Question

The antimalarial drug quinine, \(\mathrm{C}_{20} \mathrm{H}_{24} \mathrm{O}_{2} \mathrm{N}_{2},\) is a diprotic base with a water solubility of \(1.00 \mathrm{g} / 1900 \mathrm{mL}\) of solution. (a) Write equations for the ionization equilibria corresponding to \(\mathrm{p} K_{\mathrm{b}_{1}}=6.0\) and \(\mathrm{p} K_{\mathrm{b}_{2}}=9.8\) (b) What is the \(\mathrm{pH}\) of saturated aqueous quinine?

Short answer

The ionization of quinine, a diprotic base, proceeds in two steps, each step involving the formation of a conjugate acid and a hydroxide ion. The pH of a saturated solution of quinine can be calculated by initial concentration and the Kb values derived from given pKb values. Detailed calculation involves solving for concentration of hydroxide ions and applying the pH and pOH relationship.

Step-by-step solution

Write Ionization Equilibria

For a diprotic base like quinine, there are two ionization steps. In the first step, quinine (\(C_{20}H_{24}O_{2}N_{2}\)) accepts a proton to form a conjugate acid (\(C_{20}H_{24}O_{2}N_{2}H\)) and a hydroxide ion (\(OH^-\)). Let's represent this below - \[ C_{20}H_{24}O_{2}N_{2} + H_2O \leftrightarrow C_{20}H_{24}O_{2}N_{2}H + OH^- \]In the second ionization step, the conjugate acid from the first step (\(C_{20}H_{24}O_{2}N_{2}H\)) again accepts a proton to form another conjugate acid (\(C_{20}H_{24}O_{2}N_{2}H_2\)) and a hydroxide ion (\(OH^-\)). Let's represent this as well - \[ C_{20}H_{24}O_{2}N_{2}H + H_2O \leftrightarrow C_{20}H_{24}O_{2}N_{2}H_2 + OH^- \]

Calculate Concentration

The second step is to calculate the concentration of quinine in the solution which is the mass of solute (quinine) divided by the volume of solution. Given that the solubility of this base is 1g in 1900 mL of water, this can be calculated as follows: \[ C = \frac{m}{V} = \frac{1\,g}{1900\,mL} = 0.000526\,mol/L \]

Conversion of pKb to Kb

Given the pKb values, we can find out the Kb values using the formula Kb = \(10^{-pKb}\). For \(pKb_1 = 6.0\),\[ Kb_1 = 10^{-6.0} \]And for \(pKb_2 = 9.8\),\[ Kb_2 = 10^{-9.8} \]

Calculation of pH

Considering quinine as a weak base, it may not be fully ionized in water. Hence, we can apply the equilibrium law and write the expressions for Kb as shown below -\[ Kb_1 =\frac {[OH^-][C_{20}H_{24}O_{2}N_{2}H]}{[C_{20}H_{24}O_{2}N_{2}]} \]With the approximation that at equilibrium, [OH-] = [C20H24O2N2H] and using the given values, we can solve for [OH-]. The concentration of OH- can be used to find pOH, and then we can find pH as \[ pH = 14 - pOH \]

Key concepts

Chemical Equilibrium

Chemical equilibrium is a state in a chemical reaction where the rates of the forward and reverse reactions are equal, resulting in no overall change in the concentrations of reactants and products over time. This concept is vital when we consider reactions such as ionization, where a base like quinine can accept protons from water to form hydroxide ions and its conjugate acid.

For quinine, a diprotic base, this process occurs in two stages with each having its own equilibrium constant. The equilibrium constants are represented by Kb values, which are inversely related to pKb values given by the formula \( Kb = 10^{-pKb} \). The larger the Kb (or the smaller the pKb), the stronger the base, which implies a greater degree of ionization at equilibrium.

Understanding this equilibrium allows us to predict the concentration of various species in solution at saturation, which is necessary to calculate the pH of the solution. It's essential to remember that at equilibrium, the concentration of products to reactants remains constant, and this ratio is represented by the equilibrium constant (Kb in the case of bases).

Acid-Base Reactions

Acid-base reactions involve the transfer of protons (H+ ions) between reactants. For bases like quinine, the reaction includes the base accepting protons from water molecules, resulting in the formation of its conjugate acid and hydroxide ions (\(OH^-\)).

The ionization equilibria of a diprotic base such as quinine involves two distinct steps, each of which can be characterized by its own pKb value. The first step involves the base \(C_{20}H_{24}O_{2}N_{2}\) accepting a proton to become \(C_{20}H_{24}O_{2}N_{2}H\), and the second step involves this conjugate acid accepting yet another proton to form \(C_{20}H_{24}O_{2}N_{2}H_2\).

For each step, the ionization equilibrium can be depicted as quinine interacting with water to form hydroxide ions and the conjugate acid. Such reactions greatly affect the pH of solutions and understanding each step's equilibrium is key to comprehending how the pH of the solution can be determined.

pH Calculation

pH calculation is a measure of the acidity or basicity of an aqueous solution. For the calculation of pH, the ionization of water and the dissociation constants of the base in question are considered. pH is calculated using the concentration of hydrogen ions (\(H^+\)) or hydroxide ions (\(OH^-\)) in solution, which are related to the equilibrium constants of the acid-base reactions.

In the case of quinine, we are dealing with a weak base that partially ionizes in water, and the pH can be found by first calculating the concentration of \(OH^-\) using the equilibrium expressions for Kb. Once we have the \(OH^-\) concentration, we can calculate the pOH (the negative logarithm of the \(OH^-\) concentration). Because pH and pOH are related by the equation \(pH + pOH = 14\), knowing one allows us to calculate the other.

It is important to take a stepwise approach, considering the contributions to \(OH^-\) from both steps of ionization, but often the second ionization has a negligible effect due to the large difference in pKb values (a higher pKb means a weaker base and less contribution to \(OH^-\) concentration). Subsequently, the pH of the saturated solution of quinine can be determined, aiding in understanding the nature of quinine as a base in aqueous solutions.