Chapter 13, Problem 123E

An unknown salt is either NaCN, NaC₂H₃O₂, NaF, NaCl, or NaOCl. When 0.100 mole of the salt is dissolved in 1.00 L of solution, the pH of the solution is 8.07. What is the identity of the salt?

Interpretation Introduction

Interpretation: Identity the given unknown salt is to be determined.

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,

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

Answer to Problem 123E

Answer

The given salt is $\text{NaF}$.

Explanation of Solution

Explanation

To determine: If the given salt is $\text{NaCN}$.

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

The dominant equilibrium reaction is,

${\text{CN}}^{\text{-}}{}_{\left(\text{aq}\right)}\rightleftharpoons {\text{HCN}}_{\left(\text{aq}\right)}+{\text{OH}}^{-}{}_{\left(\text{aq}\right)}$

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,

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

The $K_{\text{b}}$ value is $\underset{\_}{1.613\times 10^{-5}}$.

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

The value of $K_{\text{a}}$ for $\text{HCN}$ is $6.2\times {10}^{-10}$.

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.

$\begin{array}{c}{K}_{\text{b}}=\frac{1.0\times {10}^{-14}}{6.2\times {10}^{-10}}\\ =\underset{\_}{1.613\times 10^{-5}}\end{array}$

The $\left[{\text{OH}}^{-}\right]$ is $\underset{\_}{1.27\times 10^{-3}M}$.

The change in concentration of ${\text{CN}}^{-}$ 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,

$\begin{array}{cccccc}& {\text{CN}}^{-}\left(aq\right)& \rightleftharpoons & \text{HCN}\left(aq\right)& +& {\text{OH}}^{-}\left(aq\right)\\ \text{Inititial}\text{concentration}& 0.100& & 0& & 0\\ \text{Change}& -x& & +x& & +x\\ \text{Equilibrium}\text{concentration}& 0.100-x& & x& & x\end{array}$

The equilibrium concentration of $\left[{\text{CN}}^{-}\right]$ is $\left(0.100-x\right)\text{M}$.

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

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

The calculated value of $K_{\text{b}}$ is $1.613\times {10}^{-5}$.

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

$\begin{array}{c}1.613\times {10}^{-5}=\frac{\left[x\right]\left[x\right]}{\left[0.100-x\right]}\\ 1.613\times {10}^{-5}=\frac{{\left[x\right]}^2}{\left[0.100-x\right]}\end{array}$

Solve the above expression.

$\begin{array}{c}1.613\times {10}^{-5}=\frac{{\left[x\right]}^2}{\left[0.100-x\right]}\\ \left[x\right]=\underset{\_}{1.27\times 10^{-3}M}\end{array}$

Therefore, the $\left[{\text{OH}}^{-}\right]$ is $\underset{\_}{1.27\times 10^{-3}M}$.

The $\text{pOH}$ of the given solution is $\underset{\_}{2.9}$.

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.

$\begin{array}{c}\text{pOH}=-\log \left[1.27\times {10}^{-3}\right]\\ =\underset{\_}{2.9}\end{array}$

The $\text{pH}$ of the given solution is $\underset{\_}{11.1}$. Therefore, it is not the given salt.

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.

$\begin{array}{c}\text{pH}=14-\text{2}\text{.9}\\ =\underset{\_}{11.1}\end{array}$

The $\text{pH}$ of the given unknown salt is $8.07$. Therefore, it is not the given salt.

Interpretation Introduction

Interpretation: Identity the given unknown salt is to be determined.

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: If the given salt is ${\text{NaC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$.

Explanation of Solution

Explanation

The equilibrium constant expression for the given reaction is, $K_{\text{b}}=\frac{\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_2\text{H}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_2^{-}\right]}$

The dominant equilibrium reaction is,

${\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_2^{-}\left(aq\right)\rightleftharpoons {\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_2\text{H}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$

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,

$K_{\text{b}}=\frac{\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_2\text{H}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_2^{-}\right]}$ (1)

The $K_{\text{b}}$ value is $\underset{\_}{5.5\times 10^{-10}}$.

