Chapter 15, Problem 15.59QP

(a)

Interpretation Introduction

Interpretation: The $\text{p}{K}_{\text{b}}$ value of morphine is given to be $5.79$. The $\text{pH}$ of a $1.8\times {10}^{-3}\text{M}$ solution of morphine is to be calculated. Similarly, The $\text{p}{K}_{\text{a}}$ value of codeine is given to be $8.21$. The $\text{pH}$ of a $2.7\times {10}^{-4}\text{M}$ solution of codeine 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 $\text{pH}$ value of a $1.8\times {10}^{-3}\text{M}$ solution of morphine.

Answer to Problem 15.59QP

Solution

The $\text{pH}$ value of $1.8\times {10}^{-3}\text{M}$ solution of morphine is $\underset{\_}{9.73}$.

Explanation of Solution

Explanation

Given

The concentration of morphine is $1.8\times {10}^{-3}\text{M}$.

The $\text{p}{K}_{\text{b}}$ value of morphine is $5.79$.

The $K_{\text{b}}$ of solution of morphine 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).

$\begin{array}{c}{K}_{\text{b}}={10}^{-5.79}\\ =1.6\times {10}^{-6}\end{array}$

The $\left[{\text{OH}}^{-}\right]$ concentration is calculated by the formula,

$\left[{\text{OH}}^{-}\right]=\sqrt{K_{\text{b}}+\left[\text{base}\right]}$ (2)

Substitute the value of $K_{\text{b}}$ and $\left[\text{base}\right]$ in equation (2).

$\begin{array}{c}\left[{\text{OH}}^{-}\right]=\sqrt{\left(1.6\times {10}^{-6}\text{M}\right)\times 1.8\times {10}^{-3}\text{M}}\\ =5.36\times {10}^{-5}\text{M}\end{array}$

The concentration of $\left[{\text{H}}_3{\text{O}}^{+}\right]$ is calculated by the formula,

$\left[{\text{H}}_3{\text{O}}^{+}\right]=\frac{K_{\text{w}}}{\left[{\text{OH}}^{-}\right]}$ (3)

The value of $K_{\text{w}}$ is ${10}^{-14}$.

Substitute the value of $K_{\text{w}}$ and $\left[{\text{OH}}^{-}\right]$ in equation (3).

$\begin{array}{c}\left[{\text{H}}_3{\text{O}}^{+}\right]=\frac{{10}^{-14}}{5.36\times {10}^{-5}}\\ =1.86\times {10}^{-10}\end{array}$

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

$\text{pH}=-\text{log}\left[{\text{H}}_3{\text{O}}^{+}\right]$ (4)

Substitute the value of $\left[{\text{H}}_3{\text{O}}^{+}\right]$ in equation 4.

$\begin{array}{c}\text{pH}=-\text{log}\left(1.86\times {10}^{-10}\right)\\ =\underset{\_}{9.73}\end{array}$

Therefore, the $\text{pH}$ value of $1.8\times {10}^{-3}\text{M}$ solution of morphine is $\underset{\_}{9.73}$.

(b)

Interpretation Introduction

To determine: The $\text{pH}$ of a $2.7\times {10}^{-4}\text{M}$ solution of codeine.

Answer to Problem 15.59QP

Solution

The $\text{pH}$ value of $2.7\times {10}^{-4}\text{M}$ solution of codeine is $\underset{\_}{9.32}$.

Explanation of Solution

Explanation

Given

The concentration of codeine is $2.7\times {10}^{-4}\text{M}$.

The $\text{p}{K}_{\text{a}}$ value of codeine is $8.21$.

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

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

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

$\begin{array}{c}{K}_{\text{a}}={10}^{-8.21}\\ =6.2\times {10}^{-9}\end{array}$

Thus, $K_{\text{b}}$ of solution of codeine is calculated by the formula,

$\begin{array}{c}{K}_{\text{b}}=\frac{K_{\text{w}}}{K_{\text{a}}}\\ =\frac{1\times {10}^{-14}}{6.2\times {10}^{-9}}\\ =1.6\times {10}^{-6}\end{array}$

Thus, the value of $K_{\text{b}}$ of codeine is $1.6\times {10}^{-6}$.

The $\left[{\text{OH}}^{-}\right]$ concentration is calculated by the formula,

$\left[{\text{OH}}^{-}\right]=\sqrt{K_{\text{b}}+\left[\text{base}\right]}$ (2)

Substitute the value of $K_{\text{b}}$ and $\left[\text{base}\right]$ in equation (2).

$\begin{array}{c}\left[{\text{OH}}^{-}\right]=\sqrt{\left(1.6\times {10}^{-6}\text{M}\right)\times 2.7\times {10}^{-4}\text{M}}\\ =2.07\times {10}^{-5}\text{M}\end{array}$

The concentration of $\left[{\text{H}}_3{\text{O}}^{+}\right]$ is calculated by the formula,

$\left[{\text{H}}_3{\text{O}}^{+}\right]=\frac{K_{\text{w}}}{\left[{\text{OH}}^{-}\right]}$ (3)

The value of $K_{\text{w}}$ is ${10}^{-14}$.

Substitute the value of $K_{\text{w}}$ and $\left[{\text{OH}}^{-}\right]$ in equation (3).

$\begin{array}{c}\left[{\text{H}}_3{\text{O}}^{+}\right]=\frac{{10}^{-14}}{2.07\times {10}^{-5}}\\ =4.83\times {10}^{-10}\end{array}$

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

$\text{pH}=-\text{log}\left[{\text{H}}_3{\text{O}}^{+}\right]$ (4)

Substitute the value of $\left[{\text{H}}_3{\text{O}}^{+}\right]$ in equation 4.

$\begin{array}{c}\text{pH}=-\text{log}\left(4.83\times {10}^{-10}\right)\\ =\underset{\_}{9.32}\end{array}$

Therefore, the $\text{pH}$ value of $2.7\times {10}^{-4}\text{M}$ solution of codeine is $\underset{\_}{9.32}$.

Conclusion

a. The $\text{pH}$ value of $1.8\times {10}^{-3}\text{M}$ solution of morphine is $\underset{\_}{9.73}$.

b. The $\text{pH}$ value of $2.7\times {10}^{-4}\text{M}$ solution of codeine is $\underset{\_}{9.32}$.