Question

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 $p{K}_a$ of formic acid $=3.74$) is:

• A. $8.747$
• B. $8.219$
• C. $4.743$
• D. $6.066$

Answer 1

Answer: (B)

$8.219$

At the equivalence point, the salt sodium formate is present. It is a salt of a weak acid and strong base. Its pH is given by the expression:

$pH=7+0.5pKa+0.5logC$

Where $C$ is the concentration of salt. Formic acid is monobasic acid.
The titration of $25$ ml of $0.10$ M formic acid will require $25$ ml $0.1$ M NaOH solution.

Total volume will be $25+25=50$ ml.

The salt concentration $C=0.10\times \frac{25}{25+25}=0.05$ M.

$pH=7+0.5pKa+0.5logC$

$pH=7+0.5\times 3.74+0.5log\times 0.05$

$pH=7+1.87-0.65=8.219$