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)

Respuesta :

Answer:

The value of the [tex]K_a[/tex] of the acid is :

[tex]K_a=4.109\times 10^{-8}[/tex]

Explanation:

The pH of the weak acid solution = 3.879

Concentration of hydrogen ions = [tex][H^+][/tex]

[tex]pH=-\log[H^+][/tex]

[tex]3.879=-\log[H^+][/tex]

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

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

[tex]HClO(aq)\rightarrow H^+(aq)+ClO^-(aq)[/tex]

Initially:

0.426 M     0       0

At equilibrium

(0.426-0.0001321)M             0.0001321 M               0.0001321 M

The expression of [tex]K_a[/tex] will be given as;

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

[tex]K_a=\frac{0.0001321 M\times 0.0001321 M}{(0.426-0.0001321) M}[/tex]

[tex]K_a=4.109\times 10^{-8}[/tex]

The value of the [tex]K_a[/tex] of the acid is :

[tex]K_a=4.109\times 10^{-8}[/tex]