Question

Calculate the pH at the equivalence point in the titration of 25 mL of $0.10M$ formic acid with a $0.1MNaOH$ solution (given that $p{K}_a$ of formic acid = 3.74).
$log0.1=-1$

• A. 4.74
• B. 8.74
• C. 8.37
• D. 6.06

Answer 1

The correct option is C 8.37

Given, $p{K}_a\left(HCOONa\right)=3.74,\left[HCOONa\right]=0.10M,\left[NaOH\right]=0.1$
At the equivalence point, 0.10 M of $HCOONa$ (Sodium formate) is formed.
We know the relation between $pH$, concentration, $p{K}_w$, and $p{K}_a$ for salt of weak acid and strong base is given by
$pH=\frac{1}{2}p{K}_w+\frac{1}{2}p{K}_a+\frac{1}{2}\log C=\frac{1}{2}\times 14+\frac{1}{2}\times 3.74+\frac{1}{2}\times \log 0.1$
$=7+\frac{1}{2}\times 3.74+\frac{1}{2}\times \log 0.1$
$=7+1.87-0.5$
$=8.37$