOH(aq)⇌HX+(aq)+CHX3CHX2COOX−(aq)\ce{CH3CH2COOH(aq) <=> H+(aq) + CH3CH2COO-(aq)}, so Ka=[CHX3CHX2COOX−][HX+][CHX3CHX2COOH]K_\text{a} = \dfrac{[\ce{CH3CH2COO-}][\ce{H+}]}{[\ce{CH3CH2COOH}]} with units mol dm⁻³.
  2. Ka=[CHX3NHX2][HX+][CHX3NHX3X+]K_\text{a} = \dfrac{[\ce{CH3NH2}][\ce{H+}]}{[\ce{CH3NH3+}]} (units mol dm⁻³).
  3. Acid A: −lg⁡(3.2×10−4)-\lg(3.2\times10^{-4}) = 3.49. Acid B: −lg⁡(4.5×10−6)-\lg(4.5\times10^{-6}) = 5.35. Acid A has the larger KaK_\text{a} and the smaller pKa\text{p}K_\text{a}, so it is the stronger acid.
  4. Ka=10−2.87K_\text{a} = 10^{-2.87} = 1.35 × 10⁻³, so 1.3 × 10⁻³ mol dm⁻³.
  5. KaK_\text{a} is an equilibrium constant, so it depends only on temperature. On dilution the concentrations fall, the equilibrium shifts right until the ratio of products to reactant equals KaK_\text{a} again, so the degree of dissociation rises but the value of KaK_\text{a} is unchanged.
  6. Water is the solvent, present in huge excess, so its concentration is effectively constant and is absorbed into KaK_\text{a}.