Chapter 14, Problem 97E

(a)

Interpretation Introduction

Interpretation: $\left[{\text{OH}}^{\text{-}}\right]\text{,}\left[{\text{H}}^{\text{+}}\right]\text{and}\text{pH}$ for the given set has 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,

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

Answer to Problem 97E

Answer

(a)

The $\left[{\text{OH}}^{-}\right]$ is $\underset{\_}{8.94\times 10^{-3}M}$. The $\text{pH}$ value is $\underset{\_}{11.95}$. The $\left[{\text{H}}^{+}\right]$ is $\underset{\_}{1.12\times 10^{-12}}$.

Explanation of Solution

To determine: The $\text{pH,}\left[{\text{H}}^{+}\right]$ and $\left[{\text{OH}}^{-}\right]$ of ${\left({\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\right)}_{\text{3}}\text{N}$.

The equilibrium constant expression for the given reaction is,

$K_{\text{b}}=\frac{\left[{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\right)}_{\text{3}}{\text{NH}}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\right)}_{\text{3}}\text{N}\right]}$

The reaction involved is,

${\left({\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\right)}_{\text{3}}{\text{N}}_{\left(aq\right)}+{\text{H}}_{\text{2}}{\text{O}}_{\left(l\right)}\rightleftharpoons {\left({\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\right)}_{\text{3}}{\text{NH}}^{+}{}_{\left(aq\right)}+{\text{OH}}^{-}{}_{\left(aq\right)}$

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,

$K_{\text{b}}=\frac{\left[{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\right)}_{\text{3}}{\text{NH}}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\right)}_{\text{3}}\text{N}\right]}$

The change in concentration of ${\left({\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\right)}_{\text{3}}\text{N}$ is assumed to be $x$.

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

$\begin{array}{cccccc}& {\left({\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\right)}_{\text{3}}{\text{N}}_{\left(aq\right)}& \rightleftharpoons & {\left({\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\right)}_{\text{3}}{\text{NH}}^{+}{}_{\left(aq\right)}& +& {\text{OH}}^{-}{}_{\left(aq\right)}\\ \text{Inititial}\text{concentration}& 0.20& & 0& & 0\\ \text{Change}& -x& & +x& & +x\\ \text{Equilibrium}\text{concentration}& 0.20-x& & x& & x\end{array}$

The equilibrium concentration of $\left[{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\right)}_{\text{3}}\text{N}\right]$ is $\left(0.20-x\right)\text{M}$.

The equilibrium concentration of $\left[{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\right)}_{\text{3}}{\text{NH}}^{+}\right]$ is $x\text{M}$.

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

The $K_{\text{b}}$ value is given to be $4.0\times {10}^{-4}$.

Substitute the value of $K_{\text{b}}$, $\left[{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\right)}_{\text{3}}\text{N}\right]$, $\left[{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\right)}_{\text{3}}{\text{NH}}^{+}\right]$ and $\left[{\text{OH}}^{-}\right]$

$\begin{array}{c}4.0\times {10}^{-4}=\frac{\left[x\right]\left[x\right]}{\left[0.20-x\right]}\\ 4.0\times {10}^{-4}=\frac{{\left[x\right]}^2}{\left[0.20-x\right]}\end{array}$

The value of $x$ will be very small as compared to $0.20$. Hence, it is ignored from the term $\left[0.20-x\right]$.

Simplify the above expression.

$\begin{array}{c}4.0\times {10}^{-4}=\frac{{\left[x\right]}^2}{\left[0.20\right]}\\ {\left[x\right]}^2=\left(8.0\times {10}^{-5}\right)\\ \left[x\right]=\underset{\_}{8.94\times 10^{-3}M}\end{array}$

Therefore, the $\left[{\text{OH}}^{-}\right]$ is $\underset{\_}{8.94\times 10^{-3}M}$.

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.

$\begin{array}{c}\text{pOH}=-\log \left[8.94\times {10}^{-3}\right]\\ =\underset{\_}{2.05}\end{array}$

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

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

$\begin{array}{c}\text{pH}+\text{2}\text{.05}=14\\ \text{pH}=\underset{\_}{11.95}\end{array}$

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

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

Rearrange the above expression to calculate the value of $\left[{\text{H}}^{+}\right]$.

