Chapter 16, Problem 16.148QP

Interpretation Introduction

Interpretation:

The concentration of all species in a 0.100 M ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ solution has to be calculated

Concept Information:

Acid ionization constant ${\text{K}}_{\text{a}}$:

The equilibrium expression for the reaction ${\text{HA}}_{(aq)}\rightleftarrows {\text{H}}^{+}{}_{(aq)}+{\text{A}}^{\text{-}}{}_{(aq)}$ is given below.

$K_a=\frac{[{\text{H}}^{\text{+}}{\text{][A}}^{\text{-}}\text{]}}{[\text{HA]}}$

Where $K_a$ is acid ionization constant, $[{\text{H}}^{\text{+}}\text{]}$  is concentration of hydrogen ion, $[{\text{A}}^{\text{-}}\text{]}$  is concentration of acid anion, $[\text{HA]}$ is concentration of the acid

Diprotic and polyprotic acids:

Acids having two or more hydrogen atoms are termed as diprotic or polyprotic acids.  These acids lose one proton at a time by undergoing successive ionizations.

For diprotic acids, the successive ionization constants are designated as $K_{a1}and{K}_{a2}$

For triprotic acids, the successive ionization constants are designated as $K_{a1},K_{a2}and{K}_{a3}$

Answer to Problem 16.148QP

Answer

The concentration of all species in a 0.100 M ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ solution

$\begin{array}{l}\text{[H+]}{\text{= [H}}_{\text{2}}{\text{PO}}_{\text{4}}{}^{\text{-}}\text{]}\text{=}\text{0}\text{.0239 M}\\ {\text{ [H}}_{\text{3}}{\text{PO}}_{\text{4}}\text{]}\text{=}\text{0}\text{.076 M}\\ [{\text{HPO}}_{\text{4}}{}^{\text{2-}}]\text{=}\text{0}\text{.076 M}\\ {\text{ [HPO}}_{\text{4}}{}^{\text{2-}}\text{]}\text{=}\text{6}\text{.2}\times {\text{10}}^{\text{-8}}M\\ [{\text{PO}}_{\text{4}}{}^{\text{3-}}\text{]}\text{=}\text{1}\text{.2}\times {\text{10}}^{\text{-18}}M\end{array}$

Explanation of Solution

Given data:

The concentration of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ solution is 0.100 M

Species of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$

${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ is a triprotic acid.

The species that are present in ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ are ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}{\text{, H}}^{\text{+}},{\text{ H}}_{\text{2}}{\text{PO}}_{\text{4}}{}^{\text{-}}{\text{, HPO}}_{\text{4}}{}^{\text{2-}}{\text{ and PO}}_{\text{4}}{}^{\text{3-}}$.

The concentration of these species will be represented as ${\text{[H}}_{\text{3}}{\text{PO}}_{\text{4}}\text{]}$, ${\text{[H}}^{\text{+}}]$, ${\text{[H}}_{\text{2}}{\text{PO}}_{\text{4}}{}^{\text{-}}\text{]}$, ${\text{[HPO}}_{\text{4}}{}^{\text{2-}}\text{]}$ and ${\text{[PO}}_{\text{4}}{}^{\text{3-}}\text{]}$ and can be calculated as follows.

First ionization of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$

	${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}{}_{\text{(aq)}}\stackrel{}{\to }{\text{ H}}^{\text{+}}{}_{\text{(aq)}}\text{ +}{\text{H}}_2{\text{PO}}_{\text{4}}{{}^{\text{-}}}_{\text{(aq)}}$
Initial $(M)$	0.100 -x (0.100-x)	$0.00$	$0.00$
Change $(M)$		+x	+x
Equilibrium $(M)$		x	x

The ${\text{K}}_{\text{a1}}$ for phosphoric acid is $7.5\times {10}^{-3}$

$\begin{array}{l}{\text{K}}_{\text{a1}}=\frac{{\text{[H}}^{\text{+}}{\text{][H}}_{\text{2}}{\text{PO}}_{\text{4}}{}^{\text{-}}\text{]}}{{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}}\\ 7.5\times {10}^{-3}=\frac{{(x)}^2}{0.100-x}\end{array}$

In this case we probably cannot say that $0.100-x\approx ~0.100$ due to the magnitude of ${\text{K}}_{\text{a}}$

We obtain the quadratic equation:

