Question

pH calculation upon dilution of a strong acid solution is generally done by equating $n_{H^{+}}$ in original solution and diluted solution. However, if strong acid solution is very dilute, then $H^{+}$ from water are also to be considered. Take log3.7 = 0.568 and answer the following questions.

A 1 litre solution of pH = 4 (solution of strong acid) is added to the $\frac{7}{3}$ litre of water. What is the pH of resulting solution?

• A. 4.52
• B. 4.365
• C. 4.4
• D. 4.432

Answer 1

The correct option is A

4.52

Initial pH = 4
$\left[H^{+}\right]={10}^{-4}\Rightarrow n_{H^{+}}={10}^{-4}{V}_{new}=1+\frac{7}{3}=\frac{10}{3}\Rightarrow [H^{+}{]}_{new}=\frac{{10}^{-4}\times 3}{10}=3\times {10}^{-5}$
$\left[H^{+}\right]=3\times {10}^{-5},$ so pH = 5-log3 = 5 - 0.48 = 4.52