The pH of a 0.60 M aqueous solution of formic acid,HCHO₂,at 25.0 °C is 1.98.What is the value of K_a for formic acid?

Question

Quizplus · Accepted Answer

Formic acid, HCHO2, ionizes in water as follows: HCHO2(aq) + H2O(l) ⇌ H3O+(aq) + CHO2-(aq)
The equilibrium constant for this reaction is Ka = [H3O+][CHO2−]/[HCHO2].
The pH of the solution is 1.98, so [H3O+] = 10^(-1.98) = 1.12 × 10^(-2) M.
Since the initial concentration of formic acid is 0.60 M and the amount that ionizes is x, the concentration of HCHO2 remaining is (0.60 - x) and the concentration of H3O+ and CHO2- formed is x.
So, by the equilibrium expression,
Ka = [H3O+][CHO2-]/[HCHO2] = x^2/(0.60 - x)
At equilibrium, the concentration of formic acid remaining is (0.60 - x), which can be assumed to be ≈ 0.60 M (since x will be small).
So, Ka ≈ x^2/0.60, but x = [H3O+] = 1.12 × 10^(-2) M.
Hence, Ka = (1.12 × 10^(-2))^2/0.60 ≈ 1.8 × 10^(-4). 
Therefore, the correct answer is choice B.

The PH of a 0