Chemistry 102: Equilibrium: Calculate the equilibrium concentration

Question

Formic acid, HCOOH, ionizes in water according to the following equation. The equilibrium constant is K = 1.8 × 10^–4.

HCOOH(aq)+H2​O(l)HCOO−(aq)+H3​O+(aq)

Calculate the equilibrium concentration of H₃O⁺ in a 0.985 M solution.   

Answer 1

HCOOH + H2O <==> H3O⁺ + HCOO^-

K = [H3O⁺][HCOO-] / [HCOOH]

1.8x10^-4 = (x)(x) / 0.985 - x (assume x is small relative to 0.985 and ignore it in the denominator)

1.8x10^-4 = x² / 0.985

x² = 1.773x10-4

x = .0133 M (this is only ~1.3% of 0.985 so the above assumption was valid)

[H₃O⁺] = 0.0133 M