Consider the titration of a 60.0 mL of 0.365 M weak acid HA (Ka = 4.2 x 10⁻⁶) with 0.400 M KOH.

Question

After 30.0 mL of KOH have been added, what would the pH of the solution be?

Answer 1

The addition of a strong base (KOH) to a weak acid (HA) creates a buffer containing the weak acid plus the conjugate base (or salt) of that acid (A-). Thus, HA + KOH ==> KA + H2O.

So, we need to find the moles of HA left over after the addition of KOH and also the moles of A- formed from the addition of KOH.

Initial moles HA = 60.0 ml x 1 L/1000 ml x 0.365 mol/L = 0.0219 moles

Moles KOH added = 30.0 ml x 1 L/1000 ml x 0.400 mol/L = 0.012 moles

HA + KOH ===> KA + H2O

0.0219...0.012...........0...............Initial

-0.012...-0.012........+0.012.........Change

0.0099....0..............0.012...........Equilibrium

We can use the Henderson Hasselbalch equation to solve for pH:

pH = pKa + log [A^-]/[HA] and pKa = -log Ka = -log 4.2x10^-6 = 5.38

pH = 5.38 + log (0.012/0.0099)

pH = 5.38 + 0.084

pH = 5.46