Question

A 20.2 mL sample of a weak monoprotic acid, HX, requires 50.0 mL of 0.060 M NaOH to reach the equivalence point. After the addition of 30.0 mL of NaOH, the pH is 5.50. What is the Ka of HX?

Answer 1

millimols HX initially = 50 x 0.06 = 3.00
mmols NaOH added for pH 5.50 = 30.0 x 0.06 = 1.8

............HX + NaOH ==> NaX + H2O
initial....3.00....0.......0......0
add...............1.8...............
change.....-1.8..-1.8.......+1.8...+1.8
equil.......1.2.....0......1.8......1.8

Plug 5.50 for pH and the other info into the Henderson-Hasselbalch equation. Solve for pKa and convert top Ka.

Answer 2

6.5