Question

Calculate $pH$ change when $0.01$ mol $C{H}_{3}KCOONa$ solution is added to $1L$ of $0.01MC{H}_{3}COOH$ solution.

$K_a\left(C{H}_{3}COOH\right)=1.8\times {10}^{-5},p{K}_a=4.74$

• A. $3.37$
• B. $1.37$
• C. $4.74$
• D. $8.01$

Answer 1

The correct option is B $1.37$

$pH$ of $0.01MC{H}_{3}COOH$

$pH=\frac{1}{2}[p{K}_a-\log C]$

$=\frac{1}{2}(4.74-\log 0.01)$

$=\frac{1}{2}(4.74+2)$

$=3.37$

When $0.01molC{H}_{3}COONa$ is added to it, it is now a buffer and $\left[C{H}_{3}COONa\right]=0.01M$.

Now, for buffer solution,

$pH=p{K}_a+\log \frac{\left[C{H}_{3}COO\right]}{\left[C{H}_{3}COOH\right]}$

$=4.74+\log \frac{0.01}{0.01}=4.74$

$\therefore$ Change in $pH=4.74-3.37=1.37$