$x^2+(7.5\times {10}^{-3})x-(7.5\times {10}^{-4})=0$

The positive root is x = 0.00239 M. We have:

$\begin{array}{l}{\text{[H+]= [H}}_{\text{2}}{\text{PO}}_{\text{4}}{}^{\text{-}}\text{]=}\text{0}\text{.0239 M}\\ {\text{ [H}}_{\text{3}}{\text{PO}}_{\text{4}}\text{]}\text{=}\text{(0}\text{.100-0}\text{.0239) M}\\ \text{=}\text{0}\text{.076 M}\end{array}$

For the second ionization:

	${\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}{{}^{-}}_{\text{(aq)}}\stackrel{}{\to }{\text{ H}}^{\text{+}}{}_{\text{(aq)}}\text{ +}{\text{HPO}}_{\text{4}}{{}^{\text{2-}}}_{\text{(aq)}}$
Initial $(M)$	0.0239 -y (0.0239-y)	$0.00$	$0.00$
Change $(M)$		+y	+y
Equilibrium $(M)$		(0.0239+y)	y

The ${\text{K}}_{\text{a2}}$ for phosphoric acid is $6.2\times {10}^{-8}$

$\begin{array}{l}{\text{ K}}_{\text{a2}}=\frac{{\text{[H}}^{\text{+}}{\text{][HPO}}_{\text{4}}{}^{\text{2-}}\text{]}}{{\text{H}}_2{\text{PO}}_{\text{4}}{}^{-}}\\ 6.2\times {10}^{-8}=\frac{(0.0239+y)(y)}{0.0239-y}\\ \approx \frac{(0.0239)(y)}{(0.0239)}\\ y=6.2\times {10}^{-8}M\end{array}$

Thus,

  $\begin{array}{l}{\text{[H+]= [H}}_{\text{2}}{\text{PO}}_{\text{4}}{}^{\text{-}}\text{]=0}\text{.0239 M}\\ {\text{ [HPO}}_{\text{4}}{}^{\text{2-}}\text{]}\text{= y}\\ \text{=6}\text{.2}\times {\text{10}}^{\text{-8}}M\end{array}$

For the third ionization:

	${\text{HPO}}_{\text{4}}{{}^{2-}}_{\text{(aq)}}\stackrel{}{\to }{\text{ H}}^{\text{+}}{}_{\text{(aq)}}\text{ +}{\text{PO}}_{\text{4}}{{}^{\text{3-}}}_{\text{(aq)}}$
Initial $(M)$	$\text{6}\text{.2}\times {\text{10}}^{\text{-8}}$ -z $\text{6}\text{.2}\times {\text{10}}^{\text{-8}}-z$	$0.00$	$0.00$
Change $(M)$		+z	+z
Equilibrium $(M)$		(0.0239+z)	z

The ${\text{K}}_{\text{a3}}$ for phosphoric acid is $4.8\times {10}^{-13}$

$\begin{array}{l}{\text{ K}}_{\text{a3}}=\frac{{\text{[H}}^{\text{+}}{\text{][PO}}_{\text{4}}{}^{\text{3-}}\text{]}}{{\text{HPO}}_{\text{4}}{}^{2-}}\\ \text{ 4}\text{.8}\times {10}^{-13}=\frac{(0.0239+z)(z)}{0.0239-z}\\ \approx \frac{(0.0239)(z)}{(6.2\times {10}^{-8})}\\ \text{ z}=1.2\times {10}^{-18}M\end{array}$

The equilibrium concentrations are:

$\begin{array}{l}{\text{[H+]= [H}}_{\text{2}}{\text{PO}}_{\text{4}}{}^{\text{-}}\text{]=0}\text{.0239 M}\\ {\text{ [H}}_{\text{3}}{\text{PO}}_{\text{4}}\text{]}\text{=0}\text{.076 M}\\ [{\text{HPO}}_{\text{4}}{}^{\text{2-}}]\text{=0}\text{.076 M}\\ {\text{ [HPO}}_{\text{4}}{}^{\text{2-}}\text{]}\text{=6}\text{.2}\times {\text{10}}^{\text{-8}}M\\ [{\text{PO}}_{\text{4}}{}^{\text{3-}}\text{]}\text{=1}\text{.2}\times {\text{10}}^{\text{-18}}M\end{array}$

Conclusion

The concentration of all species in a 0.100 M ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ solution was calculated.