Calculate the pH of a solution obtained by mixing 50ml of` 0.2`M HCl with `49.9` mL of `0.2`m NaOH solution.

Answer

3.699

Explanation

49.9 ml of 0.2 M NaOH will neutralise 49.9 ml of 0.2 M HCl. Hence, HCl left unneutralised = 0.1 ml of 0.2 M . Volume after mixing = 99.9 ml `~~` 100 ml. Hence , applying `N_(1)V_(1)=N_(2)V_(2), 0.1xx0.2=N_(2)xx100 or N_(2)=2xx10^(-4)`. In the final solution, `[HCl]=2xx10^(-4)M`, i.e., `[H_(3)O^(+)]=2xx10^(-4)M`.
Hence, `pH = - log (2xx10^(-4))=4-0.301=3.699`