Chapter 13, Problem 120E

(a)

Interpretation Introduction

Interpretation: The $\text{pH}$ value for each of the given solutions to be calculated.

Concept introduction: The $\text{pH}$ of a solution is defined as a figure that expresses the acidity of the alkalinity of a given solution.

The $\text{pOH}$ of a solution is calculated by the formula, $\text{pOH}=-\log \left[{\text{OH}}^{-}\right]$

The sum, $\text{pH}+\text{pOH}=14$

At equilibrium, the equilibrium constant expression is expressed by the formula,

$K_{\text{a}}=\frac{\text{Concentration}\text{of}\text{products}}{\text{Concentration}\text{of}\text{reactants}}$

The value of $K_{\text{w}}$ is calculated by the formula,

$K_{\text{w}}=K_{\text{a}}\cdot K_{\text{b}}$

To determine: The $\text{pH}$ value for each of the given solution of $0.12\text{M}$ ${\text{KNO}}_2$.

Answer to Problem 120E

Answer

The $\text{pH}$ of the given solution of $0.12\text{M}$ ${\text{KNO}}_2$ is $\underset{\_}{8.239}$.

Explanation of Solution

Explanation

The equilibrium constant expression for the given reaction is, $K_{\text{b}}=\frac{\left[{\text{HNO}}_2\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{NO}}_2^{-}\right]}$

The dominant equilibrium reaction is,

${\text{NO}}_2^{-}\left(aq\right)\rightleftharpoons {\text{HNO}}_2\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$

At equilibrium, the equilibrium constant expression is expressed by the formula,

$K_{\text{b}}=\frac{\text{Concentration}\text{of}\text{products}}{\text{Concentration}\text{of}\text{reactants}}$

Where,

• $K_{\text{b}}$ is the base ionization constant.

The equilibrium constant expression for the given reaction is,

$K_{\text{b}}=\frac{\left[{\text{HNO}}_2\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{NO}}_2^{-}\right]}$ (1)

The $K_{\text{b}}$ value is $\underset{\_}{2.5\times 10^{-11}}$.

The value, $K_{\text{w}}=K_{\text{a}}\cdot K_{\text{b}}$

The value of $K_{\text{a}}$ for ${\text{HNO}}_{\text{2}}$ is $4.0\times {10}^{-4}$.

The value of $K_{\text{b}}$ is calculated by the formula,

$K_{\text{b}}=\frac{K_{\text{w}}}{K_{\text{a}}}$

Substitute the value of $K_{\text{a}}$ in the above expression.

$\begin{array}{c}{K}_{\text{b}}=\frac{1.0\times {10}^{-14}}{4.0\times {10}^{-4}}\\ =\underset{\_}{2.5\times 10^{-11}}\end{array}$

The $\left[{\text{OH}}^{-}\right]$ is $\underset{\_}{1.73\times 10^{-6}M}$.

The change in concentration of ${\text{NO}}_2^{-}$ is assumed to be $x$.

The liquid components do not affect the value of the rate constant.

The ICE table for the stated reaction is,

$\begin{array}{cccccc}& {\text{NO}}_2^{-}\left(aq\right)& \rightleftharpoons & {\text{HNO}}_2\left(aq\right)& +& {\text{OH}}^{-}\left(aq\right)\\ \text{Inititial}\text{concentration}& 0.12& & 0& & 0\\ \text{Change}& -x& & +x& & +x\\ \text{Equilibrium}\text{concentration}& 0.12-x& & x& & x\end{array}$

The equilibrium concentration of $\left[{\text{NO}}_2^{-}\right]$ is $\left(0.12-x\right)\text{M}$.

The equilibrium concentration of $\left[{\text{HNO}}_2\right]$ is $x\text{M}$.

The equilibrium concentration of $\left[{\text{OH}}^{-}\right]$ is $x\text{M}$.

The calculated value of $K_{\text{b}}$ is $2.5\times {10}^{-11}$.

Substitute the value of $K_{\text{b}}$, $\left[{\text{NO}}_2^{-}\right]$, $\left[{\text{HNO}}_{\text{2}}\right]$ and $\left[{\text{OH}}^{-}\right]$ in equation (1).

