Question

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Determine the pH of

(a) a 0.15 M NH3 solution.

(b) a solution that is 0.15 M NH3 and 0.40 M NH4Cl.

Kb for NH3 is 1.8 × 10−5

Answer 1

a)

construct ICE table

   NH3 (Aq) + H2O (l) <---> NH4+ (aq) + OH- (aq)

I    0.15                                 0                  0

C   -x                                    +x                 +x

E 0.15 -x                             +x                  +x

Kb = [NH+] [OH-] / [NH3]

1.8 x 10^-5 = [x] [x]\ / [0.15 -x]

x² + x 1.5 * 10^-5 - 2.7 x 10^-6 = 0

solve the quadratic equation

x = [OH-] = 0.00163 M

pOH = -log[OH-] = -log[0.00163] = 2.78

pH = 14-pOH = 14-2.78 = 11.21

Part B

find the pKb using Kb

pKb = -log(1.8 x 10^-5) = 4.74

pOH = pKb = log[NH4Cl / NH3]

pOH = 4.74 + log[0.4 / 0.15]

pOH = 4.74 + 0.426

pOH = 5.166

pH = 14-pOH = 14 - 5.166 = 8.83