Question

In the laboratory, a general chemistry student measured the pH of a 0.426 M aqueous solution of hypochlorous acid to be 3.897. Use the information she obtained to determine the Ka for this acid. Ka(experiment)

Answer 1

Answer:

The value of the $K_a$ of the acid is :

$K_a=4.109\times 10^{-8}$

Explanation:

The pH of the weak acid solution = 3.879

Concentration of hydrogen ions = $[H^+]$

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

$3.879=-\log[H^+]$

$[H^+]=10^{-3.876}=0.0001321 M$

Concentration of hypochlorous acid solution = [HClO]= 0.426 M

$HClO(aq)\rightarrow H^+(aq)+ClO^-(aq)$

Initially:

0.426 M     0       0

At equilibrium

(0.426-0.0001321)M             0.0001321 M               0.0001321 M

The expression of $K_a$ will be given as;

$K_a=\frac{[H^+][ClO^-]}{[HClO]}$

$K_a=\frac{0.0001321 M\times 0.0001321 M}{(0.426-0.0001321) M}$

$K_a=4.109\times 10^{-8}$

The value of the $K_a$ of the acid is :

$K_a=4.109\times 10^{-8}$