Q18.187CP - Chemistry: Molecular Nature Of Matter And Change

Question

The antimalarial properties of quinine (${\mathbf{\text{C}}}_{\mathbf{\text{20}}}{\mathbf{\text{H}}}_{\mathbf{\text{24}}}{\mathbf{\text{N}}}_{\mathbf{\text{2}}}{\mathbf{\text{O}}}_{\mathbf{\text{2}}}$) saved thousands of lives during construction of the Panama Canal. This substance is a classic example of the medicinal wealth of tropical forests. Both N atoms are basic, but the N (coloured) of the 3amine group is far more basic (pKb =5.1 ) than the N within the aromatic ring system (pKb =9.7).

 (a) A saturated solution of quinine in water is only $\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{ }\mathbf{\text{M}}$. What is the pH of this solution?

 (b) Show that the aromatic N contributes negligibly to the pH of the solution. 

(c) Because of its low solubility, quinine is given as the salt quinine hydrochloride $(C_{20}{H}_{24}{N}_2{O}_2\times \mathrm{HCl})$,which is 120 times more soluble than quinine. What is the pH of M quinine hydrochloride?

(d) An antimalarial concentration in water is 15 % quinine hydrochloride by mass (d 1.0 g/mL). What is the pH?

Step-by-Step Solution

Answer

1.  $\mathrm{pH}=10.052$
2. The ions produced by the aromatic amine do not affect the overall total of ions significantly
3. The pH of $0.33 \mathrm{M}$ quinine hydrochloride is $4.70$.
4. $\mathrm{pH}=5.14$
   

Step 1: To find the pH

(a)

First write the reaction equation between quinine and water.

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

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

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

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

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.

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

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 $1.6\times {10}^{-3}$ in the denominator. Then substitute the ${\text{K}}_{\text{b}}$ to solve for x.

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

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

Step 2: To solve for pOH

Solve for the pOH.

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

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 $10.052$. 

Step 3:To Show that the aromatic N

(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.

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

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 $1.6\times {10}^{-3}$ the denominator. Then substitute the ${\text{K}}_{\text{b}}$ to solve for x.

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

amine,

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

Solve for the pOH.

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

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: $10.052\gg 10.055$

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

Step 4: To find the pH of M quinine hydrochloride

(c)

First, the salt will dissolve in water.

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

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

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

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.

$$\begin{gathered} {\mathrm{K}}_{\mathrm{w}}={\mathrm{K}}_{\mathrm{b}}\times {\mathrm{K}}_{\mathrm{a}} \\ {\mathrm{K}}_{\mathrm{a}}=\frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{b}}} \end{gathered}$$

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

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

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

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

Lastly, solve for the pH

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

Hence the pH of $0.33 \mathrm{M}$ quinine hydrochloride is $4.70$.

Step 5: To find the pH

(d)

First, solve for the concentration of quinine hydrochloride.

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

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

Solve for the $\left[{\text{HC}}_{20}{\text{H}}_{24} {\text{N}}_2{\text{O}}_2^{+}\right]$

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

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 ${\text{K}}_{\text{a}}$

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

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

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 $4.2\times {10}^{-2}\text{M}$ the denominator. Then substitute the ${\text{K}}_{\text{a}}$ to solve for x.

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

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

Lastly, solve for the pH.

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

Hence the pH is $5.14$