Given a 0.100 M solution of HClO (with Ka = 2.9 × 10^{-8}), what is the concentration of H_3O^+ in the solution?

• A. 0.100 M
• B. 5.4 × 10^{-5} M
• C. 1.7 × 10^{-4} M
• D. 2.9 × 10^{-8} M

Answer: (C)

Verified step by step guidance

1. Write the dissociation equilibrium for hypochlorous acid (HClO): \[\mathrm{HClO} + \mathrm{H_2O} \rightleftharpoons \mathrm{H_3O^+} + \mathrm{ClO^-}\]
2. Set up an ICE table (Initial, Change, Equilibrium) for the concentrations, where the initial concentration of HClO is 0.100 M, and initial concentrations of \( \mathrm{H_3O^+} \) and \( \mathrm{ClO^-} \) are approximately zero.
3. Express the equilibrium concentrations in terms of \( x \), the concentration of \( \mathrm{H_3O^+} \) formed: - \( [\mathrm{HClO}]_{eq} = 0.100 - x \) - \( [\mathrm{H_3O^+}]_{eq} = x \) - \( [\mathrm{ClO^-}]_{eq} = x \)
4. Write the expression for the acid dissociation constant \( K_a \): \[K_a = \frac{[\mathrm{H_3O^+}][\mathrm{ClO^-}]}{[\mathrm{HClO}]} = \frac{x^2}{0.100 - x}\] Since \( K_a \) is very small, assume \( 0.100 - x \approx 0.100 \) to simplify the calculation.
5. Solve for \( x \) using the simplified equation: \[x = \sqrt{K_a \times 0.100}\] This \( x \) represents the concentration of \( \mathrm{H_3O^+} \) in the solution.