Question

What volume of $\mathbf{0}\mathbf{.}\mathbf{0100}\mathbf{M}\mathbf{NaOH}$ must be added to$\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{L}$ of $\mathbf{0}\mathbf{.}\mathbf{0500}\mathbf{M}\mathbf{HOCl}$ to achieve a $\mathbf{pH}$of $\mathbf{8}\mathbf{.}\mathbf{00}$?

Short answer

Here,3.90L volume of 0.0100 M NaOH is needed to achieve a pH of 8.00.

Step-by-step solution

Reaction for NaOH and HOCl.

$\mathrm{NaOH}+\mathrm{HOCl}\to \mathrm{NaOCl}+{\mathrm{H}}_2\mathrm{O}$

On the hasselbalch equation

$\begin{array}{l}\mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \frac{\mathrm{salt}}{\mathrm{acid}}\\ 8.0=7.45+\log \frac{\mathrm{NaOCl}}{\mathrm{HOCl}}\\ \frac{\mathrm{NaOCl}}{\mathrm{HOCl}}=3.548\end{array}$

Concentration and Volume of the reaction

Concentration of NaOHis 0.010M.

Concentration of HOCLis 0.050M.

Assume the volume of NaOHas xL.

Volume ofHOCLis 1.0L.

Therefore,

$\begin{array}{l}\frac{0.01\mathrm{x}}{0.05}-0.01\mathrm{x}=3.548\\ \mathrm{x}=3.90\mathrm{L}\end{array}$