Question

The pH of 0.5 m aqueous solution of HF ( Ka=2×10-⁴) is

Answer 1

Answer: pH = 2

Explanation:

$HF\rightleftharpoons H^++F^-$

I     0.5             0      0

E  $0.5-0.5\alpha$  $0.5\alpha$ $0.5\alpha$

$K_a=\frac{[H^+][F^-]}{[HF]}$

$2\times 10^{-4}=\frac{[0.5\alpha][0.5\alpha]}{[0.5-0.5\alpha]}$

$\alpha=2\times 10^{-2}$

$[H^+]=0.5\alpha=0.5\times 2\times 10^{-2}=10^{-2}$

$pH=-log[H^+]$

$pH=-log[10^{-2}]=2$

Answer 2

Explanation:

Since, it is given that initial concentration of HF is 0.5 M. The reaction equation will be as follows.

                                  $HF \rightleftharpoons H^{+} + F^{-}$

Initial concentration : 0.5       0         0

Equilibrium concentration : 0.5 - 0.5x      0.5x          0.5x

                 

                 $K_a = \frac{[H^+][F^-]}{[HF]}$

         $2\times 10^{-4} = \frac{[0.5x][0.5x]}{[0.5-0.5x]}$

                        x = $2\times 10^{-2}$

As, $[H^+]$ = 0.5x. Therefore, it will be equal to:

              $0.5 \times 2 \times 10^{-2} = 10^{-2}$

 As it is known that pH = $-log[H^+]$

                                pH = $-log[10^{-2}]$

                                       = 2

Hence, we can conclude that pH of given 0.5 m aqueous solution of HF is 2.