Question

Find the change in pH value of buffer solution when $50\text{ml}\text{of}\frac{M}{10}$ $HCl$ is added to a buffer containing $100\text{ml}\text{of}\frac{M}{10}C{H}_{3}COOH\text{and}100\text{ml}\text{of}\frac{M}{10}C{H}_{3}COONa$
$(p{K}_a=4.74,log3=0.4771)$

• A. pH increases by 0.47
• B. pH decreases by 0.47
• C. pH remains constant
• D. pH increases by 0.47

Answer 1

The correct option is B pH decreases by 0.47

$\text{milliequivalent of}C{H}_{3}COOH=100\text{ml}\times \frac{M}{10}=10$
$\text{milliequivalent of}C{H}_{3}COONa=100\text{ml}\times \frac{M}{10}=10$
$\text{milliequivalent of}HCl=50\text{ml}\times \frac{M}{10}=5$

$pH=pKa+log\frac{\left[\text{Salt}\right]}{\left[\text{Acid}\right]}$

$\text{Initial}pH=4.74+log\frac{\frac{1}{10}}{\frac{1}{10}}$
$pH=4.74+log1$
$pH=4.74$

After addition of strong acid
$\left[H^{+}\right]$ of weak acid $\uparrow$ increases and
[Salt] decreases.

$\left[H^{+}\right]$ will be
$\Rightarrow \frac{1}{10}\times 100+\frac{1}{10}\times 50=M\times 150$
$\Rightarrow \frac{15}{150}=M=\frac{1}{10}$
[Salt] will be
$\Rightarrow \frac{1}{10}\times 100-\frac{1}{10}\times 50=M\times 150$
$\Rightarrow \frac{5}{150}=M=0.1=\frac{1}{30}$
$\text{Final}pH=4.74+log\frac{\frac{1}{30}}{\frac{1}{10}}$
$\text{Final}pH=4.74+log\frac{1}{3}$
$\text{Final}pH=4.74+log1-log3$
$\text{Final}pH=4.26$
Hence, it is concluded that the final pH of buffer will be decreased by 0.47. Therefore, the correct answer is option (b).