Question

Quinine \(\left(\mathrm{C}_{20} \mathrm{H}_{24} \mathrm{~N}_{2} \mathrm{O}_{2}\right)\) is the most important alkaloid derived from cinchona bark. It is used as an antimalarial drug. For quinine, \(\mathrm{p} K_{\mathrm{b}_{1}}=5.1\) and \(\mathrm{p} K_{\mathrm{b}_{2}}=9.7\left(\mathrm{p} K_{\mathrm{b}}=-\log K_{\mathrm{b}}\right) .\) Only \(1 \mathrm{~g}\) quinine will dissolve in \(1900.0 \mathrm{~mL}\) of solution. Calculate the \(\mathrm{pH}\) of a saturated aqueous solution of quinine. Consider only the reaction \(\mathrm{Q}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{QH}^{+}+\mathrm{OH}^{-}\) described by \(\mathrm{p} K_{\mathrm{b}_{1}}\), where \(\mathrm{Q}=\) quinine.

Short answer

To calculate the pH of a saturated aqueous solution of quinine, first convert its solubility to molarity, which is approximately \(1.63 \times 10^{-3} M\). Then, find the Kb value of the reaction, which is \(7.94 \times 10^{-6}\). Using the equilibrium expression, find the concentration of OH- ions (\(3.59 \times 10^{-5} M\)). Calculate the pOH (\(4.45\)) and finally, find the pH by subtracting the pOH from 14. The pH of the saturated aqueous solution of quinine is approximately \(9.55\).

Step-by-step solution

Convert mass of quinine to moles

To determine the solubility of quinine in moles per liter, we first need to convert the given mass of quinine to moles. The molecular formula of quinine is C20H24N2O2, with a molar mass of approximately 324 g/mol. Divide the mass of quinine by its molecular weight to find the number of moles. \[1 \text{g} \times \frac{1 \text{mol}}{324 \text{g}} = 3.09 \times 10^{-3} \text{mol}\]

Calculate molar solubility

Now, divide the moles of quinine by the volume of the solution to find the solubility in molarity (moles per liter). \[3.09 \times 10^{-3} \text{mol} \div 1.9 \text{L} = 1.63 \times 10^{-3} \text{M}\] #Step 2: Convert pKb to Kb#

Convert pKb1 to Kb

We are given the value of pKb1 and we need to convert it to Kb to use in further calculations. The relationship between pKb and Kb is given by: \[pK_{b}=-\log K_{b}\] Taking the antilog, we get: \[K_{b} = 10^{-pK_{b}}\] Plug in the value of pKb1 to find Kb: \[K_{b} = 10^{-5.1} = 7.94 \times 10^{-6}\] #Step 3: Find the equilibrium expression for the reaction#

Write the equilibrium expression

The equilibrium expression for the given reaction (Q + H2O <==> QH+ + OH-) is: \[K_{b} = \frac{[\mathrm{QH}^+][\mathrm{OH}^-]}{[\mathrm{Q}]}\] Since the question tells us to consider only the first reaction, the initial change in concentration will be equal to the molarity of quinine. #Step 4: Calculate the concentration of OH- ions#

Calculate the equilibrium concentrations

To find the equilibrium concentrations of QH+ and OH-, we'll set up our equilibrium table: Q + H2O <==> QH+ + OH- Initial: 1.63x10^-3 M -- 0 0 Change: -x -- +x +x Equilibrium: 1.63x10^-3 - x -- x x Since Kb is very small, the change in quinine concentration is very small, and we can assume that the equilibrium concentration of Q is still approximately 1.63x10^-3 M.

Substitute and solve for x

Substituting the values into the equilibrium expression, and solving for x, we get: \[7.94 \times 10^{-6} = \frac{x^2}{1.63 \times 10^{-3}}\] \[x^2 = 1.29 \times 10^{-8}\] \[x = \sqrt{1.29 \times 10^{-8}} = 3.59 \times 10^{-5}\] Here, x represents the concentration of OH- ions in the solution, so [OH-] = 3.59 × 10^-5 M. #Step 5: Calculate the pH of the solution#

Calculate pOH

Now that we have found the concentration of OH- ions, we can calculate the pOH of the solution. The pOH is given by the negative logarithm of the OH- concentration: \[pOH = -\log[\mathrm{OH}^-] = -\log(3.59 \times 10^{-5}) = 4.45\]

Calculate pH

Finally, we can calculate the pH of the solution using the relation: \[pH = 14 - pOH = 14 - 4.45 = 9.55\] So, the pH of the saturated aqueous solution of quinine is 9.55.

Key concepts

Alkaloids

Alkaloids are naturally occurring compounds containing nitrogen atoms, often derived from plants. Quinine is a famous example, extracted from the bark of the cinchona tree. Alkaloids are known for their diverse pharmacological effects. Quinine acts as an antimalarial drug by disrupting the life cycle of the malaria parasite in human blood. This makes alkaloids significant in medicine and research. Their chemical complexity and natural origins also make them an intriguing subject for further studies.

Quinine Solubility

Quinine is sparingly soluble, meaning it does not dissolve easily in water. In this exercise, quinine's solubility is given as 1 gram per 1900 mL of solution. This translates to a molar solubility calculation, where the mass of quinine converts to moles using its molar mass (324 g/mol).
The molar solubility, stand-alone, is not indicative of a substance's effectiveness. Solubility influences the concentration of active compounds in a solution, affecting their biological activity. Understanding solubility is crucial in evaluating how a drug like quinine will behave in the body.

Base Dissociation Constant

The base dissociation constant, represented as Kb, indicates a base's strength in water. The smaller the value of Kb, the weaker the base. In this problem, quinine has a pKb1 of 5.1, where pKb is the negative logarithm of Kb: \[pK_{b} = -\log K_{b}\]
The relationship helps in converting between pKb and Kb with the formula:\[K_{b} = 10^{-pK_{b}}\]
This conversion allows us to set up equilibrium calculations, which describe how well quinine dissociates in solution. A lower Kb means quinine scarcely forms ions, fitting its role as a weak base.

Equilibrium Expression Calculation

Equilibrium expressions involve using initial concentrations and changes to find equilibrium states in chemical reactions. For quinine, this involves the reaction:\[\mathrm{Q} + \mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{QH}^{+} + \mathrm{OH}^{-}\]
The equilibrium expression is:\[K_{b} = \frac{[\mathrm{QH}^+][\mathrm{OH}^-]}{[\mathrm{Q}]}\]
We substitute initial concentrations and changes in concentration (\(x\)) into this expression. Solving provides the concentration of hydroxide ions, \([\mathrm{OH}^-]\), which further leads to calculations of pOH and then pH. This systematic approach helps in understanding the balance of ions in a solution, key for accurately determining pH.