$\left[{\text{H}}^{+}\right]={10}^{-\text{pH}}$

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

$\begin{array}{c}\left[{\text{H}}^{+}\right]={10}^{-11.95}\\ =\underset{\_}{1.12\times 10^{-12}}\end{array}$

(b)

Interpretation Introduction

Interpretation: $\left[{\text{OH}}^{\text{-}}\right]\text{,}\left[{\text{H}}^{\text{+}}\right]\text{and}\text{pH}$ for the given set has 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,

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

Answer to Problem 97E

Answer

(b)

The $\left[{\text{OH}}^{-}\right]$ is $\underset{\_}{4.69\times 10^{-5}M}$. The $\text{pH}$ value is $\underset{\_}{9.67}$. The $\left[{\text{H}}^{+}\right]$ is $\underset{\_}{2.14\times 10^{-10}}$.

Explanation of Solution

To determine: The $\text{pH,}\left[{\text{H}}^{+}\right]$ and $\left[{\text{OH}}^{-}\right]$ of ${\text{HONH}}_2$.

The equilibrium constant expression for the given reaction is,

$K_{\text{b}}=\frac{\left[{\text{HONH}}_3^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{HONH}}_2\right]}$

The reaction involved is,

${\text{HONH}}_2{}_{\left(aq\right)}+{\text{H}}_{\text{2}}{\text{O}}_{\left(l\right)}\rightleftharpoons {\text{HONH}}_3^{+}{}_{\left(aq\right)}+{\text{OH}}^{-}{}_{\left(aq\right)}$

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,

$K_{\text{b}}=\frac{\left[{\text{HONH}}_3^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{HONH}}_2\right]}$ (1)

The change in concentration of ${\text{HONH}}_2$ is assumed to be $x$.

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

$\begin{array}{cccccc}& {\text{HONH}}_2{}_{\left(aq\right)}& \rightleftharpoons & {\text{HONH}}_3^{+}{}_{\left(aq\right)}& +& {\text{OH}}^{-}{}_{\left(aq\right)}\\ \text{Inititial}\text{concentration}& 0.20& & 0& & 0\\ \text{Change}& -x& & +x& & +x\\ \text{Equilibrium}\text{concentration}& 0.20-x& & x& & x\end{array}$

The equilibrium concentration of $\left[{\text{HONH}}_2\right]$ is $\left(0.20-x\right)\text{M}$.

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

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

The given $K_{\text{b}}$ value is given to be $1.1\times {10}^{-8}$.

Substitute the value of $K_{\text{b}}$, $\left[{\text{HONH}}_2\right]$, $\left[{\text{HONH}}_3^{+}\right]$ and $\left[{\text{OH}}^{-}\right]$ in equation (1).

$\begin{array}{c}1.1\times {10}^{-8}=\frac{\left[x\right]\left[x\right]}{\left[0.20-x\right]}\\ 1.1\times {10}^{-8}=\frac{{\left[x\right]}^2}{\left[0.20-x\right]}\end{array}$

The value of $x$ will be very small as compared to $0.20$. Hence, it is ignored from the term $\left[0.20-x\right]$.

Simplify the above expression.

$\begin{array}{c}1.1\times {10}^{-8}=\frac{{\left[x\right]}^2}{\left[0.20\right]}\\ {\left[x\right]}^2=\left(2.2\times {10}^{-9}\right)\\ \left[x\right]=\underset{\_}{4.69\times 10^{-5}M}\end{array}$

Therefore, the $\left[{\text{OH}}^{-}\right]$ is $\underset{\_}{4.69\times 10^{-5}M}$.

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.

$\begin{array}{c}\text{pOH}=-\log \left[4.69\times {10}^{-5}\right]\\ =\underset{\_}{4.33}\end{array}$

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

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

$\begin{array}{c}\text{pH}+\text{4}\text{.33}=14\\ \text{pH}=\underset{\_}{9.67}\end{array}$

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

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

Rearrange the above expression to calculate the value of $\left[{\text{H}}^{+}\right]$.

$\left[{\text{H}}^{+}\right]={10}^{-\text{pH}}$

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

$\begin{array}{c}\left[{\text{H}}^{+}\right]={10}^{-9.67}\\ =\underset{\_}{2.14\times 10^{-10}}\end{array}$