Question

Calculate the pH of a solution prepared by mixing$\mathbf{500}\mathbf{.}\mathbf{0}\mathbf{mL}\mathbf{of}\mathbf{0}\mathbf{.}\mathbf{50}\mathbf{M}{\mathbf{Na}}_3{\mathbf{PO}}_4\mathbf{and}\mathbf{500}\mathbf{.}\mathbf{0}\mathbf{mL}\mathbf{of}\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{M}{\mathbf{H}}_2{\mathbf{SO}}_4$

Short answer

The pH of a solution prepared by mixing ${\text{500.0\hspace{0.17em}mL\hspace{0.17em}of\hspace{0.17em}0.50\hspace{0.17em}M\hspace{0.17em}Na}}_{\text{3}}{\text{PO}}_{\text{4}}{\text{\hspace{0.17em}and\hspace{0.17em}500.0\hspace{0.17em}mL\hspace{0.17em}of\hspace{0.17em}0.10M\hspace{0.17em}H}}_{\text{2}}{\text{SO}}_{\text{4}}$is$12.34$

Step-by-step solution

Definition of pH

A scale used to specify the acidity or basicity of an aqueous solution is defined as $\mathbf{pH}$

Calculation of pH

The mmol of ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$is $\text{Volume\hspace{0.17em}in\hspace{0.17em}mL\hspace{0.17em}×\hspace{0.17em}Molarity}$

${\text{Number\hspace{0.17em}of\hspace{0.17em}mmol\hspace{0.17em}of\hspace{0.17em}Na}}_{\text{3}}{\text{PO}}_{\text{4}}=250\text{\hspace{0.17em}mmol}$

The mmol of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$is $\text{Volume\hspace{0.17em}in\hspace{0.17em}mL\hspace{0.17em}×\hspace{0.17em}Molarity}$

${\text{Number\hspace{0.17em}of\hspace{0.17em}mmol\hspace{0.17em}of\hspace{0.17em}H}}_{\text{2}}{\text{SO}}_{\text{4}}=50\text{\hspace{0.17em}mmol}$

Total volume of the solution is $1000\text{mL}$

The expression for equilibrium base constant is$\frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{a}}}=\frac{\left[{\mathrm{OH}}^{-}\right]\left[{\mathrm{HPO}}_4^{2-}\right]}{\left[{\mathrm{PO}}_4^{3-}\right]}$

The quadratic equation is solved as

${\mathrm{x}}^2+0.12\mathrm{x}-3.1\times {10}^{-3}=0$

$\mathrm{x}=2.2\times {10}^{-2}$

$\left[{\mathrm{OH}}^{-}\right]=2.2\times {10}^{-2}\mathrm{M}$

$\begin{array}{l}\mathrm{pOH}=-\log \left[{\mathrm{OH}}^{-}\right]\\ \mathrm{pOH}=-\log (2.2\times {10}^{-2}\mathrm{M})\\ \mathrm{pOH}=1.66\end{array}$

Since

$\begin{array}{c}\mathrm{pH}+\mathrm{pOH}=14\\ \mathrm{pH}+1.66=14\\ \mathrm{pH}=14-1.66\\ \mathrm{pH}=12.34\end{array}$

The $\text{pH}$of the solution is$12.34$