Chapter 16, Problem 16.59QP

(a)

Interpretation Introduction

To determine: The $\text{pH}$ of a $0.115\text{M}$ solution of morphine.

Interpretation: The $\text{pH}$ of a $0.115\text{M}$ solution of morphine is to be calculated.

Concept introduction: The $\text{pH}$ of a compound is the log of the concentration of hydrodium ion released by the compound upon dissociation.

Answer to Problem 16.59QP

The $\text{pH}$ of a $0.115\text{M}$ solution of morphine is $\underset{\_}{10.63}$.

Explanation of Solution

The compound is morphine.

The molar concentration of morphine is $0.115\text{M}$.

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

The ionization reaction of morphine is,

$\text{Morphine}\left(aq\right)\rightleftharpoons {\text{Morphine}}^{+}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$

The ICE Table for ionization of morphine is given below.

$\begin{array}{cccc}& \left[\text{Morphine}\right]\text{M}& \left[{\text{Morphine}}^{+}\right]\text{M}& \left[{\text{OH}}^{-}\right]\text{M}\\ \text{Intial}& 0.115& 0& 0\\ \text{Change}& -x& +x& +x\\ \text{Equlibrium}& 0.115-x& x& x\end{array}$

The equilibrium constant for the above reaction is,

$K_{\text{b}}=\frac{\left[{\text{Morphine}}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[\text{Morphine}\right]}$

Where,

• $K_{\text{b}}$ is equilibrium constant for the reaction.
• $\left[{\text{Morphine}}^{+}\right]$ is concentration of morphine ion.
• $\left[{\text{OH}}^{-}\right]$ is concentration of hydroxide ion.
• $\left[\text{Morphine}\right]$ is concentration of morphine.
• $x$ is assumed concentration of hydroxide ion.

Substitute the value of equilibrium concentration from to table to above expression.

$\begin{array}{c}{K}_{\text{b}}=\frac{\left[{\text{Morphine}}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[\text{Morphine}\right]}\\ {10}^{-5.79}=\frac{{\left(x\right)}^2}{\left(0.115-x\right)}\\ 1.62\times {10}^{-6}=\frac{{\left(x\right)}^2}{\left(0.115-x\right)}\\ x=4.32\times {10}^{-4}\text{M}\end{array}$

Hence, the concentration of hydroxide ion is $4.32\times {10}^{-4}\text{M}$.

The $\text{pH}$ is given by the formula,

$\begin{array}{c}\text{pH}=14-\text{pOH}\\ \text{=}14-\left(-\log \left[{\text{OH}}^{-}\right]\right)\end{array}$

Substitute the calculated concentration of hydroxide ion in the above expression.

$\begin{array}{c}\text{pH}=14-\left(-\log \left[{\text{OH}}^{-}\right]\right)\\ =14-\left(-\log \left[4.32\times {10}^{-4}\text{M}\right]\right)\\ =14-\left(3.36\right)\\ =\underset{\_}{10.63}\end{array}$

Hence, the calculated $\text{pH}$ is $\underset{\_}{10.63}$.

(b)

Interpretation Introduction

To determine: The $\text{pH}$ of a $3.42\times {10}^{-4}\text{M}$ solution of codeine.

Interpretation: The $\text{pH}$ of a $3.42\times {10}^{-4}\text{M}$ solution of codeine is to be calculated.

Concept introduction: The $\text{pH}$ of a compound is the log of the concentration of hydrodium ion released by the compound upon dissociation.

Answer to Problem 16.59QP

The $\text{pH}$ of a $3.42\times {10}^{-4}\text{M}$ solution of codeine is $\underset{\_}{9.37}$.

Explanation of Solution

The compound is codeine.

The molar concentration of codeine is $3.42\times {10}^{-4}\text{M}$.

