Question

Consider the reaction of NH3 with water:
NH₃ + H2O ↔ NH₄⁺+ OH^-
(Kb=Kw/Ka)

and also
NH₄+ H2O ↔ NH₃ + H3O⁺ (Ka = 5.70×10-10)

Including the ICE Tables:

a) Find the pH of 50.00 mL of 0.200M NH3 after the addition of 45.00 mL of 0.200M HCl.

b) Find the pH of 50.00 mL of 0.200M NH3 after the addition of 50.00 mL of 0.200M HCl.

c) Find the pH of 50.00 mL of 0.200M NH3 after the addition of 55.00 mL of 0.200M HCl.  

Answer 1

a)

mmoles of NH3 = 50 x 0.2 = 10

mmoles of HCl = 45 x 0.2 = 9.0

NH3    +   HCl    -------------->   NH4Cl

10              9.0                            0

-9               -9                             +9

1                0                              9

pH = pKa + log [conjugate base / acid]

   = 9.24 + log [1 / 9]

pH = 8.29

b)

mmoles of NH3 = 50 x 0.2 = 10

mmoles of HCl = 45 x 0.2 = 10

NH3    +   HCl    -------------->   NH4Cl

10              10                           0

-10               -10                           +10

0                0                              10

this is equivalence point . here salt only remains.

salt concentration = 10 / 50 + 50 = 0.1 M

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

    = 7 - 1/2 (4.76 + log 0.1)

pH = 5.12

c) Find the pH of 50.00 mL of 0.200M NH3 after the addition of 55.00 mL of 0.200M HCl.  

mmoles of NH3 = 50 x 0.2 = 10

mmoles of HCl = 55 x 0.2 = 11

NH3    +   HCl    -------------->   NH4Cl

10             11                            0

-10               -10                            +10

0                 1                           10

here strong acid .remains.

[H+] = 1 / (50 + 55) = 9.52 x 10^-3 M

pH = -log (9.5 x 10^-3)

pH = 2.02