Chapter 15, Problem 15.58QP

(a)

Interpretation Introduction

Interpretation: The $\text{p}{K}_{\text{b}}$ value of quinoline is given to be $9.15$. The $\text{pH}$ of $0.0752\text{M}$ solution and hydroxide ion concentration of solution of quinoline is to be calculated.

Concept introduction: The $\text{pH}$ is the term used for expressing the extent of ${\text{H}}^{+}$ in the solution. The $\text{p}{K}_{\text{a}}$ and $\text{p}{K}_{\text{b}}$ values are more helpful to determine the nature of the substances. These both terms related to acidic and the basic nature of any compound.

To determine: The $\text{pH}$ of $0.0752\text{M}$ solution of quinoline.

Answer to Problem 15.58QP

Solution

The $\text{pH}$ of quinoline is $\underset{\_}{8.86}$.

Explanation of Solution

Explanation

Given

The $\text{p}{K}_{\text{b}}$ value of quinoline is $9.15$.

The ionization of quinoline is shown by the equation,

Quinoline $\rightleftharpoons$ Quinoline $\left({\text{H}}^{+}\right)$ $+$ ${\text{OH}}^{-}$

The $K_{\text{b}}$ of solution of quinoline is calculated by the formula,

$K_{\text{b}}={10}^{-\text{p}{K}_{\text{b}}}$ (1)

Substitute the value of $\text{p}{K}_{\text{b}}$ in equation (1).

$\begin{array}{c}{K}_{\text{b}}={10}^{-9.15}\\ =7.08\times {10}^{-10}\end{array}$

The calculation of concentration of Quinoline $\left({\text{H}}^{+}\right)$ and $[{\text{OH}}^{-}]$ is shown by the $\text{ICE}$ table.

$\begin{array}{cccc}& \text{Quinoline}& \text{Quinoline}\left({\text{H}}^{+}\right)& \left[{\text{OH}}^{-}\right]\\ \text{Initial}& 0.0752\text{M}& 0& 0\\ \text{Change}& -x& x& x\\ \text{Equilbrium}& \left(0.0752-x\right)\text{M}& x& x\end{array}$

The equilibrium constant is calculated by the formula,

$K_{\text{b}}=\frac{\left[{\text{Quinoline}}^{+}\right]\times \left[{\text{OH}}^{-}\right]}{\text{Quinoline}}$ (2)

Substitute the values of equilibrium constant, concentration of ions and compound in equation (2).

$\begin{array}{c}7.08\times {10}^{-10}\text{M}=\frac{\left(x\right)\times \left(x\right)}{\left(0.0752\text{M}\right)-x}\\ \approx \frac{x^2}{0.0752\text{M}}\end{array}$ (3)

In equation (3), $x$ is very small as compared to $\left(0.0752\text{M}\right)$. So, it is neglected.

$\begin{array}{c}{x}^2=\left(7.08\times {10}^{-10}\text{M}\right)\times \left(0.0752M\right)\\ x=\sqrt{5.32\times {10}^{-11}}\\ =7.3\times {10}^{-6}\text{M}\end{array}$

Thus, the concentration of both ions $\left[{\text{Quinoline}}^{+}\right]$ and $\left[{\text{OH}}^{-}\right]$ in the solution is

$7.3\times {10}^{-6}\text{M}$.

The $\text{pOH}$ is calculated by the formula,

$\text{pOH}=-\text{log}\left[{\text{OH}}^{-}\right]$ (4)

Substitute the value of $\left[{\text{OH}}^{-}\right]$ in equation (4).

$\begin{array}{c}\text{pOH}=-\text{log}\left(7.3\times {10}^{-6}\text{M}\right)\\ =5.14\end{array}$

The $\text{pH}$ is calculated by using $\text{pOH}$ shown as,

$\begin{array}{c}\text{pH}=14-\text{pOH}\\ =\text{14}-5.14\\ =\underset{\_}{8.86}\end{array}$

Therefore, the $\text{pH}$ of quinoline is $\underset{\_}{8.86}$.

(b)

Interpretation Introduction

To determine: The hydroxide ion concentration of given solution.

Answer to Problem 15.58QP

Solution

The concentration of $\left[{\text{OH}}^{-}\right]$ in the solution is $\underset{\_}{7.3\times 10^{-6}M}$.

Explanation of Solution

Explanation

Given

The $\text{p}{K}_{\text{b}}$ value of quinoline is $9.15$.

The ionization of quinoline is shown by the equation,

Quinoline $\rightleftharpoons$ Quinoline $\left({\text{H}}^{+}\right)$ $+$ ${\text{OH}}^{-}$

The $K_{\text{b}}$ of solution of quinoline is calculated by the formula,

$K_{\text{b}}={10}^{-\text{p}{K}_{\text{b}}}$ (1)

Substitute the value of $\text{p}{K}_{\text{b}}$ in equation (1).

$\begin{array}{c}{K}_{\text{b}}={10}^{-9.15}\\ =7.08\times {10}^{-10}\end{array}$

The calculation of concentration of Quinoline $\left({\text{H}}^{+}\right)$ and $[{\text{OH}}^{-}]$ is shown by the $\text{ICE}$ table.

$\begin{array}{cccc}& \text{Quinoline}& \text{Quinoline}\left({\text{H}}^{+}\right)& \left[{\text{OH}}^{-}\right]\\ \text{Initial}& 0.0752\text{M}& 0& 0\\ \text{Change}& -x& x& x\\ \text{Equilbrium}& \left(0.0752-x\right)\text{M}& x& x\end{array}$

The equilibrium constant is calculated by the formula,

$K_{\text{b}}=\frac{\left[{\text{Quinoline}}^{+}\right]\times \left[{\text{OH}}^{-}\right]}{\text{Quinoline}}$ (2)

Substitute the values of equilibrium constant, concentration of ions and compound in equation (2).

$\begin{array}{c}7.08\times {10}^{-10}\text{M}=\frac{\left(x\right)\times \left(x\right)}{\left(0.0752\text{M}\right)-x}\\ \approx \frac{x^2}{0.0752\text{M}}\end{array}$ (3)

In equation (3), $x$ is very small as compared to $\left(0.0752\text{M}\right)$. So, it is neglected.

$\begin{array}{c}{x}^2=\left(7.08\times {10}^{-10}\text{M}\right)\times \left(0.0752\text{M}\right)\\ x=\sqrt{5.32\times {10}^{-11}}\\ =\underset{\_}{7.3\times 10^{-6}M}\end{array}$

Thus, the concentration of both ions $\left[{\text{Quinoline}}^{+}\right]$ and $\left[{\text{OH}}^{-}\right]$ in the solution is

$\underset{\_}{7.3\times 10^{-6}M}$.

Conclusion

1. a) The $\text{pH}$ of $0.0752\text{M}$ solution of quinoline is $\underset{\_}{8.86}$.
2. b) The concentration of $\left[{\text{OH}}^{-}\right]$ in the solution is $\underset{\_}{7.3\times 10^{-6}M}$.