Chapter 13, Problem 74E

(a)

Interpretation Introduction

Interpretation: The percent dissociation of the acid in each of the given solutions is to be calculated.

Concept introduction: 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 percent dissociation of an acid is calculated by the formula,

$\text{Percent}\text{dissociation}=\frac{\text{Equilibrium}\text{concentration}\text{of}\left[{\text{H}}^{+}\right]}{\text{Initial}\text{concentration}\text{of}\text{the}\text{acid}}\times 100$

To determine: The percent dissociation for a $0.20\text{M}$ solution of nitric acid $\left({\text{HNO}}_{\text{3}}\right)$.

Answer to Problem 74E

Answer

The percent dissociation for the given $0.20\text{M}$ solution of nitric acid $\left({\text{HNO}}_{\text{3}}\right)$ is $\underset{\_}{100\%}$.

Explanation of Solution

Explanation

${\text{HNO}}_{\text{3}}$ is a strong acid. A strong acid completely dissociates in water.

Therefore, the $\left[{\text{H}}^{\text{+}}\right]$ is equal to the initial concentration of the acid.

Therefore, the $\left[{\text{H}}^{\text{+}}\right]$ is $\underset{\_}{0.20M}$.

The equilibrium concentration of $\left[{\text{H}}^{\text{+}}\right]$ is $0.20\text{M}$.

The initial concentration of ${\text{HNO}}_{\text{3}}$ is $0.20\text{M}$.

The percent dissociation of an acid is calculated by the formula,

$\text{Percent}\text{dissociation}=\frac{\text{Equilibrium}\text{concentration}\text{of}\left[{\text{H}}^{+}\right]}{\text{Initial}\text{concentration}\text{of}\text{the}\text{acid}}\times 100$

Substitute the value of the concentration of $\left[{\text{H}}^{\text{+}}\right]$ and the initial concentration of ${\text{HNO}}_{\text{3}}$ in the above expression.

$\begin{array}{c}\text{Percent}\text{dissociation}=\left(\frac{0.20}{\text{0}\text{.20}}\times 100\right)\%\\ =\underset{\_}{100\%}\end{array}$

(b)

Interpretation Introduction

Interpretation: The percent dissociation of the acid in each of the given solutions is to be calculated.

Concept introduction: 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 percent dissociation of an acid is calculated by the formula,

$\text{Percent}\text{dissociation}=\frac{\text{Equilibrium}\text{concentration}\text{of}\left[{\text{H}}^{+}\right]}{\text{Initial}\text{concentration}\text{of}\text{the}\text{acid}}\times 100$

To determine: The percent dissociation for a $0.20\text{M}$ solution of nitrous acid $\left({\text{HNO}}_2\right)$.

Answer to Problem 74E

Answer

The percent dissociation for the given $0.20\text{M}$ solution of nitrous acid $\left({\text{HNO}}_{\text{2}}\right)$ is $\underset{\_}{4.45\%}$.

Explanation of Solution

Explanation

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

${\text{HNO}}_2$ is a comparatively stronger acid than ${\text{H}}_{\text{2}}\text{O}$.

The dominant equilibrium reaction for the given case is,

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

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,

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

The $\left[{\text{H}}^{\text{+}}\right]$ is $\underset{\_}{8.9\times 10^{-3}M}$.

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

The ICE table for the stated reaction is,

$\begin{array}{cccccc}& {\text{HNO}}_2\left(aq\right)& \rightleftharpoons & {\text{H}}^{+}\left(aq\right)& +& {\text{NO}}_2^{-}\left(aq\right)\\ \text{Inititial}\text{concentration}& 0.20& & 0& & 0\\ \text{Change}& -x& & +x& & +x\\ \text{Equilibrium}\text{concentration}& 0.20-x& & x& & x\end{array}$

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

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

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

The $K_{\text{a}}$ for ${\text{HNO}}_2$ is $4.0\times {10}^{-4}$.

