Chapter 20, Problem 20.57P

Interpretation Introduction

Interpretation:

The $\text{pH}$ value and concentration for ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(aq\right)$ ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\left(aq\right)$${\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}{}^{-}\left(aq\right)$${\text{HPO}}_{\text{4}}{}^{2-}\left(aq\right)$${\text{PO}}_{\text{4}}{}^{3-}\left(aq\right)$ and ${\text{OH}}^{\text{-}}\left(aq\right)$ in $0.100\text{ M}$ phosphoric acid solution has to be calculated.

Answer to Problem 20.57P

Therefore, the $\text{pH}$ value is $\text{1}\text{.6}$.

The concentration of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\left(aq\right)$ is $0.074\text{M}$.

The concentration of ${\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}{}^{-}\left(aq\right)$ is $\text{2}\text{.6}\times {\text{10}}^{\text{-2}}\text{M}$

The concentration of ${\text{HPO}}_{\text{4}}{}^{2-}\left(aq\right)$ is $\text{4}\text{.0}\times {\text{10}}^{\text{-5}}\text{M}$

The concentration of ${\text{PO}}_{\text{4}}{}^{3-}\left(aq\right)$ is $\text{4}\text{.4}\times {\text{10}}^{\text{-9}}\text{M}$

The concentration of ${\text{OH}}^{\text{-}}\left(aq\right)$ is $3.98\times {10}^{-13}\text{M}$

Explanation of Solution

The equilibrium reaction after first dissociation is as follows,

    ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\left(aq\right){\text{ + H}}_{\text{2}}\text{O}\left(l\right)\rightleftharpoons {\text{ H}}_{\text{2}}{\text{PO}}_{\text{4}}{}^{-}\left(aq\right){\text{ + H}}_{\text{3}}{\text{O}}^{+}\left(aq\right)$

The ICE table for first dissociation is as follows,

    $\begin{array}{l}{\text{ H}}_{\text{3}}{\text{PO}}_{\text{4}}\left(aq\right){\text{ + H}}_{\text{2}}\text{O}\left(l\right)\rightleftharpoons {\text{ H}}_{\text{2}}{\text{PO}}_{\text{4}}{}^{-}\left(aq\right){\text{ + H}}_{\text{3}}{\text{O}}^{+}\left(aq\right)\\ \\ Initial\text{ 0}\text{.100 0 0}\\ Change\text{ -x +x +x}\\ Equilibrium\text{ 0}\text{.100-x }\text{ +x +x}\\ \\ {\text{K}}_{\text{a1}}\text{ = }\frac{\left[{\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}{}^{-}\right]\left[{\text{ H}}_{\text{3}}{\text{O}}^{+}\right]}{\left[{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\right]}\end{array}$

    $\begin{array}{c}{\text{K}}_{\text{a1}}\text{ = }\frac{\left[{\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}{}^{-}\right]\left[{\text{ H}}_{\text{3}}{\text{O}}^{+}\right]}{\left[{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\right]}\\ \\ 6.9\times {10}^{-3}=\frac{x^2}{0.100\text{- }x}\\ \\ 6.9\times {10}^{-3}=\frac{x^2}{0.100}\\ \\ x=\text{ 2}\text{.6}\times {\text{10}}^{\text{-2}}\\ \\ \left[{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\right]=0.074\text{M}\\ \\ \left[{\text{ H}}_{\text{3}}{\text{O}}^{+}\right]=\left[{\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}{}^{-}\right]=\text{2}\text{.6}\times {\text{10}}^{\text{-2}}\text{M}\end{array}$

The ICE table for obtained ${\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}{}^{-}$ ion after second dissociation is shown below,

    $\begin{array}{l}{\text{ H}}_{\text{2}}{\text{PO}}_{\text{4}}{}^{-}\left(aq\right){\text{ + H}}_{\text{2}}\text{O}\left(l\right)\rightleftharpoons {\text{ HPO}}_{\text{4}}{}^{2-}\left(aq\right){\text{ + H}}_{\text{3}}{\text{O}}^{+}\left(aq\right)\\ \\ Initial\text{ 2}\text{.6}\times {\text{10}}^{\text{-2}}\text{ 0 0}\\ Change\text{ -x +x +x}\\ Equilibrium\text{ 2}\text{.6}\times {\text{10}}^{\text{-2}}\text{-x }\text{ +x +x}\\ \\ {\text{K}}_{\text{a2}}\text{ = }\frac{\left[{\text{HPO}}_{\text{4}}{}^{2-}\right]\left[{\text{ H}}_{\text{3}}{\text{O}}^{+}\right]}{\left[{\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}{}^{-}\right]}\end{array}$

