What is the expected pH of a 0.806M solution of formic acid in water?

Question

What is the expected pH of a 0.806M solution of formic acid in water? Ka(HCO2H) = 1.798x10^-4

Answer 1

HCO₂H <==> H⁺ + HCO₂^-

Ka = [H⁺][HCO₂^-] / [HCO₂H]

1.798x10^-4 = (x)(x) / 0.806 - x and if we assume x is small relative to 0.806 we can ignore in denominator

1.798x10^-4 = x² / 0.806

x² = 1.449x10^-4

x = 1.204x10^-2 M = [H⁺] and this is only ~ 1.5% of 0.806 so above assumption was valid, i.e. <5%

pH = -log [H⁺]

pH = -log 1.204x10^-2

pH = 1.92