Question

The pH of $0.1M$ solution of a weak acid $\left(H A\right)$ is 4.50. It is neutralised with certain ammount of $NaOH$ solution to decrease the acid content to half of initial value. Calculate the pH of the resulting solution.

Answer 1

$pH=pKa+log \frac{\left[\right. S a l t \left]\right.}{\left[\right. A c i d \left]\right.}$

$pH=pKa$

when weak acid is half neutralized by strong base.

For weak acid = $\text{pH} \, \text{=} \, \frac{\text{1}}{\text{2}} \left[\text{pKa} \, - \, \text{log} \, \text{C}\right]$

$\text{4} \text{.5} \, \text{=} \, \frac{\text{1}}{\text{2}} \text{[pKa} - \text{log10}^{- \text{1}} \text{]}$

$\text{9} - \text{1} \, \text{=} \, \text{pKa} \Rightarrow \text{pKa} \, \text{=} \, \text{8}$