Question

Calculate the pH of 50.00 mL of a solution in which the analytical concentration of HClO4 is 0.1000 M and the following volumes of 0.2000 M KOH are added:

               a.            0.00 mL KOH

               b.            15.00 mL KOH

               c.            25.00 mL KOH

               d.            45.00 mL KOH

               e.            50.00 mL KOH

Answer 1

millimoles of HClO4 = 50 x 0.1 = 5.0

a) pH = - log [H⁺]

pH = - log [0.1]

pH = 1.0

b) millimoles of KOH added = 15 x 0.2 = 3

5 - 3 = 2 millimoles acid left

[HCl4] = [H⁺] = 3 / 65 =0.046 M

pH = - log [H⁺]

pH = - log [0.046]

pH = 1.34

c) millimoles of KOH added = 25 x 0.2 = 5

5 - 5 = 0

at equivalence point

pH = 7.0

d) millimoles of KOH added = 45 x 0.2 = 9

9 - 5 = 4 millimoles KOH left

[KOH] = 4 / 95 = 0.042 M

pOH = - log [OH^-]

pOH = - log [0.042]

pOH = 1.38

pH = 14 - 1.38

pH = 12.62

e ) millimoles of KOH = 50 x 0.2 = 10

10 - 5 = 5 millimoles KOH left

[KOH] = 5 / 100 = 0.05 M

pOH = - log [OH^-]

pOH = - log [0.05]

pOH = 1.30

pH = 14 - 1.30

pH = 12.7