General Chemistry 2

Question

The pH of an aqueous solution of 0.1890 M carbonic acid, H₂CO₃ (aq), is

Answer 1

H₂CO₃ ==> H⁺ + HCO₃^- ... dissociation of carbonic acid. Ka = 4.3x10^-7 (Wikipedia)

Ka = 4.3x10^-7 = [H⁺][HCO₃^-] / [H₂CO₃] = (x)(x) / 0.1890 - x

If we assume x is small relative to 0.1890 M, we can ignore it in the denominator.

4.3x10^-7 = x² / 0.1890

x² = 8.127x10^-8

x = 2.85x10^-4 (this is small relative to 0.1890 so our above assumption was valid)

This is the [H⁺]

pH = -log [H⁺]

pH = -log 2.85x10^-4

pH = 3.55