NH3(aq) + H2O(l) → ← NH4+(aq) + OH–(aq)
Kb = [NH4+]e[OH–]e [NH3]e
Ka (NH4+) = 5.6×10–10
Kb (NH3) = Kw/Ka = 1.0×10–14/5.6×10–10 = 1.8×10–5
Initial0.10 0 0
Change–x +x +x
Equilibrium0.10 – x x x
Can we approximate? 0.10/1.8×10–5 = 5500 > 100 Yes!
1.8×10–5 = (x)(x) 0.10
x2 = 1.8×10–6
x = 1.3×10–3
[OH–]e = x = 1.3×10–3 M
pOH = –log[OH–] = –log(1.3×10–3) = 2.89
pH = 14.00 – pOH = 14.00 – 2.89 = 11.11