What is the [OH-] of a buffer that consists of 0.630 M NH3 and 0.969 M NH4Cl? pKb of ammonia is 4.75?

Question

What is the [OH-] of a buffer that consists of 0.630 M NH3 and 0.969 M NH4Cl? pKb of ammonia is 4.75?

Answer 1

First lets write the equation:

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

Next lets convert the pKb into Kb

pKb = 4.75

10^-4.75 = Kb

1.778*10^-5 = Kb

[NH3] = .63M [NH4+] = .969M

Use the equation above to write the equilibrium equilibrium expression.

Kb = ([NH4+][OH-])/[NH3]

1.778*10^-5 = ((.969M)[OH-])/(.63M)

[OH-] = 1.16*10^-5M