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?

pH = 2.96

Explanation:

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

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

the answer is true.

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