Chapter 5, Problem 5E.1E

(a)

Interpretation Introduction

Interpretation:

The fraction of intact $H_{2}P{O}_3$ molecules in an aqueous solution of phosphorous acid at $pH=5.0$ has to be calculated.

Concept Introduction:

Acid equilibrium constant:

The equilibrium constant $K_a$ for the acid dissociation reaction is,

    $\begin{array}{l}HA\stackrel{}{\to }{H}^{+}+A^{-}\\ K_a=\frac{[H^{-}][A^{-}]}{[HA]}\end{array}$

$\text{pH:}$

The concentration of hydrogen ion in given solution is given as $\text{pH}$.

The negative logarithm of base ten hydrogen ion concentrations is known as $\text{pH}$.

    $\text{pH}\text{=}{\text{-log[H}}^{+}\text{]}$

Explanation of Solution

Given

Phosphorous acid $\text{solution}\text{pH}\text{=}\text{5}\text{.0}$

Phosphorous acid, $H_{2}P{O}_3$

  $\begin{array}{l}{\text{K}}_{a1}\text{=1}\text{.0 }\times {\text{10}}^{-2}p{K}_a=2.00\\ {\text{K}}_{a2}\text{=2}\text{.6}\times {\text{10}}^{-7}p{K}_a=6.59\end{array}$

  $\begin{array}{l}{\text{H}}_2{\text{PO}}_3{}_{\text{(aq)}}+H_{2}O\stackrel{}{\to }{\text{HPO}}_3{{}^{-}}_{\text{(aq)}}+{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}p{K}_a=2.00\\ K_{a1=}\frac{[{\text{HPO}}_3{{}^{-}}_{\text{(aq)}}][{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}]}{{\text{H}}_2{\text{PO}}_3{}_{\text{(aq)}}}=1.0\times {10}^{-2}\\ \\ {\text{HPO}}_3{{}^{-}}_{\text{(aq)}}+H_{2}O\stackrel{}{\to }{\text{PO}}_3{{}^{2-}}_{\text{(aq)}}+{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}p{K}_a=6.59\\ K_{a2=}\frac{[{\text{PO}}_3{{}^{2-}}_{\text{(aq)}}][{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}]}{{\text{HPO}}_3{{}^{-}}_{\text{(aq)}}}=2.6\times {10}^{-7}\end{array}$

From the general rule,

The acid form is dominant for $pH<p{K}_a$

The conjugate pair have equal concentrations at $pH=p{K}_a$

The base form is dominant for $pH>p{K}_a$

At $\text{5}\text{.0}\text{pH}$ concentration of hydrogen ion is,

    $\begin{array}{l}{\text{[H}}_3{\text{O}}^{+}\text{]}\text{=}{10}^{\text{-pH}}\\ ={10}^{-5}\\ =0.00001\end{array}$

At the given pH, equilibrium constant of acid is,

    $\begin{array}{l}{\text{K}}_{\text{a1}}\text{=}\frac{{\text{[HPO}}_{\text{3}}{{}^{\text{-}}}_{\text{(aq)}}{\text{][H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}\text{]}}{{\text{H}}_{\text{2}}{\text{PO}}_{\text{3}}{}_{\text{(aq)}}}\text{=}\text{1}{\text{.0×10}}^{\text{-2}}\\ \\ \text{1}{\text{.0×10}}^{\text{-2}}\text{=}\frac{{\text{[0}\text{.00001]}}^{\text{2}}}{{\text{H}}_{\text{2}}{\text{PO}}_{\text{3}}{}_{\text{(aq)}}}\\ \\ {\text{H}}_{\text{2}}{\text{PO}}_{\text{3}}{}_{\text{(aq)}}\text{=}\frac{{\text{[0}\text{.00001]}}^{\text{2}}}{\text{1}{\text{.0×10}}^{\text{-2}}}\text{=}\frac{{\text{1×10}}^{\text{-10}}}{\text{1}{\text{.0×10}}^{\text{-2}}}\\ \\ \text{=}{\text{1×10}}^{\text{-8}}\text{M}\text{.}\end{array}$

Hence, the fraction of intact $H_{2}P{O}_3$ molecules in an aqueous solution of phosphorous acid at $pH=5.0$ is ${\text{1×10}}^{\text{-8}}\text{M}$.

(b)

Interpretation Introduction

Interpretation:

The fraction of intact $H_{2}P{O}_3$ molecules in an aqueous solution of phosphorous acid at $pH=9.0$ has to be calculated.

Concept Introduction:

Acid equilibrium constant:

The equilibrium constant $K_a$ for the acid dissociation reaction is,

    $\begin{array}{l}HA\stackrel{}{\to }{H}^{+}+A^{-}\\ K_a=\frac{[H^{-}][A^{-}]}{[HA]}\end{array}$

$\text{pH:}$

The concentration of hydrogen ion in given solution is given as $\text{pH}$.

