Question

What is the pH of a 0.286 M solution of hypochlorous acid (Ka 2.9 x 10-8)?

Answer 1

Answer:

4.0406

Explanation:

Given that:

$K_{a}=2.9\times 10^{-8}$

Concentration = 0.286 M

Consider the ICE take for the dissociation of acetic acid as:

                                      HClO    ⇄     H⁺ +        ClO⁻

At t=0                            0.286                 -              -

At t =equilibrium        (0.286-x)                x           x            

The expression for dissociation constant of acetic acid is:

$K_{a}=\frac {\left [ H^{+} \right ]\left [ {ClO}^- \right ]}{[HClO]}$

$2.9\times 10^{-8}=\frac {x^2}{0.286-x}$

x is very small, so (0.286 - x) ≅ 0.286

Solving for x, we get:

x = 9.1071×10⁻⁵  M

pH = -log[H⁺] = -log(9.1071×10⁻⁵) = 4.0406