We interrogate the equilibrium:
NH3(aq)+H2O(l)⇌NH+4+−OH,
for which Kb=1.8×10−5=[NH+4][HO−][NH3(aq)]
Initially, [NH3]=0.157⋅mol⋅L−1, and we assume some ammonia accepts a proton from water to give stoichiometric NH+4 and HO−. If we let the amount of protonation be x, then we write:
Kb=1.8×10−5=x20.157−x
This is a quadratic in x, which we could solve exactly, but we make that approximation that 0.157−x≅0.157, and thus........
x1≅√1.8×10−5×0.157=1.68×10−3
This is indeed small compared to 0.157, but we could plug it in again, now that we have an approx. for x.
x2=1.67×10−3, and since the values have converged, I am willing to accept this value for our answer.
Now x=[NH+4]=[HO−] by our definition. What is the pH of this solution?
I hope this agrees with your model answer............If you want something re-explained or rephrased, ax, and someone will help you.
