Chapter 14, Problem 74E

Using the K_a values in Table 14.2, calculate the percent dissociation in a 0.20-M solution of each of the following acids.

  a. nitric acid (HNO₃)

  b. nitrous acid (HNO₂)

  c. phenol (HOC₆H₅)

  d. How is percent dissociation of an acid related to the K_a value for the acid (assuming equal initial concentrations of acids)?

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,

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

Answer to Problem 74E

(a)

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

(b)

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

(c)

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 $\underset{\_}{0.0028\%}$.

(d)

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

Explanation of Solution

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

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

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.

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

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,

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

Answer to Problem 74E

(a)

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

(b)

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

(c)

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 $\underset{\_}{0.0028\%}$.

(d)

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

Explanation of Solution

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

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,

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

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,

${\text{K}}_{\text{a}}$ 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,

$\begin{array}{cccccc}& {\text{HNO}}_{\text{2}}{}_{\left(\text{aq}\right)}& \rightleftharpoons & {\text{H}}^{\text{+}}{}_{\left(\text{aq}\right)}& \text{+}& {\text{NO}}_{\text{2}}^{\text{-}}{}_{\left(\text{aq}\right)}\\ \text{Inititial}\text{concentration}& \text{0}\text{.20}& & \text{0}& & \text{0}\\ \text{Change}& \text{-x}& & \text{+x}& & \text{+x}\\ \text{Equilibrium}\text{concentration}& \text{0}\text{.20-x}& & \text{x}& & \text{x}\end{array}$

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

Substitute the value of ${\text{K}}_{\text{a}}$, $\left[{\text{HNO}}_2\right]$, $\left[{\text{H}}^{+}\right]$ 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 $0.20\text{M}$. 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 $\left[{\text{H}}^{+}\right]$ that is equal to $x$ is $\underset{\_}{8.9\times 10^{-3}M}$.

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

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 $\left({\text{HNO}}_{\text{2}}\right)$ in the above expression.

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

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,

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

Answer to Problem 74E

(a)

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

(b)

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

(c)

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 $\underset{\_}{0.0028\%}$.

(d)

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

Explanation of Solution

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

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,

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

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,

$\begin{array}{cccccc}& {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{OH}}_{\left(\text{aq}\right)}& \rightleftharpoons & {\text{H}}^{\text{+}}{}_{\left(\text{aq}\right)}& \text{+}& {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}^{\text{-}}{}_{\left(\text{aq}\right)}\\ \text{Inititial}\text{concentration}& \text{0}\text{.20}& & \text{0}& & \text{0}\\ \text{Change}& \text{-x}& & \text{+x}& & \text{+x}\\ \text{Equilibrium}\text{concentration}& \text{0}\text{.20-x}& & \text{x}& & \text{x}\end{array}$

The equilibrium concentration of $\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{OH}\right]$ is $\left(\text{0}\text{.20-x}\right)\text{M}$.

The equilibrium concentration of $\left[{\text{H}}^{\text{+}}\right]$ 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 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 $\underset{\_}{0.0028\%}$.

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)

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,

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

Answer to Problem 74E

(a)

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

(b)

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

(c)

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 $\underset{\_}{0.0028\%}$.

(d)

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

Explanation of Solution

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

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 ${\text{HNO}}_{\text{2}}$ is $4.0\times {10}^{-4}$.

The percent dissociation of ${\text{HNO}}_{\text{2}}$ 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.