What is the pH?

Question

What is the pH of a solution prepared by mixing 50.00 mL of 0.10 M NH₃ with 10.00 mL of 0.10 M NH₄Cl? Assume that the volume of the solutions are additive and that K_b = 1.8 × 10^-5 for NH₃.

Answer 1

This is a buffer solution made up of a weak base, NH3 and its weak conjugate acid salt NH4Cl

The easiest solution knowing this is a buffer solution is to use the Henderson - Hasselbalch equation:

pH = pKa (-log of the Ka) + log of base/acid

To get the Ka value since your given the Kb as NH3 is a base you have 2 methods

1) you could find the pKb (-log Kb) once the pKb is found using this relationship pKb + pKa=14 and solve for the pKa and use the Henderson - Hasselbalch equation.

2) Or another method that can be used to find the Ka is: (Ka)(Kb) = 1 x 10-14 and solve for Ka and of course use the Henderson-Hasselbalch equation.

Both methods give you the Ka that you need.

I'll use relationship number (2) and solving for the Ka you get 5.55 x 10-10 and then calculate the pKa which we will do after this next step.

Next step is to convert the Molarity of both NH3, and NH4Cl to moles as the volumes are different for NH3 and NH4Cl

It's fine to use Molarity in the Henderson-Hasselbalch equation provided the volumes of both species are the same and using Molarity or moles in that situation would give you the same correct answer.

moles of NH3 = 0.05 L X 0.1 mol/L = 0.005 moles

moles of NH4Cl = 0.01 L X 0.1 mol/L = 0.001 moles

Now using the Henderson-Hasselbalch equation

pH = -log 5.55 x 10-10 + log 0.005/0.001

pH = 9.256 + 0.699 = 9.955

Just as an aside, if the acid species is greater than the base the pH would be less than the pKa

if the base species is greater than the acid the pH would be greater than the pKa as it is in this case

Finally if the base and the acid are equal the pH would equal the pka.