Substitute the value of $K_{\text{a}}$, $\left[{\text{HNO}}_2\right]$, $\left[{\text{H}}^{+}\right]$ and $\left[{\text{HNO}}_2^{-}\right]$ in equation (1).

$\begin{array}{c}{K}_{\text{a}}=\frac{\left[{\text{H}}^{+}\right]\left[{\text{NO}}_2^{-}\right]}{\left[{\text{HNO}}_2\right]}\\ 4.0\times {10}^{-4}=\frac{\left[x\right]\left[x\right]}{\left[0.20-x\right]}\\ 4.0\times {10}^{-4}=\frac{{\left[x\right]}^2}{\left[0.20-x\right]}\end{array}$

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.

$\begin{array}{c}4.0\times {10}^{-4}=\frac{{\left[x\right]}^2}{\left[0.20\right]}\\ {\left[x\right]}^2=\left(8.0\times {10}^{-5}\right)\\ \left[x\right]=\underset{\_}{8.9\times 10^{-3}M}\end{array}$

Therefore, the $\left[{\text{H}}^{\text{+}}\right]$ that is equal to $x$ is $\underset{\_}{8.9\times 10^{-3}M}$.

The calculated concentration of $\left[{\text{H}}^{\text{+}}\right]$ is $\text{8}{\text{.9×10}}^{-3}\text{M}$.

The initial concentration of ${\text{HNO}}_{\text{2}}$ is $0.20\text{M}$.

The percent dissociation of an acid is calculated by the formula,

$\text{Percent}\text{dissociation}=\frac{\text{Equilibrium}\text{concentration}\text{of}\left[{\text{H}}^{+}\right]}{\text{Initial}\text{concentration}\text{of}\text{the}\text{acid}}\times 100$

Substitute the value of the concentration of $\left[{\text{H}}^{\text{+}}\right]$ and the initial concentration of ${\text{HNO}}_{\text{2}}$ in the above expression.

$\begin{array}{c}\text{Percent}\text{dissociation}=\left(\frac{\text{8}{\text{.9×10}}^{-3}}{\text{0}\text{.20}}\times 100\right)\%\\ =\underset{\_}{4.45\%}\end{array}$

(c)

Interpretation Introduction

Interpretation: The percent dissociation of the acid in each of the given solutions is to be calculated.

Concept introduction: 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 percent dissociation of an acid is calculated by the formula,

$\text{Percent}\text{dissociation}=\frac{\text{Equilibrium}\text{concentration}\text{of}\left[{\text{H}}^{+}\right]}{\text{Initial}\text{concentration}\text{of}\text{the}\text{acid}}\times 100$

To determine: The percent dissociation for a $0.20\text{M}$ solution of phenol $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{OH}\right)$.

Answer to Problem 74E

Answer

The percent dissociation for the given $0.20\text{M}$ solution of phenol $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{OH}\right)$ is $\underset{\_}{0.0028\%}$.

Explanation of Solution

Explanation

The equilibrium constant expression for the given reaction is, $K_{\text{a}}=\frac{\left[{\text{H}}^{+}\right]\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}^{-}\right]}{\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{OH}\right]}$

${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{OH}$ is a comparatively stronger acid than ${\text{H}}_{\text{2}}\text{O}$.

The dominant equilibrium reaction for the given case is,

${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{OH}\left(aq\right)\rightleftharpoons {\text{H}}^{+}\left(aq\right)+{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}^{-}\left(aq\right)$

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,

$K_{\text{a}}=\frac{\left[{\text{H}}^{+}\right]\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}^{-}\right]}{\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{OH}\right]}$ (1)

The $\left[{\text{H}}^{\text{+}}\right]$ is $\underset{\_}{5.66\times 10^{-6}M}$.

The change in concentration of ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{OH}$ is assumed to be $x$.

The ICE table for the stated reaction is,

$\begin{array}{cccccc}& {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{OH}\left(aq\right)& \rightleftharpoons & {\text{H}}^{+}\left(aq\right)& +& {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}^{-}\left(aq\right)\\ \text{Inititial}\text{concentration}& 0.20& & 0& & 0\\ \text{Change}& -x& & +x& & +x\\ \text{Equilibrium}\text{concentration}& 0.20-x& & x& & x\end{array}$

The equilibrium concentration of $\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{OH}\right]$ is $\left(0.20-x\right)\text{M}$.

