Chapter 13, Problem 74QAP

Consider pyridine, C₅H₅N, a pesticide and deer repellent. Its conjugate acid, C₅H₅NH⁺, has $K_a=6.7\times {10}^{-6}$.

(a) Write a balanced net ionic equation for the reaction that shows the basicity of aqueous solutions of pyridine.

(b) Calculate K_b for the reaction in (a).

(c) Find the pH of a solution prepared by mixing 2.74 g of pyridine in enough water to make 685 mL of solution.

Interpretation Introduction

(a)

Interpretation:

The base dissociation reaction of pyridine needs to be determined.

Concept introduction:

The dissociation reaction of a weak base is represented as follows:

$\text{BOH}\rightleftarrows {\text{B}}^{+}+{\text{OH}}^{-}$

The expression for the base dissociation constant will be as follows:

$K_b=\frac{\left[{\text{B}}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[\text{BOH}\right]}$

Here, $\left[{\text{B}}^{+}\right]$ is concentration of conjugate acid, $\left[{\text{OH}}^{-}\right]$ is concentration of hydroxyl ion and $\left[\text{BOH}\right]$ is the concentration of weak base.

Answer to Problem 74QAP

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

Explanation of Solution

The basic nature of pyridine, ${\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}$ in water can be represented as follows:

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

Here, ${\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}$ acts as base and can accept hydrogen ion from water to form cation ${\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}{\mathrm{H}}^{+}$ and hydroxide ion ${\text{OH}}^{-}$.

For the given base, the expression for base dissociation constant can be calculated as follows:

$K_b=\frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}$

Interpretation Introduction

(b)

Interpretation:

The base ionization constant for the pyridine needs to be determined.

Concept introduction:

The dissociation reaction of a weak base is represented as follows:

$\text{BOH}\rightleftarrows {\text{B}}^{+}+{\text{OH}}^{-}$

The expression for the base dissociation constant will be as follows:

$K_b=\frac{\left[{\text{B}}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[\text{BOH}\right]}$

Here, $\left[{\text{B}}^{+}\right]$ is concentration of conjugate acid, $\left[{\text{OH}}^{-}\right]$ is concentration of hydroxyl ion and $\left[\text{BOH}\right]$ is the concentration of weak base.

The relation between the acid dissociation constant and weak dissociation constant is as follows:

$K_w=K_a\times K_b$

Here, $K_w$ is the ionic product of water and its value is constant that is ${10}^{-14}$.

Answer to Problem 74QAP

$1.50\times {10}^{-9}$

Explanation of Solution

The conjugate acid of the weak base ${\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}$ is ${\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}{\mathrm{H}}^{+}$. The acid dissociation constant for the conjugate acid is $6.7\times {10}^{-6}$.

The relation between the acid dissociation constant and weak dissociation constant is as follows:

$K_w=K_a\times K_b$

Here, $K_w$ is the ionic product of water and its value is constant that is ${10}^{-14}$.

On rearranging, the base dissociation constant can be calculated as follows:

$K_b=\frac{K_w}{K_a}$

Putting the values,

$K_b=\frac{{10}^{-14}}{6.7\times {10}^{-6}}=1.50\times {10}^{-9}$

Thus, the value of $K_b$ is $1.50\times {10}^{-9}$.

Interpretation Introduction

(c)

Interpretation:

The pH of the given solution of pyridine needs to be determined.

Concept introduction:

The dissociation reaction of a weak base is represented as follows:

$\text{BOH}\rightleftarrows {\text{B}}^{+}+{\text{OH}}^{-}$

The expression for the base dissociation constant will be as follows:

$K_b=\frac{\left[{\text{B}}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[\text{BOH}\right]}$

Here, $\left[{\text{B}}^{+}\right]$ is concentration of conjugate acid, $\left[{\text{OH}}^{-}\right]$ is concentration of hydroxyl ion and $\left[\text{BOH}\right]$ is the concentration of weak base.

The pOH of a solution is calculated as follows:

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

From pOH, pH of a solution is calculated as follows:

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

Answer to Problem 74QAP

$\text{8}\text{.94}$

Explanation of Solution

The mass of the pyridine is 2.74 g and volume of solution is 685 mL.

The molar mass of pyridine is 79.1 g/mol. The number of moles of pyridine can be calculated as follows:

$n=\frac{m}{M}$

Putting the values,

$n=\frac{2.74\text{ g}}{79.1\text{ g/mol}}=0.03464\text{ mol}$

From the number of moles and volume, molarity of solution can be calculated as follows:

$M=\frac{n}{V\left(\text{in L}\right)}$

Putting the values,

$M=\frac{\left(0.03464\text{ mol}\right)}{\left(685\text{ mL}\right)\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)}=0.0506\text{ mol/L}\left(\frac{1\text{ M}}{1\text{ mol/L}}\right)=0.0506\text{ M}$

Thus, the molarity of pyridine is $0.0506\text{ M}$.

The base ionization reaction of the ${\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}$ is as follows:

${\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{N+H}}_{\text{2}}\text{O}\rightleftharpoons {\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}+{\text{OH}}^{-}$

The concentration of all species can be calculated from the ICE table as follows:

$\begin{array}{l}{\text{ C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{N+H}}_{\text{2}}\text{O}\rightleftharpoons {\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}+{\text{OH}}^{-}\\ I\text{ 0}\text{.0506 - - -}\\ C-x+x+x\\ E\text{ 0}\text{.0506-}x+x+x\end{array}$

Now, putting the values in the base dissociation constant as follows:

$K_b=\frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}=\frac{\left(x\right)\left(x\right)}{\left(0.0506-x\right)}$

Or,

$1.50\times {10}^{-9}=\frac{x^2}{0.0506-x}$

Rearranging,

$7.6\times {10}^{-11}-1.50\times {10}^{-9}x=x^2$

Or,

$x^2+1.50\times {10}^{-9}x-7.6\times {10}^{-11}=0$

This can be related to the quadratic equation as follows:

$a{x}^2+bx+c=0$

The value of x can be calculated using the following relation:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Putting the values,

$\begin{array}{l}x=\frac{-1.50\times {10}^{-9}\pm \sqrt{{\left(1.50\times {10}^{-9}\right)}^2-4\left(1\right)\left(-7.6\times {10}^{-11}\right)}}{2\left(1\right)}\\ =+8.72\times {10}^{-6},-8.72\times {10}^{-6}\end{array}$

Since, the value of x cannot be negative thus, it is equal to $8.72\times {10}^{-6}$.

From the ICE table:

$\left[{\text{OH}}^{-}\right]=8.72\times {10}^{-6}$

Thus, pOH of solution can be calculated as follows:

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

Putting the value,

$\text{pOH=}-\text{log}\left(8.72\times {10}^{-6}\right)=5.06$

From the pOH of the solution, pH can be calculated as follows:

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

Putting the values,

$\text{pH}=14-5.06\text{=8}\text{.94}$

Therefore, the pH of the solution is $\text{8}\text{.94}$.