   # Quinine (C 20 H 24 N 2 O 2 ) is the most important alkaloid derived from cinchona bark. It is used as an antimalarial drug. For quinine, p K b 1 = 5.1 and p K b 2 = 9.7 (p K b = −log K b ) . Only 1 g quinine will dissolve in 1900.0 mL of solution. Calculate the pH of a saturated aqueous solution of quinine. Consider only the reaction Q+H 2 O ⇌ QH + + OH − described by p K b 1 , where Q = quinine. ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 13, Problem 154AE
Textbook Problem
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## Quinine (C20H24N2O2) is the most important alkaloid derived from cinchona bark. It is used as an antimalarial drug. For quinine, p K b 1 = 5.1 and p K b 2 = 9.7 (pKb = −log Kb). Only 1 g quinine will dissolve in 1900.0 mL of solution. Calculate the pH of a saturated aqueous solution of quinine. Consider only the reaction Q+H 2 O ⇌ QH + + OH − described by p K b 1 , where Q = quinine.

Interpretation Introduction

Interpretation: The value of pKb1 , pKb2 , mass and volume of quinine solution is given. By using these values, the pH value of quinine solution is to be calculated.

Concept introduction: The equilibrium constant Kb is the ratio of product of equilibrium concentrations of product to the equilibrium concentration of reactant base.

The number of moles is calculated using the formula,

Numberofmoles=Massofcompound×1molofcompoundMolarmassofcompound

### Explanation of Solution

Explanation

To determine: The pH value of quinine solution.

The number of moles of quinine is 3.08×103mol_ .

Given

Mass of quinine is 1.0g .

The molar mass of quinine is 324.417g/mol .

Formula

The number of moles is calculated using the formula,

Numberofmoles=Massofcompound×1molofcompoundMolarmassofcompound

Substitute the values of mass and molar mass of quinine in the above expression.

Numberofmoles=Massofcomound×1molofcompoundMolarmassofcompound=1.0g×1molofquinine324.417g=3.08×103mol_

The concentration of quinine is 1.62×103M_ .

Given

Volume of quinine solution is 1900.0mL .

The number of moles of quinine is 3.08×103mol .

The conversion of milliliter (mL) into liter (L) is done as,

1mL=103L

Hence, the conversion of 1900.0mL into liter is,

1900.0mL=(1900.0×103)L=1.9000L

Formula

The concentration is calculated using the formula,

Concentration=NumberofmolesVolumeofsolution

Substitute the values of number of moles and volume of quinine solution in the above expression.

Concentrationofquinine=NumberofmolesVolumeofsolution=3.0×103mol1.9000L=1.62×103M_

The stated reaction is,

Q+H2OQH++OH

The concentration of [OH] is 1.10×104_ .

The concentration of quinine is 1.62×103M .

It is assumed that the change in concentration of [OH] is x .

Make the ICE table for the above reaction.

Q+H2O(l)QH++OH(aq)Initial:1

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