QuestionJuly 15, 2025

Find the pH of a 0.0153 M solution of hypochlorous acid. (The value of K_(a) for hypochlorous acid is 2.9times 10^-8 Express your answer using two decimal places. pH=square

Find the pH of a 0.0153 M solution of hypochlorous acid. (The value of K_(a) for hypochlorous acid is 2.9times 10^-8 Express your answer using two decimal places. pH=square
Find the pH of a 0.0153 M solution of hypochlorous acid. (The value of K_(a) for hypochlorous acid is 2.9times 10^-8
Express your answer using two decimal places.
pH=square

Solution
4.1(231 votes)

Answer

4.68 Explanation 1. Write the dissociation equation Hypochlorous acid (HOCl) dissociates as HOCl \rightleftharpoons H^+ + OCl^-. 2. Set up the expression for K_a The expression is K_a = \frac H^+][OCl^- HOCl . 3. Assume initial concentrations and changes Initial: [HOCl] = 0.0153 \, M, [H^+] = [OCl^-] = 0. Change: [HOCl] = 0.0153 - x, [H^+] = [OCl^-] = x. 4. Substitute into K_a expression 2.9 \times 10^{-8} = \frac{x^2}{0.0153 - x}. 5. Simplify assumption Assume x \ll 0.0153, so 0.0153 - x \approx 0.0153. Then, 2.9 \times 10^{-8} = \frac{x^2}{0.0153}. 6. Solve for x x^2 = 2.9 \times 10^{-8} \times 0.0153. Calculate x = \sqrt{4.437 \times 10^{-10}} \approx 2.106 \times 10^{-5}. 7. Calculate pH pH = -\log_{10}(2.106 \times 10^{-5}).

Explanation

1. Write the dissociation equation<br /> Hypochlorous acid ($HOCl$) dissociates as $HOCl \rightleftharpoons H^+ + OCl^-$. <br /><br />2. Set up the expression for $K_a$<br /> The expression is $K_a = \frac{[H^+][OCl^-]}{[HOCl]}$.<br /><br />3. Assume initial concentrations and changes<br /> Initial: $[HOCl] = 0.0153 \, M$, $[H^+] = [OCl^-] = 0$. Change: $[HOCl] = 0.0153 - x$, $[H^+] = [OCl^-] = x$.<br /><br />4. Substitute into $K_a$ expression<br /> $2.9 \times 10^{-8} = \frac{x^2}{0.0153 - x}$.<br /><br />5. Simplify assumption<br /> Assume $x \ll 0.0153$, so $0.0153 - x \approx 0.0153$. Then, $2.9 \times 10^{-8} = \frac{x^2}{0.0153}$.<br /><br />6. Solve for $x$<br /> $x^2 = 2.9 \times 10^{-8} \times 0.0153$. Calculate $x = \sqrt{4.437 \times 10^{-10}} \approx 2.106 \times 10^{-5}$.<br /><br />7. Calculate pH<br /> $pH = -\log_{10}(2.106 \times 10^{-5})$.
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