Question

A monoprotic acid in a 0.1M solution ionises to 0.001%. lts ionisation constant is

Answer 1

HA = H+     + A-

% Ionization = [H+] / HA *100

0.001 = [H+] / 0.1*100

[H+]= 1.0^10^-6

ionisation constant = [H+] [A-]/[HA]  

= 1.0^10^-6 *1.0^10^-6 /0.1

= 1.0*10^-11

Answer 2

Answer: Ionization constant is $1\times 10^{-11}$

Explanation: For a monoprotic acid, the reaction follows:

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

at $t=0$           0.1        0       0

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

[HA] = 0.1M

$\alpha =0.001%=1\times 10^{-5}$

Ionization Constant for the above reaction is:

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

Putting the above values, we get

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

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

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

As $\alpha << 0.05\%\text{, therefore } (1-\alpha )\approx 1$ and can be neglected.

Hence, $K_a=0.1\times (10^{-5})^2\approx 1\times 10^{-11}$