Question

$100$ mL of $0.02M$ benzoic acid ($pKa=4.20$) is titrated using $0.02M$ $NaOH$. $pH$ after $50$ mL and $100$ mL of $NaOH$ have been added are:

• A. $3.50,7$
• B. $4.2,7$
• C. $4.2,8.1$
• D. $4.2,8.25$

Answer 1

Answer: (B)

$4.2,7$

Given $100ml$ of $0.02M$ $C_6{H}_{5}COOH$ with $p{K}_a=4.2$

So, $K_a={10}^{-4.2}=6.31\times {10}^{-5}$

It is titrated using $0.02M$ $NaOH$.

$\left(i\right)$ After addition of $50ml$ of $NaOH$:

Number of moles of $C_6{H}_{5}COOH=100\times 0.02=2mmol$

Number of moles of $NaOH=50\times 0.02=1mmol$

Here after the addition of $50ml$ of $NaOH$,we get basic Buffer solution $C_6{H}_{5}COONa$ and $C_6{H}_{5}COOH$.

Number of moles of $C_6{H}_{5}COOH=2-1=1mmol$

Number of moles of $C_6{H}_{5}COONa=1mmol$

$pH=p{K}_a+log\frac{\text{salt}}{\text{acid}}$ $=4.2+log\frac{1}{1}$

$pH=4.2$

$\left(ii\right)$ After addition of $100ml$ of $NaOH$:

Number of moles of $C_6{H}_{5}COOH=100\times 0.02=2mmol$

Number of moles of $NaOH=100\times 0.02=2mmol$

So, Number of moles of $C_6{H}_{5}COONa=2mmol$

Therefore the given point is equivalent point. The salt $C_6{H}_{5}COONa$ is weak base, so we will find basic ionization constant.

Now,

$K_b=\frac{K_w}{K_a}=\frac{{10}^{-14}}{6.31\times {10}^{-5}}=1.58\times {10}^{-10}$

Concentration of $\left[C_6{H}_{5}COONa\right]=\frac{Numberofmolesof{C}_6{H}_{5}COONa}{Totalvolumeofthesolution}=\frac{2}{200}=0.01$

$\left[O{H}^{-}\right]=\sqrt{K_b\times \left[C_6{H}_{5}COONa\right]}=\sqrt{(1.58\times {10}^{-10})\times 0.01}=1.257\times {10}^{-7}M$

$pOH=-\log \left[O{H}^{-}\right]=-\log (1.257\times {10}^{-7})=6.9$

$pH=14-pOH=14-6.9\approx 7.0$