If you have 1.50 mol/L of NH3 how do you find the pH?
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Am I to understand that this is a question from your actual final exam that you seeking help to complete? Or are you in preparation for your final? I hope it is the latter.
Ammonia reacts with water to produce a small amount of NH4+ and OH-. The Kb for this equilibrium is 1.8x10^-5. Knowing the concentration of ammonia, you can then determine the concentration of OH-. Knowing that Kw = [H+][OH-], you can compute the H+ ion concentration, and from that the pH, according to pH = -log[H+].
NH3 + H2O <==> NH4+ + OH- ........ Kb = 1.8x10^-5
Kb = [NH4+] [OH-] / [NH3]
1.8x10^-5 = x² / (1.50 - x) ....... we assume that (1.5-x) = 1.5 since x is small
x = 0.0052
[OH-] = 0.0052M
Kw = [H+] [OH-] ..... at 25C
[H+] = Kw / [OH-]
[H+] = 1.00x10^-14 / 0.0052
[H+] = 1.92x10^-12
pH = -log[H+]
pH = -log(1.92x10^-12)
pH = 11.72
Ammonia reacts with water to produce a small amount of NH4+ and OH-. The Kb for this equilibrium is 1.8x10^-5. Knowing the concentration of ammonia, you can then determine the concentration of OH-. Knowing that Kw = [H+][OH-], you can compute the H+ ion concentration, and from that the pH, according to pH = -log[H+].
NH3 + H2O <==> NH4+ + OH- ........ Kb = 1.8x10^-5
Kb = [NH4+] [OH-] / [NH3]
1.8x10^-5 = x² / (1.50 - x) ....... we assume that (1.5-x) = 1.5 since x is small
x = 0.0052
[OH-] = 0.0052M
Kw = [H+] [OH-] ..... at 25C
[H+] = Kw / [OH-]
[H+] = 1.00x10^-14 / 0.0052
[H+] = 1.92x10^-12
pH = -log[H+]
pH = -log(1.92x10^-12)
pH = 11.72