Methanoic (formic) acid has $p K_{a} = 3.75$ at 298 K. For a 0.10 M aqueous solution, calculate the percentage of the acid that is ionised, using the exact quadratic expression (do not use the approximation $1 - α \approx 1$ ). Give the answer to the nearest integer.

Numerical answer type, answer in %, tolerance 0.5.

Answer: 4 %

K_{a} = 1 0^{- 3.75} = 1.778 e - 04

. For

HA H X^{+} + A X^{-}

with

x

ionised,

x^{2} / (0.10 - x) = K_{a}

, giving

x = [- K_{a} + K_{a}^{2} + 4 K_{a} c] /2 = 4.129 e - 03 M

. Percentage ionised

= x /0.10 \times 100 = 4.13%

, i.e. about 4%.