Solution:
A) Ionization of NH3 in water:
NH3 + H2O = NH4+ + OH-
B) Kb = [NH4+] [OH-] / [NH3]
1.8 x 10^-5 = [NH4+] [OH-] / [NH3]
C) pH of the buffer solution is calculated by Hendersen equation,
pOH = pkb + log [NH4+] / [NH3+]
pOH = 4.74 + log (0.2 / 0.1)
pOH = 4.74 + log 2
pOH = 4.74 + 0.30 = 5.04
Thus,
[OH-] = 10^-pOH = 10^-5.04
[OH-] = 9.12 x 10^-6
Then,
[H+] = 1 x 10^-14 / [OH-]
= 1 x 10^-14 / 9.12 x 10^-6
[H+]= 1.1 x 10^-9
D) Calculation of OH- and H+:
NH3 + H2O = NH4+ + OH-
0.1 M -------------- 0 -------0 M (initial)
(0.1- X) ----------- X ------- X M (final)
Thus,
Kb = X . X / (0.1 - X)
Kb = X^2 / 0.1
(Since, X <<<< 0.1 because NH3 is a weak base, hence neglected from denominator)
1.8 x 10^-5 = X^2 / 0.1
X^2 = 1.8 x 10^-6
X = 1.34 x 10^-3
Therefore,
[OH-] = X = 1.34 x 10^-3 M
[H+] = 1 x 10^-14 / [OH-]
= 1 x 10^-14 / 1.34 x 10^-3
[H+] = 7.46 x 10^-12 M
pH = - log [H+]
= - log 7.46 x 10^-12
= 12 - log 7.46 = 12 - 0.87
pH = 11.13
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