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
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.
