What is the hydronium ion concentration of a 0.200 M hypochlorous acid solution with Ka=3.5 x 10-8? The equation for the dissociation of hypochlorous acid is: HOCI(aq) + H2O() = H30*(aq) + OCI"(aq). O 3.7 x x 10-5 M O 8.4 x 10-4 M O 8.4 x 10-5 M O 1.9 x 10-4 M

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**Problem: Hydronium Ion Concentration Calculation**

What is the hydronium ion concentration of a 0.200 M hypochlorous acid solution with \( K_a = 3.5 \times 10^{-8} \)? The equation for the dissociation of hypochlorous acid is:

\[ \text{HOCl(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{H}_3\text{O}^+(\text{aq}) + \text{OCl}^-(\text{aq}) \]

**Options:**
- \( \bigcirc \) \( 3.7 \times 10^{-5} \) M
- \( \bigcirc \) \( 8.4 \times 10^{-4} \) M
- \( \bigcirc \) \( 8.4 \times 10^{-5} \) M
- \( \bigcirc \) \( 1.9 \times 10^{-4} \) M

**Instructions:**
Calculate the hydronium ion concentration using the given \( K_a \) value and the initial concentration of hypochlorous acid. Select the correct answer.
Transcribed Image Text:**Problem: Hydronium Ion Concentration Calculation** What is the hydronium ion concentration of a 0.200 M hypochlorous acid solution with \( K_a = 3.5 \times 10^{-8} \)? The equation for the dissociation of hypochlorous acid is: \[ \text{HOCl(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{H}_3\text{O}^+(\text{aq}) + \text{OCl}^-(\text{aq}) \] **Options:** - \( \bigcirc \) \( 3.7 \times 10^{-5} \) M - \( \bigcirc \) \( 8.4 \times 10^{-4} \) M - \( \bigcirc \) \( 8.4 \times 10^{-5} \) M - \( \bigcirc \) \( 1.9 \times 10^{-4} \) M **Instructions:** Calculate the hydronium ion concentration using the given \( K_a \) value and the initial concentration of hypochlorous acid. Select the correct answer.
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