Chemistry 112

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

 

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