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1 April, 22:17

What is the Ka of a 0.653 m solution of hydrocyanic acid with a ph of 5.47

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  1. 1 April, 22:24
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    1.758 x 10⁻¹¹.

    Explanation:

    ∵ pH = - log[H⁺].

    ∴ 5.47 = - log[H⁺].

    log[H⁺] = - 5.47.

    ∴ [H⁺] = 3.388 x 10⁻⁶ M.

    ∵ [H⁺] = √ (Ka. C)

    ∴ [H⁺]² = Ka. C

    ∴ Ka = [H⁺]²/C = (3.388 x 10⁻⁶) ² (0.653) = 1.758 x 10⁻¹¹.
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