A 0.010 mol sample of a weak monoprotic acid HA is dissolved in water to make 1.00 L of solution. At equilibrium, the concentration of H+ is 2.0 × 10^-4 M. What is the value of the acid dissociation constant, Ka, for HA?
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Correct answer
A. 4.0 × 10^-6
Principle or equation
For a weak monoprotic acid HA, the acid dissociation constant is Ka = [H+][A-]/[HA]. At equilibrium, [H+] = [A-] and [HA] = initial concentration - [H+].
Why this answer is correct
Initial [HA] = 0.010 M. At equilibrium, [H+] = [A-] = 2.0 × 10^-4 M. [HA] = 0.010 - 2.0 × 10^-4 ≈ 0.0098 M. Then Ka = (2.0 × 10^-4)^2 / 0.0098 ≈ 4.08 × 10^-6, which rounds to 4.0 × 10^-6.
Example
For a 0.10 M weak acid with [H+] = 1.0 × 10^-3 M, Ka = (1.0 × 10^-3)^2 / (0.10 - 1.0 × 10^-3) ≈ 1.0 × 10^-5.
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