Question

What is the approximate pH of a 0.06 M solution of CH3COOH (given that Ka = 1.78 x 10-5)? 1 3 6 7 9

Answer 1

The answer would come out as 3

Answer 2

Answer : The pH of the solution is, 3

Solution :  Given,

Concentration (C) = 0.06 M

Acid dissociation constant = $k_a=1.78\times 10^{-5}$

The equilibrium reaction for dissociation of $CH_3COOH$ (weak acid) is,

                           $CH_3COOH+H_2O\rightleftharpoons CH_3COO^-+H_3O^+$

initially conc.         c                                   0                  0

At eqm.              $c(1-\alpha)$                             $c\alpha$               $c\alpha$

First we have to calculate the concentration of $[H^+]$

As, we know that

$\alpha=\sqrt{\frac{k_a}{c}}$    (for weak electrolyte)   ...........(1)

where, $\alpha$ is degree of dissociation

$[H^+]=c\alpha$       ..................(2)

By equation both the equations (1) and (2), we get

$[H^+]=\sqrt{k_a\times c}$

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

$[H^+]=\sqrt{1.78\times 10^{-5}\times 0.06}=1.033\times 10^{-3}M$

Now we have to calculate the pH.

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

$pH=-\log (1.033\times 10^{-3})$

$pH=2.8=3$

Therefore, the pH of the solution is, 3