Question

1. Calculate the pH of the solution formed when 10.0 mL of 0.200 M NaOH is added to:

A) 5.0 mL of 0.100 M of benzoic acid, C6H5COOH

B) 20.0 mL of 0.100 M of benzoic acid, C6H5COOH

C) 30.0 mL of 0.100 M of benzoic acid, C6H5COOH

Answer 1

mmol of NaOH = MV = 10*0.2 = 2

a)

mmol of acid = 5*0.1 = 0.5

mmol of base = 2-0.5 = 1.5

Vtotal = 5+10 = 15

[OH-] = mmol/V = 1.5/15 = 0.01

pH =14 + log(OH) = 14 + log(0.01) = 12

b)

mmol of acid = 0.1*20 = 2

mmol of base = 2

then, equivalence point

benzoate ion forms equilbirium

B- + H2O <-> HB + OH-

Kb = [HB][OH]/[B-]

[B-] = 2/(10+20) = 0.06666

(10^-14)/(10^-4.2) = x*x/(0.06666-x)

x = 3.25*10^-6

OH =  3.25*10^-6

pH = 14 + log(OH) = 14 + log( 3.25*10^-6 = 8.51

c)

mmol of acid = MV = 0.1*30 = 3

mmol of base = 2

mmol of acid left = 3-2 = 1

mmol of conjguate formed = 2

this is a buffer so

pH = pKa +log (conjugate/acid)

pH = 4.2 + log(2/1)

pH = 4.5010