Question

Determine the [H3O⁺] in a 0.365 M HClO solution. The Ka of HClO is 2.9 × 10^-8.

Answer 1

The concentration of hydronium ion, [H₃O⁺] =  1.3 × 10⁻⁶ M

Explanation:

HClO is a weak acid and it can be ionized and given by the equation as,

HClO + H₂O ⇄  H₃O⁺ + ClO⁻

Ka is given by the expression that the product of the concentration of products divided by the product of the concentration of the reactants.

$Ka = \frac{[H_{3} O^{+}]\times [ClO^{-} ]}{[HClO]}$  = 2.9 ×10⁻⁸

Consider [H₃O⁺] = [ClO⁻] = x

[HClO] = 0.365 - x

Since Ka is very small, consider x <<<0.365, also it should be ignored.

Ka = 2.9 ×10⁻⁸ = $\frac{x^{2} }{0.365}$

x² = 2.9 × 10⁻⁸ × 0.365

x = 1.3 × 10⁻⁶ M

Thus we can conclude 3rd Option as correct answer