$\begin{array}{c}2.5\times {10}^{-11}=\frac{\left[x\right]\left[x\right]}{\left[0.12-x\right]}\\ 2.5\times {10}^{-11}=\frac{{\left[x\right]}^2}{\left[0.12-x\right]}\end{array}$

Solve the above expression.

$\begin{array}{c}2.5\times {10}^{-11}=\frac{{\left[x\right]}^2}{\left[0.12-x\right]}\\ \left[x\right]=\underset{\_}{1.73\times 10^{-6}M}\end{array}$

Therefore, the $\left[{\text{OH}}^{-}\right]$ is $\underset{\_}{1.73\times 10^{-6}M}$.

The $\text{pOH}$ of the given solution is $\underset{\_}{5.761}$.

The $\text{pOH}$ of a solution is calculated by the formula,

$\text{pOH}=-\log \left[{\text{OH}}^{-}\right]$

Substitute the value of $\left[{\text{OH}}^{-}\right]$ in the above expression.

$\begin{array}{c}\text{pOH}=-\log \left[1.73\times {10}^{-6}\right]\\ =\underset{\_}{5.761}\end{array}$

The $\text{pH}$ of the given solution is $\underset{\_}{8.239}$.

The sum, $\text{pH}+\text{pOH}=14$

The value of $\text{pH}$ is calculated by the formula,

$\text{pH}=14-\text{pOH}$

Substitute the calculated value of $\text{pOH}$ in the above expression.

$\begin{array}{c}\text{pH}=14-\text{5}\text{.761}\\ =\underset{\_}{8.239}\end{array}$

(b)

Interpretation Introduction

Interpretation: The $\text{pH}$ value for each of the given solutions to be calculated.

Concept introduction: The $\text{pH}$ of a solution is defined as a figure that expresses the acidity of the alkalinity of a given solution.

The $\text{pOH}$ of a solution is calculated by the formula, $\text{pOH}=-\log \left[{\text{OH}}^{-}\right]$

The sum, $\text{pH}+\text{pOH}=14$

At equilibrium, the equilibrium constant expression is expressed by the formula,

$K_{\text{a}}=\frac{\text{Concentration}\text{of}\text{products}}{\text{Concentration}\text{of}\text{reactants}}$

The value of $K_{\text{w}}$ is calculated by the formula,

$K_{\text{w}}=K_{\text{a}}\cdot K_{\text{b}}$

To determine: The $\text{pH}$ value for each of the given solution of $0.45\text{M}$ $\text{NaOCl}$.

Answer to Problem 120E

Answer

The $\text{pH}$ of the given solution of $0.45\text{M}$ $\text{NaOCl}$ is $\underset{\_}{10.554}$.

Explanation of Solution

Explanation

The equilibrium constant expression for the given reaction is, $K_{\text{b}}=\frac{\left[\text{HOCl}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{OCl}}^{-}\right]}$

The dominant equilibrium reaction is,

${\text{OCl}}^{-}\left(aq\right)\rightleftharpoons \text{HOCl}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$

At equilibrium, the equilibrium constant expression is expressed by the formula,

$K_{\text{b}}=\frac{\text{Concentration}\text{of}\text{products}}{\text{Concentration}\text{of}\text{reactants}}$

Where,

• $K_{\text{b}}$ is the base ionization constant.

The equilibrium constant expression for the given reaction is,

$K_{\text{b}}=\frac{\left[\text{HOCl}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{OCl}}^{-}\right]}$ (1)

The $K_{\text{b}}$ value is $\underset{\_}{2.857\times 10^{-7}}$.

The value, $K_{\text{w}}=K_{\text{a}}\cdot K_{\text{b}}$

The value of $K_{\text{a}}$ for $\text{HOCl}$ is $3.5\times {10}^{-8}$.

The value of $K_{\text{b}}$ is calculated by the formula,

$K_{\text{b}}=\frac{K_{\text{w}}}{K_{\text{a}}}$

Substitute the value of $K_{\text{a}}$ in the above expression.

$\begin{array}{c}{K}_{\text{b}}=\frac{1.0\times {10}^{-14}}{3.5\times {10}^{-8}}\\ =\underset{\_}{2.857\times 10^{-7}}\end{array}$

The $\left[{\text{OH}}^{-}\right]$ is $\underset{\_}{3.58\times 10^{-4}M}$.

