Question

A buffer solution is formed by mixing 100 mL 0.01 M $C{H}_{3}COOH$ with 200 mL 0.02 M $C{H}_{3}COONa$. If this buffer solution is made to 1 lit by adding 700 mL of water, $pH$ will change by a factor of:

Answer 1

$\left[C{H}_{3}COOH\right]=\frac{100\times 0.01}{100+200}=\frac{1}{300}$

$\left[C{H}_{3}COONa\right]=\frac{200\times 0.02}{100+200}=\frac{4}{300}$

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

$=p{K}_a+\log \frac{1}{4}$

Now, if the solution is diluted, the concentration of $\left[C{H}_{3}COOH\right]$ & $\left[C{H}_{3}COONa\right]$ will change in such a way that their ratios remain the same. This is because the number of moles are the same & only the total volume (which is equal for both) is increasing.

Hence, $pH$ will remain constant on dilution.