Question

What is the ratio of [a–]/[ha] at ph 2.75? the pka of formic acid (methanoic acid, h–cooh) is 3.75?

Answer 1

The dissociation of formic acid is:

$HCOOH \rightleftharpoons HCOO^{-} + H^{+}$

The acid dissociation constant of formic acid, $k_a$ is:

$k_a = \frac{[HCOO^{-}] [H^{+}]}{HCOOH}$

Rearranging the equation:

$\frac{[HCOO^{-}]}{[HCOOH]} = \frac{k_a}{[H_+]}$

pH = 2.75

$pH = -log[H^{+}]$

$[H^{+}]= 10^{-2.75} = 1.78 \times 10^{-3}$

$pk_a = 3.75$

$k_a = 10^{-3.75} = 1.78\times 10^{-4}$

Substituting the values in the equation:

$\frac{[HCOO^{-}]}{[HCOOH]} = \frac{k_a}{[H_+]}$

$\frac{[HCOO^{-}]}{[HCOOH]} = \frac{1.78\times 10^{-4}}{1.78\times 10^{-3}}$

Hence, the ratio is $\frac{1}{10}$.