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

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.

 = 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 =

x² = 2.9 × 10⁻⁸ × 0.365

x = 1.3 × 10⁻⁶ M

Thus we can conclude 3rd Option as correct answer

not clear

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