Chapter 22, Problem 22.56P

(a)

Interpretation Introduction

Interpretation:

A balanced chemical equation between tetraphosphorus decoxide and water has to be written.

Concept Introduction:

Balanced Chemical equation:

A balanced chemical equation is an equation which contains same elements in same number on both the sides (reactant and product side) of the chemical equation thereby obeying the law of conservation of mass.

Explanation of Solution

Water is incorporated in ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$ (drying agent) to form phosphoric acid.  The balanced chemical equation tetraphosphorus decoxide and water is given by

    ${\text{P}}_{\text{4}}{\text{O}}_{\text{10(s)}}{\text{ + 6H}}_{\text{2}}{\text{O}}_{\text{(l) }}\underset{}{\overset{}{\to }}{\text{4H}}_{\text{3}}{\text{PO}}_{\text{4}}{}_{\text{(l)}}$

(b)

Interpretation Introduction

Interpretation:

The pH of the solution formed from the addition of ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$ to water of $\text{0}\text{.750L}$ has to be calculated.

Concept Introduction:

• $\text{pH}$ is the logarithm of the reciprocal of the concentration  of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$  in a solution.
• $\text{pH}$ is used to determine the acidity or basicity of an aqueous solution.
• $\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$
• $\text{pH}\text{+}\text{pOH}\text{=}\text{14}$

Explanation of Solution

Water is incorporated in ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$ (drying agent) to form phosphoric acid.  The balanced chemical equation tetraphosphorus decoxide and water is given by

    ${\text{P}}_{\text{4}}{\text{O}}_{\text{10(s)}}{\text{ + 6H}}_{\text{2}}{\text{O}}_{\text{(l) }}\underset{}{\overset{}{\to }}{\text{4H}}_{\text{3}}{\text{PO}}_{\text{4}}{}_{\text{(l)}}$

The molarity of phosphoric acid is determined as,

    $\begin{array}{l}{\text{Moles of H}}_{\text{3}}{\text{PO}}_{\text{4}}\text{ =}\left(\text{8}\text{.5g}{\text{P}}_{\text{4}}{\text{O}}_{\text{10}}\right)\left(\frac{\text{1mol}{\text{P}}_{\text{4}}{\text{O}}_{\text{10}}}{\text{283}\text{.88g}{\text{P}}_{\text{4}}{\text{O}}_{\text{10}}}\right)\left(\frac{\text{4mol}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}}{\text{1mol}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}}\right)\\ \\ \text{=}\text{0}\text{.11976892}\text{mol}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\\ \\ {\text{Molarity of H}}_{\text{3}}{\text{PO}}_{\text{4}}\text{ =}\left(\frac{\text{0}\text{.11976892}\text{mol}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}}{\text{0}\text{.750L}}\right)\text{=}\text{0}{\text{.1596919 M H}}_{\text{3}}{\text{PO}}_{\text{4}}\end{array}$

Phosphoric acid is a weak acid and only partly dissociates to form ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ in water, based on its ${\text{k}}_{\text{a}}$.  Since ${\text{K}}_{\text{a1}}{\text{ >> K}}_{\text{a2}}{\text{ >> K}}_{\text{a3}}$, hydronium ion concentration of second and third dissociation is less than first.

    $\begin{array}{l}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4(l)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l)}}\underset{}{\overset{}{\to }}{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{PO}}_{\text{4}}{{}^{\text{-}}}_{\text{(aq)}}\\ \\ {\text{K}}_{\text{a}}\text{ = 7}{\text{.2x10}}^{\text{-3}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][H}}_{\text{2}}{\text{PO}}_{\text{4}}{}^{\text{-}}\text{]}}{{\text{[H}}_{\text{3}}{\text{PO}}_{\text{4}}\text{]}}\\ \\ \text{7}{\text{.2x10}}^{\text{-3}}\text{=}\frac{\text{[x][x]}}{\text{[0}\text{.1596919-x]}}\text{Assume x is small compared to 0}\text{.1596919}\text{.}\\ \\ \text{x}\text{=}\text{0}\text{.0339084 M}\\ \\ \text{Check assumption that x is small compared to 0}\text{.1596919:}\\ \\ \frac{\text{0}\text{.0339084}}{\text{0}\text{.1596919}}\text{×}\text{100}\text{=}\text{21\%}\text{error,}\text{so}\text{the}\text{assumption}\text{is}\text{invalid}\\ \\ \text{The problem will need to be solved as a quadratic}\text{.}\\ \\ {\text{x}}^{\text{2}}\text{ = (7}{\text{.2x10}}^{\text{-3}}\text{)(0}\text{.1596919 -x)}\text{=}\text{1}\text{.14978}\text{×}{\text{10}}^{\text{-3}}\text{-}\text{7}\text{.2}\text{×}{\text{10}}^{\text{-3}}\text{x}\\ \\ {\text{x}}^{\text{2}}\text{+}\text{7}\text{.2}\text{×}{\text{10}}^{\text{-3}}\text{x}\text{-}\text{1}\text{.14978}\text{×}{\text{10}}^{\text{-3}}\text{=}\text{0}\\ \\ \text{x =}\frac{\text{-7}\text{.2}\text{×}{\text{10}}^{\text{-3}}\text{±}\sqrt[]{{\text{(7}\text{.2}\text{×}{\text{10}}^{\text{-3}}\text{)}}^{\text{2}}\text{-4(1)(-1}\text{.14978}\text{×}{\text{10}}^{\text{-3}}\text{)}}}{\text{2(1)}}\\ \\ \text{x = 3}{\text{.049897x10}}^{\text{-2}}\text{M}{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\end{array}$

The pH of the solution is calculated as,

    ${\text{pH = -log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\text{=}\text{-log}\text{(3}\text{.049897}\text{×}{\text{10}}^{\text{-2}}\text{)}\text{=}\text{1}\text{.52}$