Question

The pH of 0.10 M solution of an acid is 6. What is the acid dissociation constant of the acid?

Answer 1

Answer: Dissociation constant of the acid is $10^{-11}$.

Explanation: Assuming the acid to be monoprotic, the reaction follows:

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

pH of the solution = 6

and we know that

$pH=-log([H^+])$

$[H^+]=antilog(-pH)$

$[H^+]=antilog(-6)=10^{-6}M$

As HA ionizes into its ions in 1 : 1 ratio, hence

$[H^+]=[A^-]=10^{-6}M$

As the reaction proceeds, the concentration of acid decreases as it ionizes into its ions, hence the decreases concentration of acid at equilibrium will be:

$[HA]=[HA]-[H^+]$

$[HA]=0.1M-10^{-6}M$

$[HA]=0.09999M$

Dissociation Constant of acid, $K_a$ is given as:

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

Putting values of $[H^+],[A^-]\text{ and }[HA]$ in the above equation, we get

$K_a=\frac{(10^{-6}M).(10^{-6}M)}{0.09999M}$

$K_a=1.0001\times 10^{-11}M$

Rounding it of to one significant figure, we get

$K_a=1.0\times 10^{-11}M\approx 10^{-11}M$