Question

A 20.0 mL sample of 0.200 MHBr solution is titrated with 0.200 MNaOH solution. Calculate the pH of the solution after the following volumes of base have been added.

A) 15.0 mL

B) 19.8 mL

C) 20.1 mL

D) 35.0 mL

Answer 1

mmol of acid = MV = 20*0.2 = 4 mmol of acid

a)

mmol of base = MV = 0.2*15= 3

mmol of acid left = 4-3 = 1

VT = 20+15 = 35

[H] = 1/35

pH = -log(0.02857) = 1.544

b)

mmol of base = MV = 0.2*19.8= 3.96

mmol of acid left = 4-3.96 = 0.04

H+ = 0.04/(20+19.8) = 0.001

pH = -og(0.001) = 3

c)

mmol of base = 20.1*0.2 = 4.02

mmol of base left = 4.02-4 = 0.02

then

OH = 0.02/(20.1+20) = 0.0004987

pOH = -log(0.0004987) = 3.302

pH = 14-3.302 = 10.698

ph = 10.698

d)

mmol of OH = MV = 0.2*35 = 7

mmol left of OH= 7-4 = 3

OH= 3/(20+35) = 0.05454

pOH = -log(0.05454)= 1.26

ph = 14-pOH= 12.74