Question

For the titration of 70.0 mL of 0.200 M NH3 with 0.500 M HCl at 25 °C, determine the relative pH at each of these points.
(a) before the addition of any HCl
(b) after 28.0 mL of HCl has been added
(c) after 38.0 mL of HCl has been added

Answer 1

millimoles of NH3 = 70 x 0.2 = 14.0

pKb of NH3 = 4.74

a) before HCl added only NH3 present

for weak bases

pOH = 1/2[pKb - log C]

pOH =1/2 [4.74 - log 0.2]

pOH = 2.72

pH = 14 - 2.72

pH = 11.28

b) millimoles of HCl added = 28 x 0.5 = 14

means at equivalence

[salt] = 14 / 28+70 = 0.143 M

pOH = 1/2 [pKw + pKb + log C

pOH = 1/2[14 + 4.74 + log 0.143]

pOH = 8.95

pH = 14 - 8.95

pH = 5.05

c) millimoles of HCl added = 38 x 0.5 = 19

19 - 14 = 5.0 millimoles HCl left

[HCl] = 5 / 38 + 70 = 0.0463 M

pH = - log [H⁺] = - log [0.0463]

pH = 1.33