91Ó°ÊÓ

(a)

First write the reaction equation between quinine and water.

${\text{C}}_{20}{\text{H}}_{24}{\text{N}}_2{\text{O}}_2+{\text{H}}_2\text{O}⇌{\text{HC}}_{20}{\text{H}}_{24}{\text{N}}_2{\text{O}}_2^{+}+{\text{OH}}^{-}$

Next, construct the ICE table to obtain the equation for ${\text{K}}_{\text{b}}$.

${\text{K}}_{\text{b}}=\frac{\left[{\text{OH}}^{-}\right]\left[{\text{HC}}_{20}{\text{H}}_{24}{\text{N}}_2{\text{O}}_2^{+}\right]}{\left|{\text{C}}_{20}{\text{H}}_{24}{\text{N}}_2{\text{O}}_2\right|}$

${\text{K}}_{\text{b}}=\frac{x^2}{1.6×{10}^{-3}-x}$

First, solve for the${\text{K}}_{\text{b}}$from the given${\text{pK}}_{\text{b}}$of the tertiary amine group because it is the amine that has a significant effect on the pill.

$$\begin{gathered} {\text{pK}}_{\text{b}}=-\log \left({\text{K}}_{\text{b}}\right) \\ {\text{K}}_{\text{b}}=10{\text{pK}}_{\text{s}} \\ {\text{K}}_{\text{b}}={10}^{-5.1} \\ {\text{K}}_{\text{b}}=7.94×{10}^{-6} \end{gathered}$$

We know that $x=\left[{\mathrm{HC}}_{20}{\mathrm{H}}_{21}{\mathrm{N}}_2{\mathrm{O}}_2^{+}\right]=\left[{\mathrm{OH}}^{-}\right]$.

Since ${\text{C}}_{20}{\text{H}}_{24}{\text{N}}_2\text{O}$is a weak base, its ${\text{K}}_{\text{b}}$must be very small. So, assume that the x has no effect on $1.6×{10}^{-3}$in the denominator. Then substitute the ${\text{K}}_{\text{b}}$to solve for x.

$$\begin{gathered} {\text{K}}_{\text{b}}=\frac{x^2}{1.6×{10}^{-3}} \\ {x}^2=\left({\text{K}}_{\text{b}}\right)\left(1.6×{10}^{-3}\right) \\ x=\sqrt{\left(7.94×{10}^{-6}\right)\left(1.6×{10}^{-3}\right)} \\ x=1.13×{10}^{-4} \end{gathered}$$

Since $x=\left[{\mathrm{HC}}_{20}{\mathrm{H}}_{24}{\mathrm{N}}_2{\mathrm{O}}_2^{+}\right]=\left[{\mathrm{OH}}^{-}\right]\mathrm{then}\left[{\mathrm{OH}}^{-}\right]=1.13×{10}^{-4}\mathrm{M}$.

Solve for the pOH.

$$\begin{gathered} \mathrm{pOH}=-\log \left[{\mathrm{OH}}^{-}\right] \\ =-\log \left(1.13×{10}^{-4}\right) \\ \mathrm{pOH}=3.948 \end{gathered}$$

Lastly, solve for the pH.

$$\begin{gathered} \mathrm{pH}+\mathrm{pOH}=14 \\ \mathrm{pH}=14-\mathrm{pOH} \\ =14-3.948 \\ \mathrm{pH}=10.052 \end{gathered}$$

Hence the pH is $10.052$.

(b)

The ICE table was already constructed in the previous problem, we can proceed directly to solving${\text{K}}_{\text{b}}$from the given${\text{pK}}_{\text{b}}$of the aromatic amine.

$$\begin{gathered} {\mathrm{pK}}_{\mathrm{b}}=-\log \left({\mathrm{K}}_{\mathrm{b}}\right) \\ {\mathrm{K}}_{\mathrm{b}}={10}^{-p{K}_b} \\ {\mathrm{K}}_{\mathrm{b}}={10}^{-9.7} \\ {\mathrm{K}}_{\mathrm{b}}=2.00×{10}^{-10} \end{gathered}$$

We know that $x=\left[{\mathrm{HC}}_{20}{\mathrm{H}}_{24}{\mathrm{N}}_2{\mathrm{O}}_2^{+}\right]=\left[{\mathrm{OH}}^{-}\right]$. Since ${\text{C}}_{20}{\text{H}}_{24}{\text{N}}_2{\text{O}}_2$is a weak base, its ${\text{K}}_{\text{b}}$must be very small. So, assume that the x has no effect on $1.6×{10}^{-3}$the denominator. Then substitute the localid="1663268370941" ${\text{K}}_{\text{b}}$to solve for x.

