What is the hydroxide ion concentration and pH of 0.250M NH3 (aq)? Kb = 1.8×10^-5
For NH3
NH3 + H2O <==> NH4+ + OH-
with x amount of reaction
Kb = [NH4+][OH-]/[NH3]
1.8 x 10^-5 = x^2/0.250
x = [OH-] = 2.12 x 10^-3 M
pOH = -log[OH-] = 2.67
pH = -log[H+] = 11.33
What is the hydroxide ion concentration and pH of 0.250M NH3 (aq)? Kb = 1.8×10^-5
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