    $\begin{array}{c}{\text{K}}_{\text{a2}}\text{ = }\frac{\left[{\text{HPO}}_{\text{4}}{}^{2-}\right]\left[{\text{ H}}_{\text{3}}{\text{O}}^{+}\right]}{\left[{\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}{}^{-}\right]}\\ \\ 6.2\times {10}^{-8}=\frac{x^2}{2.6\times {10}^{-2}\text{- }x}\\ \\ 6.2\times {10}^{-8}=\frac{x^2}{2.6\times {10}^{-2}}\\ \\ x=\text{ 4}\text{.0}\times {\text{10}}^{\text{-5}}\\ \\ \left[{\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}{}^{-}\right]=0.026\text{M}\\ \\ \left[{\text{ H}}_{\text{3}}{\text{O}}^{+}\right]=\left[{\text{HPO}}_{\text{4}}{}^{2-}\right]=\text{4}\text{.0}\times {\text{10}}^{\text{-5}}\text{M}\end{array}$

The resulting ${\text{HPO}}_{\text{4}}{}^{2-}\left(aq\right)$ ion after its third dissociation is as follows,

    $\begin{array}{l}{\text{ HPO}}_{\text{4}}{}^{2-}\left(aq\right){\text{ + H}}_{\text{2}}\text{O}\left(l\right)\rightleftharpoons {\text{ PO}}_{\text{4}}{}^{3-}\left(aq\right){\text{ + H}}_{\text{3}}{\text{O}}^{+}\left(aq\right)\\ \\ Initial\text{ 4}\text{.0}\times {\text{10}}^{\text{-5}}\text{ 0 0}\\ Change\text{ -x +x +x}\\ Equilibrium\text{ 4}\text{.0}\times {\text{10}}^{\text{-5}}\text{-x }\text{ +x +x}\\ \\ {\text{K}}_{\text{a3}}\text{ = }\frac{\left[{\text{PO}}_{\text{4}}{}^{3-}\right]\left[{\text{ H}}_{\text{3}}{\text{O}}^{+}\right]}{\left[{\text{HPO}}_{\text{4}}{}^{2-}\right]}\end{array}$

    $\begin{array}{c}{\text{K}}_{\text{a3}}\text{ = }\frac{\left[{\text{PO}}_{\text{4}}{}^{3-}\right]\left[{\text{ H}}_{\text{3}}{\text{O}}^{+}\right]}{\left[{\text{HPO}}_{\text{4}}{}^{2-}\right]}\\ \\ 4.8\times {10}^{-13}=\frac{x^2}{4.0\times {10}^{-5}\text{- }x}\\ \\ 4.8\times {10}^{-13}=\frac{x^2}{4.0\times {10}^{-5}}\\ \\ x=\text{ 4}\text{.4}\times {\text{10}}^{\text{-9}}\\ \\ \left[{\text{HPO}}_{\text{4}}{}^{2-}\right]\text{ = 4}\text{.0}\times {\text{10}}^{\text{-5}}\text{M}\\ \\ \left[{\text{ H}}_{\text{3}}{\text{O}}^{+}\right]=\left[{\text{PO}}_{\text{4}}{}^{3-}\right]=\text{4}\text{.4}\times {\text{10}}^{\text{-9}}\text{M}\end{array}$

The total concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ obtained after 3 dissociation is $\text{0}\text{.026M}$.

The ${\text{OH}}^{\text{-}}$ value present in solution is calculated as shown below,

    $\begin{array}{c}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=0.026\text{ M}\\ \\ \text{pH = -log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \\ =\text{ -log}\left[0.026\right]\\ \\ =\text{ 1}\text{.6}\end{array}$

    $\begin{array}{c}\text{pOH = 14 - 1}\text{.6}\\ \\ \text{= 12}\text{.4}\\ \\ \text{pOH = -log}\left[{\text{OH}}^{\text{-}}\right]\\ \\ \text{12}\text{.4 = -log}\left[{\text{OH}}^{\text{-}}\right]\\ \\ \left[{\text{OH}}^{\text{-}}\right]=a^{\text{-12}\text{.4}}\\ \\ =3.98\times {10}^{-13}\text{M}\end{array}$