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.

Step by step solution

01

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. \[1 \text{g} \times \frac{1 \text{mol}}{324 \text{g}} = 3.09 \times 10^{-3} \text{mol}\]

02

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#

03

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#

04

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#

05

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.

06

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#

07

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: \[pOH = -\log[\mathrm{OH}^-] = -\log(3.59 \times 10^{-5}) = 4.45\]

08

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.

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