$$\begin{gathered} K_b=\frac{x^2}{1.6×{10}^{-3}} \\ {x}^2=\left(K_b\right)\left(1.6×{10}^{-3}\right) \\ x=\sqrt{\left(K_b\right)\left(1.6×{10}^{-3}\right)} \\ x=\sqrt{\left(2.00×{10}^{-10}\right)\left(1.6×{10}^{-3}\right)} \\ x=5.65×{10}^{-7} \end{gathered}$$

amine,

Total$\left[{\mathrm{OH}}^{-}\right]=5.65×{10}^{-5}\mathrm{M}+1.13×{10}^{-4}=1.14×{10}^{-4}$

Solve for the pOH.

$$\begin{gathered} \mathrm{pOH}=-\log \left[{\mathrm{OH}}^{-}\right] \\ =-\log \left(1.14×{10}^{-4}\right) \\ \mathrm{pOH}=3.945 \end{gathered}$$

Lastly, solve for the pH.

$$\begin{gathered} \mathrm{pH}+\mathrm{pOH}=14 \\ \mathrm{pH}=14-\mathrm{pOH} \\ \mathrm{pH}=14-3.948 \\ \mathrm{pH}=10.055 \end{gathered}$$

If we compare the pH from (a) and the pH we obtained from calculating the total$\left[{\text{OH}}^{\text{-}}\right]$, they are approximately equal:$10.052≫10.055$

Hence The ions produced by the aromatic amine do not affect the overall total of ions significantly.

(c)

First, the salt will dissolve in water.

${\text{C}}_{20}{\text{H}}_{24}{\text{N}}_2{\text{O}}_2·\text{HCl}+{\text{H}}_2\text{O}→{\text{HC}}_{20}{\text{H}}_{24}{\text{N}}_2{\text{O}}_2^{+}+{\text{Cl}}^{-}$

Then ${\text{HC}}_{20}{\text{H}}_{24}{\text{N}}_2{\text{O}}_2^{+}$will react, with water.

${\text{HC}}_{20}{\text{H}}_{21}{\text{N}}_2{\text{O}}_2^{+}+{\text{H}}_2\text{O}⇌{\text{C}}_{20}{\text{H}}_{24}{\text{N}}_2{\text{O}}_2+{\text{H}}_3{\text{O}}^{+}$

Next, construct the ICF table to obtain the equation for${\text{K}}_{\text{a}}$-

$$\begin{gathered} {\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\left[{\mathrm{C}}_{20}{\mathrm{H}}_{24}{\mathrm{N}}_2{\mathrm{O}}_2\right]}{\left[{\mathrm{HC}}_{20}{\mathrm{H}}_{24}{\mathrm{N}}_2{\mathrm{O}}_2^{+}\right]} \\ {\mathrm{K}}_{\mathrm{a}}=\frac{{\mathrm{x}}^2}{0.33-\mathrm{x}} \end{gathered}$$

First, solve for the${\text{K}}_{\text{a}}$from the given${\text{K}}_{\text{b}}$of the tertiary amine.

localid="1663269283288" $$\begin{gathered} {\mathrm{K}}_{\mathrm{w}}={\mathrm{K}}_{\mathrm{b}}×{\mathrm{K}}_{\mathrm{a}} \\ {\mathrm{K}}_{\mathrm{a}}=\frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{b}}} \end{gathered}$$

$$\begin{gathered} {\mathrm{K}}_{\mathrm{a}}=\frac{1×10}{7.94×{10}^{-6}} \\ {\mathrm{K}}_{\mathrm{a}}=1.26×{10}^{-9} \end{gathered}$$

We know that localid="1663271326291" $\mathbf{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{[}{\mathbf{C}}_{\mathbf{20}}{\mathbf{H}}_{\mathbf{24}}\mathbf{}{\mathbf{N}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{2}}\mathbf{]}\mathbf{}\mathbf{=}\mathbf{}\mathbf{\left[}{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}^{\mathbf{+}}\mathbf{\right]}$.Since localid="1663271340268" ${\mathbf{\text{HC}}}_{\mathbf{20}}{\mathbf{\text{H}}}_{\mathbf{24}}\mathbf{}{\mathbf{\text{N}}}_{\mathbf{2}}{\mathbf{\text{O}}}_{\mathbf{2}}^{\mathbf{+}}$is a weak acid, its localid="1663271354507" ${\mathbf{\text{K}}}_{\mathbf{\text{a}}}$must be very small. So, assume that the x has no denominator. Then substitute theto solve localid="1663271371123" ${\mathbf{\text{K}}}_{\mathbf{\text{a}}}$ for x.

$$\begin{gathered} {\mathrm{K}}_{\mathrm{a}}=\frac{x^2}{0.33} \\ {x}^2=\left({\mathrm{K}}_{\mathrm{a}}\right)(0.33) \\ x=\sqrt{\left({\mathrm{K}}_{\mathrm{a}}\right)(0.33)} \\ x=\sqrt{\left(1.26×{10}^{-9}\right)(0.33)} \\ x=2.04×{10}^{-5} \end{gathered}$$

Since $\mathrm{x}=\left[{\mathrm{C}}_{20}{\mathrm{H}}_{24}{\mathrm{N}}_2{\mathrm{O}}_2\right]=\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right],\mathrm{then}\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=2.04×{10}^{-5}\mathrm{M}$.