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

The value of $K_{\text{a}}$ for ${\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_2\text{H}$ is $1.8\times {10}^{-5}$.

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.

$\begin{array}{c}{K}_{\text{b}}=\frac{1.0\times {10}^{-14}}{1.8\times {10}^{-5}}\\ =\underset{\_}{5.5\times 10^{-10}}\end{array}$

The $\left[{\text{OH}}^{-}\right]$ is $\underset{\_}{7.415\times 10^{-6}M}$.

The change in concentration of ${\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_2^{-}$ 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,

$\begin{array}{cccccc}& {\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_2^{-}\left(aq\right)& \rightleftharpoons & {\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_2\text{H}\left(aq\right)& +& {\text{OH}}^{-}\left(aq\right)\\ \text{Inititial}\text{concentration}& 0.100& & 0& & 0\\ \text{Change}& -x& & +x& & +x\\ \text{Equilibrium}\text{concentration}& 0.100-x& & x& & x\end{array}$

The equilibrium concentration of $\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_2^{-}\right]$ is $\left(0.100-x\right)\text{M}$.

The equilibrium concentration of $\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_2\text{H}\right]$ is $x\text{M}$.

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

The calculated value of $K_{\text{b}}$ is $5.5\times {10}^{-10}$.

Substitute the value of $K_{\text{b}}$, $\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_2^{-}\right]$, $\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_2\text{H}\right]$ and $\left[{\text{OH}}^{-}\right]$ in equation (1).

$\begin{array}{c}5.5\times {10}^{-10}=\frac{\left[x\right]\left[x\right]}{\left[0.100-x\right]}\\ 5.5\times {10}^{-10}=\frac{{\left[x\right]}^2}{\left[0.100-x\right]}\end{array}$

Solve the above expression.

$\begin{array}{c}5.5\times {10}^{-10}=\frac{{\left[x\right]}^2}{\left[0.100-x\right]}\\ \left[x\right]=\underset{\_}{7.415\times 10^{-6}M}\end{array}$

Therefore, the $\left[{\text{OH}}^{-}\right]$ is $\underset{\_}{7.415\times 10^{-6}M}$.

The $\text{pOH}$ of the given solution is $\underset{\_}{5.13}$.

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.

$\begin{array}{c}\text{pOH}=-\log \left[7.415\times {10}^{-6}\right]\\ =\underset{\_}{5.13}\end{array}$

The $\text{pH}$ of the given solution is $\underset{\_}{8.87}$. Therefore, it is not the given salt.

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.

$\begin{array}{c}\text{pH}=14-5.13\\ =\underset{\_}{8.87}\end{array}$

The $\text{pH}$ of the given unknown salt is $8.07$. Therefore, it is not the given salt.

Interpretation Introduction

Interpretation: Identity the given unknown salt is to be determined.

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: If the given salt is $\text{NaF}$.

Explanation of Solution

Explanation

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

The dominant equilibrium reaction is,

${\text{F}}^{-}\left(aq\right)\rightleftharpoons \text{HF}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$

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,

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

The $K_{\text{b}}$ value is $\underset{\_}{1.39\times 10^{-11}}$.

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

The value of $K_{\text{a}}$ for $\text{HF}$ is $7.2\times {10}^{-4}$.

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.

$\begin{array}{c}{K}_{\text{b}}=\frac{1.0\times {10}^{-14}}{7.2\times {10}^{-4}}\\ =\underset{\_}{1.39\times 10^{-11}}\end{array}$

The $\left[{\text{OH}}^{-}\right]$ is $\underset{\_}{1.179\times 10^{-6}M}$.

The change in concentration of ${\text{F}}^{-}$ 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,

$\begin{array}{cccccc}& {\text{F}}^{-}\left(aq\right)& \rightleftharpoons & \text{HF}\left(aq\right)& +& {\text{OH}}^{-}\left(aq\right)\\ \text{Inititial}\text{concentration}& 0.100& & 0& & 0\\ \text{Change}& -x& & +x& & +x\\ \text{Equilibrium}\text{concentration}& 0.100-x& & x& & x\end{array}$

The equilibrium concentration of $\left[{\text{F}}^{-}\right]$ is $\left(0.100-x\right)\text{M}$.