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

The equilibrium concentration of $\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}^{-}\right]$ is $x\text{M}$.

The $K_{\text{a}}$ for ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{OH}$ is $1.6\times {10}^{-10}$.

Substitute the value of $K_{\text{a}}$, $\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{OH}\right]$, $\left[{\text{H}}^{+}\right]$ and $\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}^{-}\right]$ in equation (1).

$\begin{array}{c}{K}_{\text{a}}=\frac{\left[{\text{H}}^{+}\right]\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}^{-}\right]}{\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{OH}\right]}\\ 1.6\times {10}^{-10}=\frac{\left[x\right]\left[x\right]}{\left[0.20-x\right]}\\ 1.6\times {10}^{-10}=\frac{{\left[x\right]}^2}{\left[0.20-x\right]}\end{array}$

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.

$\begin{array}{c}1.6\times {10}^{-10}=\frac{{\left[x\right]}^2}{\left[0.20\right]}\\ {\left[x\right]}^2=\left(3.2\times {10}^{-11}\right)\\ \left[x\right]=\underset{\_}{5.66\times 10^{-6}M}\end{array}$

Therefore, the $\left[{\text{H}}^{\text{+}}\right]$ that is equal to $x$ is $\underset{\_}{5.66\times 10^{-6}M}$.

The calculated concentration of $\left[{\text{H}}^{\text{+}}\right]$ is $\text{5}{\text{.66×10}}^{-6}\text{M}$.

The initial concentration of ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{OH}$ is $0.20\text{M}$.

The percent dissociation of an acid is calculated by the formula,

$\text{Percent}\text{dissociation}=\frac{\text{Equilibrium}\text{concentration}\text{of}\left[{\text{H}}^{+}\right]}{\text{Initial}\text{concentration}\text{of}\text{the}\text{acid}}\times 100$

Substitute the value of the concentration of $\left[{\text{H}}^{\text{+}}\right]$ and the initial concentration of ${\text{HNO}}_{\text{2}}$ in the above expression.

$\begin{array}{c}\text{Percent}\text{dissociation}=\left(\frac{\text{5}{\text{.66×10}}^{-6}}{\text{0}\text{.20}}\times 100\right)\%\\ =\underset{\_}{0.0028\%}\end{array}$ (1)

(d)

Interpretation Introduction

Interpretation: The percent dissociation of the acid in each of the given solutions is to be calculated.

Concept introduction: 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 percent dissociation of an acid is calculated by the formula,

$\text{Percent}\text{dissociation}=\frac{\text{Equilibrium}\text{concentration}\text{of}\left[{\text{H}}^{+}\right]}{\text{Initial}\text{concentration}\text{of}\text{the}\text{acid}}\times 100$

To determine: The relation of percent dissociation of an acid and the $K_{\text{a}}$ value for the acid.

Answer to Problem 74E

Answer

Percent dissociation of an acid increases with an increase in the $K_{\text{a}}$ value for the acid.

Explanation of Solution

Explanation

The percent dissociation of an acid is directly proportional to the $K_{\text{a}}$ value for the acid.

The $K_{\text{a}}$ value for ${\text{HNO}}_{\text{3}}$ is $1$.

The percent dissociation of ${\text{HNO}}_{\text{3}}$ in water is $100\%$.

The $K_{\text{a}}$ value for ${\text{HNO}}_2$ is $4.0\times {10}^{-4}$.

The percent dissociation of ${\text{HNO}}_2$ in water is $4.45\%$.

The $K_{\text{a}}$ value for ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{OH}$ is $1.6\times {10}^{-10}$.

The percent dissociation of ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{OH}$ in water is $0.0028\%$.

Therefore, it can be stated that the percent dissociation of an acid increases with an increase in the $K_{\text{a}}$ value for the acid.