Chapter 25, Problem 25.101QP

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 25.101QP

Answer

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

Explanation of Solution

Explanation

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.