Chapter 22, Problem 22.106QP

A 10.0-g sample of white phosphorus was burned in an excess of oxygen. The product was dissolved in enough water to make 500 mL of solution. Calculate the pH of the solution at 25°C.

Interpretation Introduction

Interpretation:

The $\text{pH}$ of the solution formed obtained by burning white phosphorous in oxygen and dissolving in water has to be calculated.

Concept introduction:

pH: It is the logarithm of the reciprocal of the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ in a solution.

Answer to Problem 22.106QP

The $\text{pH}$ of the given solution is 1.18.

Explanation of Solution

To calculate: The $\text{pH}$ of the given solution

White phosphorus burn in air and produces phosphorus pentoxide.  Phosphorus pentoxide reacts with water and produces phosphoric acid.

$\begin{array}{l}{\text{P}}_{\text{4(s)}}+{\text{5O}}_{\text{2(g)}}\to {\text{P}}_{\text{4}}{\text{O}}_{\text{10(s)}}\\ {\text{P}}_{\text{4}}{\text{O}}_{\text{10(s)}}+{\text{6H}}_{\text{2}}{\text{O}}_{\text{(l)}}\to {\text{4H}}_{\text{3}}{\text{PO}}_{\text{4(aq)}}\end{array}$

Calculate the molarity of phosphoric acid

$\begin{array}{l}\text{10}\text{g}{\text{P}}_{\text{4}}\text{×}\frac{\text{1}\text{mol}{\text{P}}_{\text{4}}}{\text{123}\text{.9}\text{g}{\text{P}}_{\text{4}}}\text{×}\frac{\text{1}\text{mol}{\text{P}}_{\text{4}}{\text{O}}_{\text{10}}}{\text{1}\text{mol}{\text{P}}_{\text{4}}}\text{×}\frac{\text{4}\text{mol}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}}{\text{1}\text{mol}{\text{P}}_{\text{4}}{\text{O}}_{\text{10}}}\\ \text{=}\text{0}\text{.323}\text{mol}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\\ \text{Molarity}\text{=}\frac{\text{0}\text{.323}\text{mol}}{\text{0}\text{.500}\text{L}}\\ \text{=}\text{0}\text{.646}\text{M}\end{array}$

The molarity of formed phosphoric acid was calculated by the use of given weight and volume data.

Calculate the $\text{pH}$ of the given solution

$\begin{array}{l}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}{}_{\left(\text{aq}\right)}{\text{+ H}}_{\text{2}}{\text{O}}_{\left(\text{aq}\right)}\to {\text{ H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{ + H}}_{\text{2}}{\text{PO}}_{\text{4}}{}^{\text{-}}\\ \text{Initial concentration}\left(\text{M}\right)\text{: 0}\text{.646}\text{0 0}\\ \text{Change}\text{in}\text{concentration }\left(\text{M}\right)\text{: -x }\text{+x +x}\\ \text{Equilibrium}\text{concentration}\left(\text{M}\right)\text{: 0}\text{.646-x }\text{x x}\end{array}$

$\begin{array}{l}{\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][H}}_{\text{2}}{\text{PO}}_{\text{4}}{}^{\text{-}}\text{]}}{{\text{[H}}_{\text{3}}{\text{PO}}_{\text{4}}\text{]}}\\ \text{7}\text{.5}\text{×}{\text{10}}^{\text{-3}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{(0}\text{.646}\text{-}\text{x)}}\\ \text{=}{\text{x}}^{\text{2}}\text{ + 7}\text{.5×}{\text{10}}^{\text{-3}}\text{x}\text{-}\text{4}\text{.85×}{\text{10}}^{\text{-3}}\text{=}\text{0}\end{array}$

Solve the above quadratic equation,

$\text{x}\text{=}\text{0}\text{.066}\text{M}\text{=}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$

The $\text{pH}$ of the given solution

$\begin{array}{l}\text{pH}\text{= }\text{-log}\left(\text{0}\text{.066}\right)\\ \text{=}\text{1}\text{.18}\end{array}$

The $\text{pH}$ of the given solution was calculated by the uses of molarity and acid dissociation constant of phosphoric acid.

Conclusion

The $\text{pH}$ of the given solution was calculated.