Question

(50 mL of 0.02000 M KOH Titrated with 0.1000M HBr) Calculate pH after adding the following volumes of the titrant:

(a) 3 mL

(b) 10 mL

(c) 10.5 mL

Answers: (a) pH =12.12; (b) pH = 7.00; (c) pH = 3.0. I need a step-by-step explanation on how to do this.

Answer 1

KOH + HBr ----------------> KBr + H2O

millimoles of KOH = 50 x 0.02 = 1.0

millimoles of HBr added = 3 x 0.1 = 0.3

1 - 0.3 = 0.7 millimoles KOH left

[KOH] = 0.7 / 53 = 0.013 M as KOH is strong base

pOH = - log [OH^-]

pOH = - log [0.013]

pOH = 1.88

pH = 14 - 1.88

pH = 12.12

b) millimoles of HBr = 10 x 0.1 = 1.0

means all KOH reacted with HBr

at equivalence point pH = 7.0

as it is strong base and strong acid titration

pH = 7.0

c) millimoles of HBr = 10.5 x 0.1 = 1.05

1.05 - 1 = 0.05 millimoles of HBr left

total volume = 50 + 10.5 = 60.5

[HBr] = 0.05 / 60.5 = 0.000826 M

pH = -log [H⁺]

pH = - log[0.000826]

pH = 3.0