Chapter 20, Problem 20.50P

Interpretation Introduction

Interpretation:

The equilibrium concentration values of ${\text{OH}}^{\text{-}}\left(aq\right){\text{, HNO}}_{\text{2}}\left(aq\right){\text{, NO}}_{\text{2}}{}^{\text{-}}\left(aq\right)$ and ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(aq\right)$ has to be calculated. Also the $\text{pH}$ of solution should be determined at ${\text{25}}^{\text{o}}\text{C}$.

Concept introduction:

Ionic-product constant for water: It is the hydronium ion concentration times the $O{H}^{-}$ concentration present in the solution.

$\begin{array}{c}{H}_2{O}_{\left(l\right)}+H_2{O}_{\left(l\right)}{H}_3{O}^{+}{}_{\left(aq\right)}+O{H}^{-}{}_{\left(aq\right)}\\ \\ K=\frac{\left[H_3{O}^{+}\right]\left[O{H}^{-}\right]}{\left[H_{2}O\right]\left[H_{2}O\right]}\\ \\ \left[H_3{O}^{+}\right]=\left[O{H}^{-}\right]=1.00\times {10}^{-7}M\\ \\ \therefore K_W=K\left[H_{2}O\right]\left[H_{2}O\right]=\left[H_3{O}^{+}\right]\left[O{H}^{-}\right]\\ \\ K_W=1.00\times {10}^{-14}M\\ \\ K_W=K_a.K_b\end{array}$

Answer to Problem 20.50P

The concentration value of ${\text{OH}}^{\text{-}}\left(aq\right)$ is $2.1\times {10}^{-6}\text{M}$.

The concentration value of ${\text{HNO}}_{\text{2}}\left(aq\right)$ is $2.1\times {10}^{-6}\text{M}$.

The concentration value of ${\text{NO}}_{\text{2}}{}^{\text{-}}\left(aq\right)$ is $\text{0}\text{.25M}$.

The concentration value of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(aq\right)$ is $\text{4}\text{.79}\times {\text{10}}^{\text{-9}}\text{ M}$.

The $\text{pH}$ value is $\text{8}\text{.38}$.

Explanation of Solution

Given: $\text{0}\text{.25 M}$ in ${\text{NaNO}}_{\text{2}}\left(aq\right)$.

The reaction between ${\text{NaNO}}_{\text{2}}\left(aq\right)$ and water is as follows,

    ${\text{NaNO}}_{\text{2}}\left(s\right){\text{ + H}}_{\text{2}}\text{O}\left(l\right)\to {\text{HNO}}_{\text{2}}\left(aq\right){\text{ + OH}}_{}^{\text{-}}\left(aq\right)$

The ${\text{K}}_{\text{a}}$ value for ${\text{HNO}}_{\text{2}}\left(aq\right)$ is $5.6\times {10}^{\text{-4}}\text{ M}$. Now, the ${\text{K}}_{\text{b}}$ value is determined by using equation, ${\text{K}}_{\text{w}}{\text{= K}}_{\text{a}}\times {\text{K}}_{\text{b}}$.

    $\begin{array}{c}{\text{K}}_{\text{w}}{\text{= K}}_{\text{a}}\times {\text{K}}_{\text{b}}\\ \\ {\text{K}}_{\text{b}}=\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{a}}}\\ \\ =\frac{1.00\times {10}^{-14}\text{ M}}{5.6\times {10}^{\text{-4}}\text{ M}}\\ \\ =1.79\times {10}^{-11}\end{array}$

The ICE table for reaction of ${\text{NO}}_{\text{2}}{}^{\text{-}}$ with water is shown as below,

    $\begin{array}{l}{\text{NO}}_{\text{2}}{}^{\text{-}}+{\text{ H}}_{\text{2}}\text{O }\rightleftharpoons {\text{HNO}}_{\text{2}}{\text{ + OH}}^{\text{-}}\\ \\ Initial0.25\text{M}\text{ 0 0}\\ Change\text{ -x +x +x}\\ Equilibrium0.25\text{-x x x}\end{array}$

Using calculated ${\text{K}}_{\text{b}}$ value the concentration of ${\text{HNO}}_{\text{2}}$ and ${\text{OH}}^{\text{-}}$ are determined.

    $\begin{array}{c}{\text{K}}_{\text{b}}=\frac{\left[{\text{HNO}}_{\text{2}}\right]\left[{\text{OH}}^{\text{-}}\right]}{\left[{\text{NO}}_{\text{2}}{}^{\text{-}}\right]}\\ \\ 1.79\times {10}^{-11}=\frac{\text{x × x}}{0.25\text{ - x}}\\ \\ 1.79\times {10}^{-11}=\frac{{\text{x }}^{\text{2}}}{0.25}\\ \\ \text{x }=2.1\times {10}^{-6}\end{array}$

    $\begin{array}{c}\left[{\text{HNO}}_{\text{2}}\right]=\left[{\text{OH}}^{\text{-}}\right]=2.1\times {10}^{-6}\text{M}\\ \\ \left[{\text{NO}}_{\text{2}}{}^{\text{-}}\right]\text{ = 0}\text{.25 - x = 0}\text{.25 - }2.1\times {10}^{-6}\\ \\ \left[{\text{NO}}_{\text{2}}{}^{\text{-}}\right]=\text{ 0}\text{.25M}\end{array}$

Now, the $\text{pH}$ value is calculated by using above determined hydroxide ions concentration.

    $\begin{array}{c}\text{pOH = -log }\left[2.1\times {10}^{-6}\right]\\ \\ =\text{ 5}\text{.68}\\ \\ \text{pH = 14 - pOH}\\ \\ \text{= 14 - 5}\text{.68}\\ \\ =\text{ 8}\text{.32}\end{array}$

Using above calculated $\text{pH}$ the concentration of $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$ ions are calculated.

    $\begin{array}{c}\text{pH = -log }\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]={10}^{-\text{8}\text{.32}}\\ \\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{ = 4}\text{.79}\times {\text{10}}^{\text{-9}}\text{ M}\end{array}$