Question

In an aqueous solution of a certain acid the acid is 0.10% dissociated and the pH is 4.06. Calculate the acid dissociation constant Ka of the acid. Round your answer to 2 significant digits

Answer 1

Answer : The acid dissociation constant Ka of the acid is, $8.7\times 10^{-5}$

Explanation :

First we have to calculate the concentration of hydrogen ion.

$pH=-\log [H^+]$

Given: pH = 4.06

$4.06=-\log [H^+]$

$[H^+]=8.71\times 10^{-5}M$

The dissociation of acid reaction is:

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

Initial conc.      c        0         0

At eqm.           c-cα    cα       cα

Given:

Degree of dissociation = α = 0.10 % = 0.001

$[H^+]=c\alpha$

$8.71\times 10^{-5}=c\times 0.001$

$c=0.0871M$

The expression of dissociation constant of acid is:

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

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

Now put all the given values in this expression, we get:

$K_a=\frac{(0.0871\times 0.001)\times (0.0871\times 0.001)}{(0.0871-0.0871\times 0.001)}$

$K_a=8.7\times 10^{-5}$

Thus, the acid dissociation constant Ka of the acid is, $8.7\times 10^{-5}$