Question

Chlorous acid (HClO2) is used in bleaching textiles, leather, paper, and fibers. The value of Ka for chlorous acid is 1.2 ✕ 10−2. Calculate the pH of a 0.12 M solution of chlorous acid.

Answer 1

Answer:

1.49.

Explanation:

pH is used to test for the acidity or the alkalinity of content or substance. The pH is represented mathematically by using the formula below;

pH = - log [H^+].

In the question above, we are given the value for ka which is = 1.2 ✕ 10^-2. The initial concentration of Chlorous acid (HClO2) = 0.12 M.

So, we need to know the concentration of hydrogen ion in order to calculate the pH of acid.

The equation for the dissociation of the acid is given below;

HClO2 <============> H^+ + ClO2^-.

At time,t = t, the concentration of Chlorous acid (HClO2) = 0.12 - x and the concentration of H^+ and ClO2^- is x respectively.

So, ka = [H^+][ClO2^-] / [HClO2].

1.2 ✕ 10^-2 = x^2/ 0.12 - x.

x^2 = 1.44 × 10^- 3 x - 1.2 ✕ 10^-2x.

x^2 + 1.2 ✕ 10^-2x - 1.44 × 10^- 3= 0.

Since the value of X can not be negative we will have our x= 0.03242 = [H^+].

Therefore, pH = -log[H+].

pH = -log[0.03242].

pH = 1.49.

Answer 2

Answer:

1.36

Explanation:

HClO2  ⇄  H+  +  ClO2-

           [HClO2]          [H+}         [ClO2-]

initial:  0.12       0         0

change: -x        +x       +x

equil:   0.12-x    x          x

[H+][ClO2-] / [HClO2] = Ka = 1.2 * 10^-2

x* x / 0.12 - x = 1.2 * 10^-2

x^2 = (2.4  *  10^-3) - (1.2 * 10^-2)x

x^2 +  (1.2 * 10^-2)x – (2.4 * 10^-3) = 0

Upon solving for x,

x = 0.04335

[H+] = 0.04335

pH = -log[H+]

pH = -log(0.04335)  = - (-1.36) = 1.36