6. A 0.80M solution of a weak base (A-) has a pH of 11.2. What is the Kb for A-? (10pts)
pH = 11.2
pOH = 14 - pH = 14 - 11.2 = 2.8
[OH-] = 10^(-pOH) = 10^(-2.8) = 0.001585 M
A- + H2O -----> HA + OH-
Kb = [OH-] * [HA] / [A-]
Kb = (0.001586) * ( 0.001585) / (0.80 - 0.001585)
Kb = 3.15*10^-6 .....Answer
Let me know if any doubts.
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