Question

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

Group of answer choices


1.1 × 10^-10 M


7.7 × 10^-9 M


1.3 × 10^-6 M


4.9 × 10^-4 M


8.8 × 10^-5 M

Answer 1

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

Step-by-step 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 = ([H_(3) O^(+)]* [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 =
`$(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