Question

100 mL of 0.02 M benzoic acid $(pKa=4.2)$ is titrated using $0.02MNaOH$. pH after 50 mL and 100 mL of $NaOH$ have been added are

• A. 3.50, 7.0
• B. 4.2, 7.0
• C. 4.2, 8.1
• D. 4.2, 8.25

Answer 1

The correct option is C 4.2, 8.1

$C_6{H}_{5}COOH+O{H}^{-}\to C_6{H}_{5}CO{O}^{-}+H_{2}O$
$\begin{array}{cccccccccccc}t=0& & 2& & & & 1& & & & & \\ t=t_{eq}& & 1& & & & -& & & & & 1\end{array}$

At half equivalence point $pH=p{K}_a=4.2$
$C_6{H}_{5}COOH+O{H}^{-}\to C_6{H}_{5}CO{O}^{-}+H_{2}O$
$\begin{array}{ccccccccccc}t=0& & & 2& & 2& & & & & \\ t_{eq}& & & -& & -& & & & & 2\end{array}$
$\left[C_6{H}_{5}CO{O}^{-}\right]=\frac{2}{200}=0.01M$
$pH=7+p{K}_a+\frac{1}{2}\text{log C}=7+\frac{4.2}{2}+\frac{1}{2}\text{log}\left(0.01\right)$
$=8.1$