Question

Determine the pH of a solution that is 0.15 M HClO2 (Ka = 1.1 x 10-2) and 0.15 M HClO (Ka = 2.9 × 10-8).

Answer 1

Answer:

pH = 1.39

Explanation:

Given that:

The molarity of $HClO_2$ = 0.15 M

with acid dissociation constant $K_a$ = $1.1 \times 10^{-2}$

Acid dissociation constant of HClO = $2.9 \times 10^{-8}$

Molarity of HClO = 0.15 M

The objective is to determine the pH of the solution:\

To determine the concentration of $H_3O^+$ obtained from both acids

The equation for the reaction can be expressed as :

$HClO_2 + H_2O \to H_3O^+ + ClO_2^-$

The dissociation constant for the above reaction is as follows:

$ka = \dfrac{ [H_3O^+] [ClO_2^-]} { [HClO_2]}$

$1.1 \times 10^{-2} = \dfrac{ [x] [x]} { [0.15]}$

$1.1 \times 10^{-2} = \dfrac{ [x]^2 } { [0.15]}$

$x^2 = 0.15 \times 1.1 \times 10^{-2}$

$x^2 = 0.00165$

$x=\sqrt{ 0.00165}$

x = 0.04062 M

Now to determine the concentration of  $H3O^+$ obtained from HClO

The equation for the reaction can be expressed as :

$HClO + H_2O \to H_3O^+ + ClO^-$

$ka = \dfrac{[H3O^+] [ClO^-]}{ [HClO]}$

$2.9 \times 10 ^{-8} = \dfrac{[x] [x]}{ [0.15]}$

$2.9 \times 10 ^{-8} \times [0.15] = {[x]^2}$

${[x]} ^2=4.35 \times 10^{-9}$

$x=\sqrt{4.35 \times 10^{-9}}$

$x=6.595 \times 10^{-5}$

Thus the total concentration now is :

x = $H_3O^+$ = $0.04062 + 6.595 \times 10^{-5} M$

$H_3O^+$ = 0.04068595 M

$H_3O^+$  $\simeq$ 0.04069 M

pH = -log [H3O⁺]

pH =  -log [0.04069]

pH = 1.39