Browse Questions

# Both $HCOOH$ and $CH_3COOH$ solutions have equal $pH$. If $\frac{K_1}{K_2}$ (ratio of acid ionisation constants) of these acids is 4, their molar concentration ratio will be

(a) 2

(b) 0.5

(c) 4

(d) 0.25

As $pH$ of $HCOOH$ solution = $pH$ of $CH_3COOH$ solution

$\therefore [H^+]$ in $HCOOH$ solution = $[H^+]$ in $CH_3COOH$ solution

$K_1 = \frac{[HCOO^-][H^+]}{[HCOOH]} = \frac{[H^+]^2}{[HCOOH]}$
$K_2 = \frac{[CH_3COO^-][H^+]}{[CH^3COOH]} = \frac{[H^+]^2}{[CH_3COOH]}$

$\therefore K_1[HCOOH] = K_2[CH_3COOH]$
$\Rightarrow \frac{[HCOOH]}{CH_3COOH} = \frac{K_2}{K_1} = \frac{1}{4} = 0.25$
edited Mar 25, 2014