Question

A titration is performed by adding 0.300 M KOH to 30.0 mL of 0.400 M HCl.

a) Calculate the pH before addition of any KOH.

b) Calculate the pH after the addition of 10.0, 25.0 and 39.5 mL of the base.

c) Calculate the volume of base needed to reach the equivalence point.

d) Calculate the pH at the equivalence point.

e) Calculate the pH after adding 5.00 mL of KOH past the endpoint.

f) Sketch the titration curve based on the calculated points above.

Answer 1

NOTE: KOH is strong base and HCl is strong acid, there will be NO hydrolysis

a)

pH before any addition of KOH

[HCl] = [H+] = 0.4 M

pH = -log([H+] = -log(0.4) = 0.3979400

bi)

mmol of HCl = Macid*Vacid = 30*0.4 = 12 mmol of HCl

after V = 10 mL of KOH

mmol of KOH = Mbase*Vbase = 10*0.3 = 3 mmol

mmol of HCl left = 12-3 = 9 mmol

V total = 10+30 = 40 mL

[H+] = mmol of H+ / Vtotal = 9/40 = 0.225

pH = -log(H+) = -log(0.225) = 0.64781

bii)

after V = 25 mL of KOH

mmol of KOH = Mbase*Vbase = 25*0.3 = 7.5 mmol

mmol of HCl left = 12-7.5 = 4.5 mmol

V total = 25+30 = 55 mL

[H+] = mmol of H+ / Vtotal = 4.5/55 = 0.08181

pH = -log(H+) = -log(0.08181) = 1.087

biii)

after V = 39.5 mL of KOH

mmol of KOH = Mbase*Vbase = 39.5*0.3 = 11.85 mmol

mmol of HCl left = 12-11.85 = 0.15 mmol

V total = 39.5+30 = 69.5 mL

[H+] = mmol of H+ / Vtotal = 0.15/69.5 = 0.002158

pH = -log(H+) = -log(0.002158) = 2.665

d)

volume of base required

ratio is 1:1 mol of HCl : mol of KOH

mmol of HCl = 12 mmol

then we need 12 mmol of KOH

mmol of KOH = 12

V = mmol of KOH / Mbase = 12/(0.3) =40 mL of base required

d)

pH in equivalence point must be pH = 7 since this is a neutral base

e)

after 5mL of KOH

mmol of HCl = 12

mmol of KOH = 0.3*(40+5) = 13.5 mmol

mmol of OH- left = 13.5 - 12 = 1.5 mmol

V total = 40+5+30 = 75 mL

[OH-] = mmol /V = 1.5/75 = 0.02 M

pOH = -log(0.02) = 1.6989

pH = 14-pOH = 14-1.6989

pH = 12.3011