   # Codeine (C 18 H 21 NO 3 ) is a derivative of morphine that is used as an analgesic, narcotic, or antitussive. It was once commonly used in cough syrups but is now available only by prescription because of its addictive properties. If the pH of a 1.7 × 10 −3 - M solution of codeine is 9.59, calculate K b. ### 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 155AE
Textbook Problem
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## Codeine (C18H21NO3) is a derivative of morphine that is used as an analgesic, narcotic, or antitussive. It was once commonly used in cough syrups but is now available only by prescription because of its addictive properties. If the pH of a 1.7 × 10−3-M solution of codeine is 9.59, calculate Kb.

Interpretation Introduction

Interpretation: The concentration and pH of derivative of morphine (codeine) solution is given. By using these values, the value of Kb is to be calculated.

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

### Explanation of Solution

Explanation

To determine: The pOH and concentration of OH ; the concentration of morphine; the value of Kb .

The stated reaction is,

C18H21O3N(aq)+H2O(l)C18H21O3NH+(aq)+OH(aq)

The value of pOH is 4.41_ .

Given

The initial concentration of codeine is 1.7×103M .

The pH value of reactant is 9.59 .

Formula

The relation between pOH and pH is,

pH+pOH=14

Where,

• pH is the measure of H+ ions.
• pOH is the measure of OH .

Substitute the value of is concentration of pOH in the above equation.

pH+pOH=14pOH=149.59=4.41_

The stated reaction is,

C18H21O3N(aq)+H2O(l)C18H21O3NH+(aq)+OH(aq)

The concentration of OH ions is 3.89×105M_ .

The pOH is calculated using the formula,

pOH=log10[OH]

Where,

• pOH is the measure of OH ions.
• [OH] is concentration of OH ions.

Substitute the value of pOH in the above equation.

pOH=log10[OH]4.41=log10[OH][OH]=3.89×105M_

The stated reaction is,

C18H21O3N(aq)+H2O(l)C18H21O3NH+(aq)+OH(aq)

The value of Kb is 9

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