Lastly, solve for the pH

$$\begin{gathered} \mathrm{pH}=-\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right] \\ \mathrm{pH}=-\log \left(2.04×{10}^{-5}\right) \\ \mathrm{pH}=4.70 \end{gathered}$$

Hence the pH of $0.33\mathrm{M}$quinine hydrochloride islocalid="1663269916463" $4.70$.

(d)

First, solve for the concentration of quinine hydrochloride.

$$\begin{gathered} \text{M}=\frac{1.0\text{g}}{\text{mL}×\frac{1\text{L}}{1000\text{mL}}}×\frac{1\text{mol}{\mathrm{C}}_{20}{\mathrm{H}}_{24}{\mathrm{N}}_2{\mathrm{O}}_2×\mathrm{HCl}}{360.87\text{g}\left({\mathrm{C}}_{20}{\mathrm{H}}_{24}{\mathrm{N}}_2{\mathrm{O}}_2\right)×\mathrm{HCl}} \\ =2.77\text{M} \end{gathered}$$

${\mathrm{C}}_{20}{\mathrm{H}}_{24}{\mathrm{N}}_2{\mathrm{O}}_2×\mathrm{HCl}⇌{\mathrm{HC}}_{20}{\mathrm{H}}_{24}{\mathrm{N}}_2{\mathrm{O}}_2^{+}+{\mathrm{Cl}}^{-}$

Solve for the role="math" localid="1663270350638" $\left[{\text{HC}}_{20}{\text{H}}_{24}{\text{N}}_2{\text{O}}_2^{+}\right]$

$$\begin{gathered} \left[{\mathrm{HC}}_{20}{\mathrm{H}}_{24}{\mathrm{N}}_2{\mathrm{O}}_2^{+}\right]=\left[{\mathrm{C}}_{20}{\mathrm{H}}_{24}{\mathrm{N}}_2{\mathrm{O}}_2×\mathrm{HCl}\right]×0.015 \\ =4.2×{10}^{-2}\text{M} \end{gathered}$$

Then quinine ion will react with water.

$\left.{\text{HC}}_{20}{\text{H}}_{24}{\text{N}}_2{\text{O}}_2^{+}\right]+{\text{H}}_2\text{O}={\text{C}}_{20}{\text{H}}_{24}{\text{N}}_2{\text{O}}_2+{\text{H}}_3{\text{O}}^{+}$

Next, construct the ICE table to obtain the equation for ${\text{K}}_{\text{a}}$

${\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\left[{\mathrm{C}}_{20}{\mathrm{H}}_{24}{\mathrm{N}}_2{\mathrm{O}}_2\right]}{\left[{\mathrm{HC}}_{20}{\mathrm{H}}_{24}{\mathrm{N}}_2{\mathrm{O}}_2^{+}\right]}$

${\mathrm{K}}_{\mathrm{a}}=\frac{x^2}{4.2×{10}^{-2}-x}$

We know that $\mathrm{x}=\left[{\mathrm{C}}_{20}{\mathrm{H}}_{24}{\mathrm{N}}_2{\mathrm{O}}_2\right]=\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$. Since ${\text{HC}}_{20}{\text{H}}_{24}{\text{N}}_2{\text{O}}_2^{+}$is a weak acid, its ${\text{K}}_{\text{a}}$must be very small. So, assume that the x has no effect on $4.2×{10}^{-2}\text{M}$the denominator. Then substitute the${\text{K}}_{\text{a}}$ to solve for x.

$$\begin{gathered} K_a=\frac{x^2}{4.2×{10}^{-2}} \\ {x}^2=\left(K_a\right)\left(4.2×{10}^{-2}\right) \\ x=\sqrt{\left(K_a\right)\left(4.2×{10}^{-2}\right)} \\ x=\sqrt{\left(1.26×{10}^{-9}\right)\left(4.2×{10}^{-2}\right)} \\ x=7.24×{10}^{-6} \end{gathered}$$

Since $\mathrm{x}=[{\mathrm{C}}_{20}{\mathrm{H}}_{24}{\mathrm{N}}_2{\mathrm{O}}_2]=\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$then role="math" localid="1663270949668" $\left[{\text{H}}_3{\text{O}}^{+}\right]=7.24×{10}^{-6}\text{M}$.

Lastly, solve for the pH.

$$\begin{gathered} \mathrm{pH}=-\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right] \\ \mathrm{pH}=-\log \left(7.24×{10}^{-6}\right) \\ \mathrm{pH}=5.14 \end{gathered}$$

Hence the pH is role="math" localid="1663271099862" $5.14$