Question

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

Answer 1

4.0406

Step-by-step explanation:

Given that:

`K_(a)=2.9* 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* 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