Question

Consider the titration of 50.00 mL of 0.2000 M NaOH with 0.4000 M HBr. Calculate the pH of the titration solution after the addition of the following volumes of HBr titrant:

A) 0.00 mL

B) 10.0 mL

C) 25.0 mL

D) 35.0 mL

Answer 1

mmol of base = MV = 50*0.2 = 10 mmol of base

a)

pOH = -log(OH) = -log(0.2) = 0.69897

pH = 14-0.69897= 13.30103

b)

mmol of acid = MV = 0.4*10 = 4 mmol

mmol of base = 10

10-4 = 6 mmol of base left

V = 50+10 = 60

[Oh-] = 6/60 =0.1 M

pOH = -log(0.1) = 1

pH = 14-1 = 13

c)

mmol of acid = MV = 0.4*25= 10 mmol

mmol of base = 10

0 mmol of base left

pH = 7

D) 35.0 mL

mmol acif = MV = 0.4*35.0 = 14

mmol of acid left = 14-10 = 4 mmol

pH = -log(H) = -log(4/(50+35)) = 1.3273