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

The $\text{p}{K}_{\text{b}}$ is given by the formula,

$\text{p}{K}_{\text{b}}=14-\text{p}{K}_{\text{a}}$

Substitute the given value of $\text{p}{K}_{\text{a}}$ to above expression,

$\begin{array}{c}\text{p}{K}_{\text{b}}=14-\text{p}{K}_{\text{a}}\\ =14-8.21\\ =5.79\end{array}$

The ionization reaction of codeine is,

${\text{C}}_{18}{\text{H}}_{21}{\text{NO}}_3\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\rightleftharpoons {\text{C}}_{18}{\text{H}}_{22}{\text{NO}}_3{}^{+}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$

The ICE Table for ionization of morphine is given below.

$\begin{array}{cccc}& \left[{\text{C}}_{18}{\text{H}}_{21}{\text{NO}}_3\right]\text{M}& \left[{\text{C}}_{18}{\text{H}}_{22}{\text{NO}}_3{}^{+}\right]\text{M}& \left[{\text{OH}}^{-}\right]\text{M}\\ \text{Intial}& 3.42\times {10}^{-4}\text{M}& 0& 0\\ \text{Change}& -x& +x& +x\\ \text{Equlibrium}& 3.42\times {10}^{-4}\text{M}-x& x& x\end{array}$

The equilibrium constant for the above reaction is,

$K_{\text{b}}=\frac{\left[{\text{C}}_{18}{\text{H}}_{22}{\text{NO}}_3{}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{C}}_{18}{\text{H}}_{21}{\text{NO}}_3\right]}$

Where,

• $K_{\text{b}}$ is equilibrium constant for the reaction.
• $\left[{\text{C}}_{18}{\text{H}}_{22}{\text{NO}}_3{}^{+}\right]$ is concentration of codeine ion.
• $\left[{\text{OH}}^{-}\right]$ is concentration of hydroxide ion.
• $\left[{\text{C}}_{18}{\text{H}}_{21}{\text{NO}}_3\right]$ is concentration of codeine.
• $x$ is assumed concentration of hydroxide ion.

Substitute the value of equilibrium concentration from to table to above expression.

$\begin{array}{c}{K}_{\text{b}}=\frac{\left[{\text{C}}_{18}{\text{H}}_{22}{\text{NO}}_3{}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{C}}_{18}{\text{H}}_{21}{\text{NO}}_3\right]}\\ {10}^{-5.79}=\frac{{\left(x\right)}^2}{\left(3.42\times {10}^{-4}\text{M}-x\right)}\\ 1.62\times {10}^{-6}=\frac{{\left(x\right)}^2}{\left(3.42\times {10}^{-4}\text{M}-x\right)}\\ x=2.35\times {10}^{-5}\text{M}\end{array}$

Hence, the concentration of hydroxide ion is $2.35\times {10}^{-5}\text{M}$.

The $\text{pH}$ is given by the formula,

$\begin{array}{c}\text{pH}=14-\text{pOH}\\ \text{=}14-\left(-\log \left[{\text{OH}}^{-}\right]\right)\end{array}$

Substitute the calculated concentration of hydroxide ion in the above expression.

$\begin{array}{c}\text{pH}=14-\left(-\log \left[{\text{OH}}^{-}\right]\right)\\ =14-\left(-\log \left[2.35\times {10}^{-5}\text{M}\right]\right)\\ =14-\left(4.63\right)\\ =\underset{\_}{9.37}\end{array}$

Hence, the calculated $\text{pH}$ is $\underset{\_}{9.37}$.

Conclusion

1. a. The $\text{pH}$ of a $0.115\text{M}$ solution of morphine, when the $\text{p}{K}_{\text{b}}$ of the conjugate base $5.79$ is $\underset{\_}{10.63}$.
2. b. The $\text{pH}$ of a $3.42\times {10}^{-4}\text{M}$ solution of codeine, when the $\text{p}{K}_{\text{a}}$ of the conjugate acid $8.21$ is $\underset{\_}{9.37}$.