1. Benzoic acid, C6H5COOH, is a weak acid with Ka = 6.46 × 10-5. If 40.00mL of 0.100 M benzoic acid were mixed with 10.00mL of 0.200M sodium hydroxide. Bearing in mind the assumptions mentioned in the last part of the introduction, calculate
(a) the concentration of the sodium benzoate formed
(b) the concentration of the unreacted benzoic acid in the resulting solution
(c) the pH of the resulting solution using equation (2) in the introduction.
1
Expert's answer
2012-10-09T09:24:25-0400
n of acid
n------------40.00 0.100-------1000
n = 0.004
10.00 ---------- n 1000 ----------0.200
n = 0.002 mol of NaOH in excess
n of C6H5COONa = n of acid = 0.004 mol
(a) [C6H5COONa] = 0.004 mol in (10 + 40 = 50 ml) = 0.004*1000/50 = 0.08 M
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