Question

Calculate the volume of $\mathbf{1}\mathbf{.50}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathit{M}\mathit{N}\mathit{a}\mathit{O}\mathit{H}$that must be added to 500.0 mL of$\mathbf{0}\mathbf{.200}\mathit{M}\mathit{H}\mathit{C}\mathit{l}$to give a solution that has$\mathit{p}\mathit{H}\mathbf{=}\mathbf{2}\mathbf{.15}$.

Short answer

The NaOH volume needed is around 4.36 L.

Step-by-step solution

The value of hydrogen ion concentration [H+]. 

The reaction for the neutralization is expressed below,

$NaOH\left(aq\right)+HCl\left(aq\right)\to NaCl\left(s\right)+H_{2}O\left(l\right)$

As,

$\begin{array}{c}pH=-\log \left[H^{+}\right]\\ 2.15=-\log \left[H^{+}\right]\end{array}$

Hence,

$\begin{array}{c}\left[H^{+}\right]={10}^{-pH}\\ ={10}^{-2.15}\\ =0.0071M\end{array}$

The volume of NaOH. 

The molarity of HCl can be denoted as,

$\begin{array}{c}MolarityHCl=\frac{MolesofHCl}{Volumeofsolution}\\ 0.0071M=\frac{MolesofHCl}{0.5L+VolumeofNaOH}\end{array}$

Then,

$\text{MolesHCl=InitialmolesHCl–MolesofNaOHaddstogetpH2.15}$

$\begin{array}{c}Molarity\left(HCl\right)=0.5L\times 0.2mol/L-VolumeofNaOH\times 0.015mol/L\\ 0.0071M=\frac{0.1moles-VolumeofNaOH\times 0.015mol/L}{0.5L+VolumeofNaOH}\end{array}$

Hence,

$VolumeofNaOH=4.36L$