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 $p{K}_a$ of formic acid = 3.74, log 5 = 0.7)

• A. 4.74
• B. 7.57
• C. 8.22
• D. 6.06

Answer 1

The correct option is C 8.22

At the equivalence point,
$N_1{V}_1=N_2{V}_2$
Since here n-factor is 1 for both the acid and the base,
$\therefore M_1{V}_1=M_2{V}_2$
Let, the volume of base be $V_2$
Then, $0.1\times 25=0.1\times V_2$
Hence, volume of $NaOH$ is also 25 mL
Hence, the concentration of salt = $\frac{\text{milli moles of the salt}}{\text{total volume of the salt}}=\frac{25\times 0.1}{50}=0.05M$
$pH=\frac{1}{2}p{K}_w+\frac{1}{2}p{K}_a+\frac{1}{2}\log C=\frac{1}{2}\times 14+\frac{1}{2}\times 3.74+\frac{1}{2}\log 0.05=8.22$