Question

Consider the titration of 100.0 mL of 0.100 M ammonia (NH3) with 0.100 M HCl. Calculate the pH of the solution after:

a. 0.0 mL of HCl has been added

b. 50.0 mL of HCl has been added

c. 100.0 mL of HCl has been added

d. 150.0 mL of HCl has been added

Answer 1

a) 0.00 mL addition of HCl :

pOH = 1/2 (pKb - log C)

      = 1/2 (4.74 - log 0.100)

pOH = 2.87

pH + pOH = 14

pH = 11.13

b) 50.0 addtion of 3 mL HCl

millimoles of NH3 = 100 x 0.1 = 10

millimoles of HCl = 50 x 0.1 = 5

this is half - equivalence point : here pOH = pKb

pOH = 4.74

pH = 9.26

c) after addtion of 100 mL HCl

mmoles of HCl = 10

this is equivalence point : here salt only remains.

salt concentration = 10 / 100 + 100 = 0.05 M

pH = 7 - 1/2 (pKb + log C)

     = 7 - 1/2 (4.74 + log 0.05)

pH = 5.28

d) after addition of 150 mL HCl

mmoles of HCl = 15

[H+] = 15 - 10 / 100 + 150 = 0.02 M

pH = 1.70