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

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

The calculated value of $K_{\text{b}}$ is $1.39\times {10}^{-11}$.

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

$\begin{array}{c}1.39\times {10}^{-11}=\frac{\left[x\right]\left[x\right]}{\left[0.100-x\right]}\\ 1.39\times {10}^{-11}=\frac{{\left[x\right]}^2}{\left[0.100-x\right]}\end{array}$

Solve the above expression.

$\begin{array}{c}1.39\times {10}^{-11}=\frac{{\left[x\right]}^2}{\left[0.100-x\right]}\\ \left[x\right]=\underset{\_}{1.179\times 10^{-6}M}\end{array}$

Therefore, the $\left[{\text{OH}}^{-}\right]$ is $\underset{\_}{1.179\times 10^{-6}M}$.

The $\text{pOH}$ of the given solution is $\underset{\_}{5.929}$.

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.

$\begin{array}{c}\text{pOH}=-\log \left[1.179\times {10}^{-6}\right]\\ =\underset{\_}{5.929}\end{array}$

The $\text{pH}$ of the given solution is $\underset{\_}{8.071}$. Therefore, it is the given salt.

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.

$\begin{array}{c}\text{pH}=14-5.929\\ =\underset{\_}{8.071}\end{array}$

The $\text{pH}$ of the given unknown salt is $8.07$. Therefore, it is the given salt.

Interpretation Introduction

Interpretation: Identity the given unknown salt is to be determined.

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: If the given salt is $\text{NaCl}$.

Explanation of Solution

Explanation

The given salt is not $\text{NaCl}$.

$\text{NaCl}$ is a salt made up of a strong acid, that is $\text{HCl}$ and a strong base, that is $\text{NaOH}$. The aqueous solution of $\text{NaCl}$ will have a $\text{pH}$ value of around $7$ (will be relatively neutral).

Therefore, the given salt is not $\text{NaCl}$.

Interpretation Introduction

Interpretation: Identity the given unknown salt is to be determined.

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: If the given salt is $\text{NaOCl}$.

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 $\text{HF}$ 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 $\underset{\_}{1.69\times 10^{-4}M}$.

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,

$\begin{array}{cccccc}& {\text{OCl}}^{-}\left(aq\right)& \rightleftharpoons & \text{HOCl}\left(aq\right)& +& {\text{OH}}^{-}\left(aq\right)\\ \text{Inititial}\text{concentration}& 0.100& & 0& & 0\\ \text{Change}& -x& & +x& & +x\\ \text{Equilibrium}\text{concentration}& 0.100-x& & x& & x\end{array}$

The equilibrium concentration of $\left[{\text{OCl}}^{-}\right]$ is $\left(0.100-x\right)\text{M}$.

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).

$\begin{array}{c}2.857\times {10}^{-7}=\frac{\left[x\right]\left[x\right]}{\left[0.100-x\right]}\\ 2.857\times {10}^{-7}=\frac{{\left[x\right]}^2}{\left[0.100-x\right]}\end{array}$

Solve the above expression.

$\begin{array}{c}2.857\times {10}^{-7}=\frac{{\left[x\right]}^2}{\left[0.100-x\right]}\\ \left[x\right]=\underset{\_}{1.69\times 10^{-4}M}\end{array}$

Therefore, the $\left[{\text{OH}}^{-}\right]$ is $\underset{\_}{1.69\times 10^{-4}M}$.

The $\text{pOH}$ of the given solution is $\underset{\_}{3.77}$.

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.

$\begin{array}{c}\text{pOH}=-\log \left[1.69\times {10}^{-4}\right]\\ =\underset{\_}{3.77}\end{array}$

The $\text{pH}$ of the given solution is $\underset{\_}{10.23}$. Therefore, it is not the given salt.

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.

$\begin{array}{c}\text{pH}=14-3.77\\ =\underset{\_}{10.23}\end{array}$

The $\text{pH}$ of the given unknown salt is $8.07$. Therefore, it is not the given salt.