Question

A compound has a pKa of 7.4. To 1 L of a 0.1 M solution of this compound at pH 8.0 is added 10 mL of 3.0 M hydrochloric acid. What is the pH of the resulting solution?

Answer 1

This problem can be solved by using the following Henderson-Hasselbalch equation.
pH = pKa + log([A^-]/[HA])                (this is for the solution before adding HCl)

where [A^-] = concentration of salt; [HA] = concentration of acid;

8.0 = 7.4 + log([A^-]/[HA]);

after solving this,    [A^-]/[HA] = 4.0

Molarity = moles x (1 / volume in Liters)

So, moles of the compound = 0.1 mol/L x 1 L = 0.1 moles

That means,
[A^-] + [HA] = 0.1
We derived that [A^-]/[HA] = 4.0

combinig above two, we will get
A^- = 0.08 moles
HA = 0.02 moles
Adding strong acid reduces A^- and increases HA by the same amount.
Number of moles of HCl added = Molarity of HCl x volume of HCl = (3 mol/L) x (0.01 L) = 0.03 moles HCl are added.

Since moles of HCl added will be equal to moles of A^- reduced, and HA incresed,

Moles of A- remained = 0.08 - 0.03 = 0.05 moles

Total moles of HA after addition of HCl = 0.02 + 0.03 = 0.05moles

That means, after HCl addition, [A-]/[HA] = 0.05 / 0.05 = 1.0
Resubstituting into the Henderson-Hasselbalch equation:
pH = 7.4 + log(1.0)

= 7.4, the final pH.