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3 September, 23:33

You are asked to prepare a pH=4.00 buffer starting from 1.50 L of 0.0200 M solution of benzoic acid (C6H5COOH) and an excess of sodium benzoate (C6H5COONa). You may want to reference (Pages 721 - 728) Section 17.2 while completing this problem. Part A What is the pH of the benzoic acid solution prior to adding sodium benzoate?

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  1. 4 September, 02:37
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    pH = 2.96

    Explanation:

    C6H5COOH ↔ C6H5COO - + H3O+

    ∴ Ka = 6.5 E-5 = [H3O+][C6H5COO-] / [C6H5COOH]

    C6H5COONa → C6H5COO - + Na+

    prior to adding C6H5COONa:

    ⇒ C C6H5COONa = 0 M = [Na+]

    ∴ mass balance:

    ⇒ C C6H5COOH = [C6H5COOH] + [C6H5COO-] = 0.0200 M

    ⇒ [C6H5COOH] = 0.0200 - [C6H5COO-]

    charge balance:

    ⇒ [H3O+] = [C6H5COO-]

    ⇒ [C6H5COOH] = 0.0200 - [H3O+]

    ⇒ Ka = [H3O+]² / (0.0200 - [H3O+]) = 6.5 E-5

    ⇒ [H3O+]² + 6.5 E-5[H3O+] - 1.3 E-6 = 0

    ⇒ [H3O+] = 1.108 E-3 M

    ∴ pH = - Log [H3O+]

    ⇒ pH = 2.9554 = 2.96
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