The change in concentration of ${\text{OCl}}^{-}$ is assumed to be $x$.

The liquid components do not affect the value of the rate constant.

The ICE table for the stated reaction is,

$\begin{array}{cccccc}& {\text{OCl}}^{-}\left(aq\right)& \rightleftharpoons & \text{HOCl}\left(aq\right)& +& {\text{OH}}^{-}\left(aq\right)\\ \text{Inititial}\text{concentration}& 0.45& & 0& & 0\\ \text{Change}& -x& & +x& & +x\\ \text{Equilibrium}\text{concentration}& 0.45-x& & x& & x\end{array}$

The equilibrium concentration of $\left[{\text{OCl}}^{-}\right]$ is $\left(0.45-x\right)\text{M}$.

The equilibrium concentration of $\left[\text{HOCl}\right]$ is $x\text{M}$.

The equilibrium concentration of $\left[{\text{OH}}^{-}\right]$ is $x\text{M}$.

The calculated value of $K_{\text{b}}$ is $2.857\times {10}^{-7}$.

Substitute the value of $K_{\text{b}}$, $\left[{\text{OCl}}^{-}\right]$, $\left[\text{HOCl}\right]$ and $\left[{\text{OH}}^{-}\right]$ in equation (1).

$\begin{array}{c}2.857\times {10}^{-7}=\frac{\left[x\right]\left[x\right]}{\left[0.45-x\right]}\\ 2.857\times {10}^{-7}=\frac{{\left[x\right]}^2}{\left[0.45-x\right]}\end{array}$

Solve the above expression.

$\begin{array}{c}2.857\times {10}^{-7}=\frac{{\left[x\right]}^2}{\left[0.45-x\right]}\\ \left[x\right]=\underset{\_}{3.58\times 10^{-4}M}\end{array}$

Therefore, the $\left[{\text{OH}}^{-}\right]$ is $\underset{\_}{3.58\times 10^{-4}M}$.

The $\text{pOH}$ of the given solution is $\underset{\_}{3.446}$.

The $\text{pOH}$ of a solution is calculated by the formula,

$\text{pOH}=-\log \left[{\text{OH}}^{-}\right]$

Substitute the value of $\left[{\text{OH}}^{-}\right]$ in the above expression.

$\begin{array}{c}\text{pOH}=-\log \left[3.58\times {10}^{-4}\right]\\ =\underset{\_}{3.446}\end{array}$

The $\text{pH}$ of the given solution is $\underset{\_}{10.554}$.

The sum, $\text{pH}+\text{pOH}=14$

The value of $\text{pH}$ is calculated by the formula,

$\text{pH}=14-\text{pOH}$

Substitute the calculated value of $\text{pOH}$ in the above expression.

$\begin{array}{c}\text{pH}=14-3.446\\ =\underset{\_}{10.554}\end{array}$

(c)

Interpretation Introduction

Interpretation: The $\text{pH}$ value for each of the given solutions to be calculated.

Concept introduction: The $\text{pH}$ of a solution is defined as a figure that expresses the acidity of the alkalinity of a given solution.

The $\text{pOH}$ of a solution is calculated by the formula, $\text{pOH}=-\log \left[{\text{OH}}^{-}\right]$

The sum, $\text{pH}+\text{pOH}=14$

At equilibrium, the equilibrium constant expression is expressed by the formula,

$K_{\text{a}}=\frac{\text{Concentration}\text{of}\text{products}}{\text{Concentration}\text{of}\text{reactants}}$

The value of $K_{\text{w}}$ is calculated by the formula,

$K_{\text{w}}=K_{\text{a}}\cdot K_{\text{b}}$

To determine: The $\text{pH}$ value for each of the given solution of $0.40\text{M}$ ${\text{NH}}_{\text{4}}{\text{ClO}}_{\text{4}}$.

Answer to Problem 120E

Answer

The $\text{pH}$ of the given solution of $0.40\text{M}$ ${\text{NH}}_{\text{4}}{\text{ClO}}_{\text{4}}$ is $\underset{\_}{4.827}$.

