Question

A 50.0 mL solution of 0.199 M KOH is titrated with 0.398 M HCl. Calculate the pH of the solution after the addition of the following amounts of HCl.

a) 0.00 mL HCl
b) 7.00 mL HCl
c) 12.5 mL HCl
d) 20.0 mL HCl

Answer 1

a) 0.00 mL HCl

here KOH = OH- = 0.199 M

pOH = -log[OH-] = -log (0.199) = 0.70

pH + pOH = 14

pH = 13.30

b) 7.00 mL HCl

millimoles of KOH = 0.199 x 50 = 9.95

millimoles of HCl = 0.398 x 7 = 2.786

base moles dominates to acid moles

so

[OH-] = base millimoles - acid millimoles / total volume

          = (9.95 -2.786) / (50 +7)

         = 0.1257 M

pOH = 0.90

pH = 13.1

c) 12.5 mL HCl

millimoles of HCl = 12.5 x 0.398 = 4.975

[OH-] = base millimoles - acid millimoles / total volume

          = (9.95 -4.975) / (50 +12.5)

         = 0.0796 M

pOH = 1.099

pH = 12.90

d) 20.0 mL HCl

millimoles of HCl = 20 x 0.398 = 7.96

[OH-] = base millimoles - acid millimoles / total volume

          = (9.95 -7.96) / (50 +20)

pOH = 1.546

pH = 12.45