Question

A monoprotic acid in 1.00m solution of 0.01% ionised. What is dissociation. Constant

Answer 1

Answer: The dissociation constant for the given monoprotic acid is $10^{-8}$

Explanation: Dissociation of a monoprotic acid 'HX' is given by the equation:

                        $HX\rightleftharpoons H^++X^-$

At $t=0$              c          0        0

At $t=t_{eq}$          $c-c\alpha$    $c\alpha$       $c\alpha$

Dissociation constant, $k_a$ is given by:

$k_a=\frac{[H^+][X^-]}{[HX]}$

$k_a=\frac{(c\alpha)(c\alpha)}{c-c\alpha}=\frac{c\alpha^2}{(1-\alpha)}$

Here, c = 1M

$\alpha=0.01\%=0.001=10^{-4}$

As $\alpha<<1;(1-\alpha)=1$

Putting values in above equation, we get:

$k_a=\frac{1\times (10^{-4})^2}{1}\\\\k_a=10^{-8}$