Question

A monoprotic acid in a 0.1m solution ionises to 0.001%. its ionisation constant is

Answer 1

Answer: $K_a=0.1\times 10^{-10}$

Explanation:

Equation for a monoprotic acid is:

                       $HA\rightleftharpoons H^++A^-$

at $t=0$            c        0        0

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

Ionization constant, $K_a=\frac{[H^+][A^-]}{[HA]}$

$K_a=\frac{(c\alpha)(c\alpha)}{c(1-\alpha)}$             ......(1)

$\alpha=0.001\%$

As $\alpha<0.05\%$ so, $1-\alpha \approx 1$

$K_a=\frac{c\alpha^2}{1}$

c = 0.1M

Putting values in equation 1, we get

$K_a=\frac{(0.1)(10^{-5})^2}{1}$

$K_a=0.1\times 10^{-10}$