Answer to Question #15976 in Chemistry for Marchelle

Question

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.

Answer 1

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

(b) 0.

(c) [OH^-] = 0.004 - 0.002 = 0.002 in 50 ml = 0.04

pOH = -log [OH^-] = 1.4

pH = 14 - pOH = 14 - 1.4 = 12.6