Question

The pH of a 0.25 M aqueous solution of hydrofluoric acid, HF, at 25.0 °C is 2.03. What is the value of Ka for HF?

Answer 1

Last option that is none of above is right answer.

Answer 2

Answer : The correct option is, $3.5\times 10^{-4}$

Solution : Given,

pH = 2.03

Concentration of HF = 0.25 M

First we have to calculate the concentration of $H^+$ ion.

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

$2.03=-\log [H^+]$

$[H^+]=9.3\times 10^{-3}M$

Now we have to calculate the value of $K_a$ for HF.

The equilibrium reaction will be

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

Concentration of $H^+$ = Concentration of $F^-$ = $9.3\times 10^{-3}M$

The expression for $K_a$ for HF will be,

$K_a=\frac{[H^+][F^-]}{[HF]}=\frac{(9.3\times 10^{-3})\times (9.3\times 10^{-3})}{0.25}=3.5\times 10^{-4}$

Therefore, the value of $K_a$ for HF is, $3.5\times 10^{-4}$