The negative logarithm of base ten hydrogen ion concentration is known as $\text{pH}$.

    $\text{pH}\text{=}{\text{-log[H}}^{+}\text{]}$

Explanation of Solution

Given

Phosphorous acid $\text{solution}\text{pH}\text{=}9.\text{0}$

Phosphorous acid, $H_{2}P{O}_3$

  $\begin{array}{l}{\text{K}}_{a1}\text{=1}\text{.0 }\times {\text{10}}^{-2}p{K}_a=2.00\\ {\text{K}}_{a2}\text{=2}\text{.6}\times {\text{10}}^{-7}p{K}_a=6.59\end{array}$

  $\begin{array}{l}{\text{H}}_2{\text{PO}}_3{}_{\text{(aq)}}+H_{2}O\stackrel{}{\to }{\text{HPO}}_3{{}^{-}}_{\text{(aq)}}+{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}p{K}_a=2.00\\ K_{a1=}\frac{[{\text{HPO}}_3{{}^{-}}_{\text{(aq)}}][{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}]}{{\text{H}}_2{\text{PO}}_3{}_{\text{(aq)}}}=1.0\times {10}^{-2}\\ \\ {\text{HPO}}_3{{}^{-}}_{\text{(aq)}}+H_{2}O\stackrel{}{\to }{\text{PO}}_3{{}^{2-}}_{\text{(aq)}}+{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}p{K}_a=6.59\\ K_{a2=}\frac{[{\text{PO}}_3{{}^{2-}}_{\text{(aq)}}][{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}]}{{\text{HPO}}_3{{}^{-}}_{\text{(aq)}}}=2.6\times {10}^{-7}\end{array}$

From the general rule,

The acid form is dominant for $pH<p{K}_a$

The conjugate pair have equal concentrations at $pH=p{K}_a$

The base form is dominant for $pH>p{K}_a$

At $\text{9}\text{.0}\text{pH}$ concentration of hydrogen ion is,

    $\begin{array}{l}{\text{[H}}_3{\text{O}}^{+}\text{]}\text{=}{10}^{\text{-pH}}\\ ={10}^{-9}\\ =1\times {10}^{-9}\end{array}$

At the given pH, equilibrium constant of acid is,

    $\begin{array}{l}{K}_{a2=}\frac{[{\text{PO}}_3{{}^{2-}}_{\text{(aq)}}][{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}]}{{\text{HPO}}_3{{}^{-}}_{\text{(aq)}}}=2.6\times {10}^{-7}\\ \\ \text{1}{\text{.0×10}}^{\text{-2}}\text{=}\frac{{\text{[1}\times {\text{10}}^{-9}\text{]}}^{\text{2}}}{{\text{HPO}}_3{{}^{-}}_{\text{(aq)}}}\\ \\ {\text{HPO}}_3{{}^{-}}_{\text{(aq)}}\text{=}\frac{{{\text{[1×10}}^{\text{-9}}\text{]}}^{\text{2}}}{2.6\times {10}^{-7}}\text{=}\frac{{\text{1×10}}^{\text{-18}}}{2.6\times {10}^{-7}}\\ \\ \text{=}0.38\times {10}^{-12}.\end{array}$

At the first dissociation,

    $\begin{array}{l}{\text{K}}_{\text{a1}}\text{=}\frac{{\text{[HPO}}_{\text{3}}{{}^{\text{-}}}_{\text{(aq)}}{\text{][H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}\text{]}}{{\text{H}}_{\text{2}}{\text{PO}}_{\text{3}}{}_{\text{(aq)}}}\text{=}\text{1}{\text{.0×10}}^{\text{-2}}\\ \\ \text{1}{\text{.0×10}}^{\text{-2}}\text{=}\frac{{\text{[0}\text{.38}\times {10}^{-12}\text{]}}^{\text{2}}}{{\text{H}}_{\text{2}}{\text{PO}}_{\text{3}}{}_{\text{(aq)}}}\\ \\ {\text{H}}_{\text{2}}{\text{PO}}_{\text{3}}{}_{\text{(aq)}}\text{=}\frac{{\text{[0}\text{.38}\times {10}^{-12}\text{]}}^{\text{2}}}{\text{1}{\text{.0×10}}^{\text{-2}}}\text{=}\frac{\text{1}{\text{.44×10}}^{\text{-25}}}{\text{1}{\text{.0×10}}^{\text{-2}}}\\ \\ \text{=}\text{1}{\text{.44×10}}^{\text{-23}}\text{M}\text{.}\end{array}$

Hence, the fraction of intact $H_{2}P{O}_3$ molecules in an aqueous solution of phosphorous acid at $pH=9.0$ is $\text{1}{\text{.44×10}}^{\text{-23}}\text{M}$.