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8 August, 21:40

Calculate the pH of a solution formed by mixing 200.0 mL of 0.30 M HOCl (hypochlorous acid) with 300.0 mL of 0.20 M NaOCl (sodium hypochlorite). The Ka for HOCl is 2.9 * 10-8.

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  1. 8 August, 21:49
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    The pH of the solution is 7.54

    Explanation:

    This is a buffer because we have a weak acid HOCl with its conjugate base OCl with Ka of 2.9*10⁻⁸ thus pKa = - log Ka = 7.54

    therefore using Henderson Hasselbach equation: pH=pKa+log ([OCl]/[HOCl]) we can determine the pH

    n = mole

    n of HOCl = C*V = 0.3 M * 0.2 L = 0.06 mole

    n of NaOCl = 0.2 M * 0.3 L = 0.06 mole

    Total volume = 300 + 200 = 500 mL = 0.5 L

    final concentration: [HOCl] = 0.06/0.5 = 0.12 M; [OCl] = 0.06/0.5 = 0.12

    pH=7.54+log (0.12/0.12) = 7.54

    pH = pKa because the amount of HOCl = the amount of OCl in the final solution.
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