Question

Calculate the pH at the equivalence point in the titration of 25 ml of 0.10 M formic acid with a 0.1 M NaOH solution. (Given that pKa of formic acid = 3.74).

Answer 1

At the equivalence point, 0.10 M of HCOONa (sodium formate) is formed. This problem is solved by considering the concept of hydrolysis equilibria
$pH=\frac{1}{2}pKw+\frac{1}{2}pKa+\frac{1}{2}logC$
$=\frac{1}{2}\times 14+\frac{1}{2}\times 3.74+\frac{1}{2}log0.1=8.37$