Explanation of Solution

Explanation

The equilibrium constant expression for the given reaction is, $K_{\text{a}}=\frac{\left[{\text{NH}}_3\right]\left[{\text{H}}^{+}\right]}{\left[{\text{NH}}_4^{+}\right]}$

${\text{NH}}_4^{+}$ is a stronger acid than ${\text{H}}_{\text{2}}\text{O}$.

The dominant equilibrium reaction is,

${\text{NH}}_4^{+}\left(aq\right)\rightleftharpoons {\text{NH}}_3\left(aq\right)+{\text{H}}^{+}\left(aq\right)$

At equilibrium, the equilibrium constant expression is expressed by the formula,

$K_{\text{a}}=\frac{\text{Concentration}\text{of}\text{products}}{\text{Concentration}\text{of}\text{reactants}}$

Where,

• $K_{\text{a}}$ is the acid dissociation constant.

The equilibrium constant expression for the given reaction is,

$K_{\text{a}}=\frac{\left[{\text{NH}}_3\right]\left[{\text{H}}^{+}\right]}{\left[{\text{NH}}_4^{+}\right]}$ (1)

The $K_{\text{a}}$ value is $\underset{\_}{5.6\times 10^{-10}}$.

The value, $K_{\text{w}}=K_{\text{a}}\cdot K_{\text{b}}$

The value of $K_{\text{b}}$ for ammonia is $1.8\times {10}^{-5}$.

The value of $K_{\text{a}}$ is calculated by the formula,

$K_{\text{a}}=\frac{K_{\text{w}}}{K_{\text{b}}}$

Substitute the value of $K_{\text{b}}$ in the above expression.

$\begin{array}{c}{K}_{\text{a}}=\frac{1.0\times {10}^{-14}}{1.8\times {10}^{-5}}\\ =\underset{\_}{5.6\times 10^{-10}}\end{array}$

The $\left[{\text{H}}^{+}\right]$ is $\underset{\_}{1.49\times 10^{-5}M}$.

The change in concentration of ${\text{NH}}_4^{+}$ is assumed to be $x$.

The liquid components do not affect the value of the rate constant.

The ICE table for the stated reaction is,

$\begin{array}{cccccc}& {\text{NH}}_4^{+}\left(aq\right)& \rightleftharpoons & {\text{NH}}_3\left(aq\right)& +& {\text{H}}^{+}\left(aq\right)\\ \text{Inititial}\text{concentration}& 0.4& & 0& & 0\\ \text{Change}& -x& & +x& & +x\\ \text{Equilibrium}\text{concentration}& 0.4-x& & x& & x\end{array}$

The equilibrium concentration of $\left[{\text{NH}}_4^{+}\right]$ is $\left(0.4-x\right)\text{M}$.

The equilibrium concentration of $\left[{\text{NH}}_3\right]$ is $x\text{M}$.

The equilibrium concentration of $\left[{\text{H}}^{+}\right]$ is $x\text{M}$.

The calculated value of $K_{\text{a}}$ is $5.6\times {10}^{-10}$.

Substitute the value of $K_{\text{a}}$, $\left[{\text{NH}}_4^{+}\right]$, $\left[{\text{NH}}_3\right]$ and $\left[{\text{H}}^{+}\right]$ in equation (1).

$\begin{array}{c}5.6\times {10}^{-10}=\frac{\left[x\right]\left[x\right]}{\left[0.4-x\right]}\\ 5.6\times {10}^{-10}=\frac{{\left[x\right]}^2}{\left[0.4-x\right]}\end{array}$

Solve the above expression.

$\begin{array}{c}5.6\times {10}^{-10}=\frac{{\left[x\right]}^2}{\left[0.4-x\right]}\\ \left[x\right]=\underset{\_}{1.49\times 10^{-5}M}\end{array}$

Therefore, the $\left[{\text{H}}^{+}\right]$ is $\underset{\_}{1.49\times 10^{-5}M}$.

The $\text{pH}$ of the given solution is $\underset{\_}{4.827}$.

The $\text{pH}$ of a solution is calculated by the formula,

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

Substitute the value of $\left[{\text{H}}^{\text{+}}\right]$ in the above expression.

$\begin{array}{c}\text{pH}=-\log \left[1.49\times {10}^{-5}\right]\\ =\underset{\_}{4.827}\end{array}$