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Chapter 16: Problem 74

Codeine \(\left(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{3}\right)\) is a weak organic base. A \(5.0 \times 10^{-3} \mathrm{M}\) solution of codeine has a pH of 9.95. Calculate the value of \(K_{b}\) for this substance. What is the \(\mathrm{p} K_{b}\) for this base?

Short Answer

Expert verified
The value of \( K_b \) is approximately \( 1.61 \times 10^{-6} \), and \( \text{p}K_b \) is about 5.79.

Step by step solution

01

Understanding the Problem

We need to find the base ionization constant \( K_b \) and the \( \text{p}K_b \) of codeine. We know the concentration of codeine solution is \( 5.0 \times 10^{-3} \text{ M} \), and its pH is 9.95.
02

Calculate the Concentration of OH鈦�

Since the pH is given, we first find the pOH by subtracting the pH from 14: \( \text{pOH} = 14 - 9.95 = 4.05 \). Then, convert the pOH to the concentration of OH鈦� using: \[\text{[OH鈦籡} = 10^{-\text{pOH}} = 10^{-4.05}.\]
03

Set Up the Equilibrium Expression

Codeine \( C_{18}H_{21}NO_3 \) partially dissociates in water to form \( C_{18}H_{21}NO_3^+ \) and OH鈦�. The equilibrium expression for the reaction is: \[ K_b = \frac{[C_{18}H_{21}NO_3^+][OH^-]}{[C_{18}H_{21}NO_3]}. \]
04

Calculate the Hydroxide Ion Concentration

Calculate \([\text{OH}^-]\) using the formula: \([\text{OH}^-] = 10^{-4.05} \approx 8.91 \times 10^{-5}\text{ M}.\)
05

Determine Initial and Equilibrium Concentrations

Initially, \([C_{18}H_{21}NO_3]\) is \(5.0 \times 10^{-3}\text{ M}\). At equilibrium, \([C_{18}H_{21}NO_3]\) decreases by \([OH^-]\) while \([C_{18}H_{21}NO_3^+]\) and \([OH^-]\) both increase by \([OH^-]\).
06

Calculate \(K_b\)

Substitute into the equilibrium expression: \[ K_b = \frac{(8.91 \times 10^{-5})(8.91 \times 10^{-5})}{5.0 \times 10^{-3} - 8.91 \times 10^{-5}}. \] Calculate \(K_b\) and simplify: \[ K_b \approx 1.61 \times 10^{-6}. \]
07

Calculate \(\text{p} K_b\)

Knowing \( K_b = 1.61 \times 10^{-6} \), calculate \( \text{p} K_b \) using \( \text{p} K_b = -\log_{10}(K_b) \): \[ \text{p} K_b = -\log_{10}(1.61 \times 10^{-6}) \approx 5.79. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

pH Calculations
Before diving into calculations related to bases, understanding the pH scale is key. The pH scale ranges from 0 to 14. It is used to measure the acidity or basicity of a solution.
- Values below 7 indicate an acidic solution.- Values above 7 indicate a basic solution.- A pH of 7 is neutral, meaning the solution is neither acidic nor basic.
To determine the pH of a base like codeine, recognize that it is inversely related to the concentration of hydrogen ions (\([ ext{H}^+]\)) in the solution. Specifically, we use the formula:
\[ ext{pH} = - ext{log}_{10}([ ext{H}^+])\]In the case of weak bases, instead of working directly with pH, we often calculate the \( ext{pOH}\). In the example of codeine, moving from pH to \( ext{pOH}\) involves using the relationship:
\[ ext{pOH} = 14 - ext{pH}\]Once \( ext{pOH}\) is determined, the concentration of hydroxide ions (\([ ext{OH}^-])\) can be found using:\[[ ext{OH}^-] = 10^{- ext{pOH}}\]This gives insight into the strength of the base in question.
Equilibrium Expressions
When studying chemical reactions involving weak bases, understanding equilibrium expressions is fundamental. For codeine, which partially ionizes in solution, we describe the reaction using an equilibrium expression.
Consider the ionization of codeine in water:- Codeine (\(C_{18}H_{21}NO_3\)) reacts with water, producing \(C_{18}H_{21}NO_3^+\) and \([ ext{OH}^-]\) ions.
The extent of this ionization is quantified using the base ionization constant, \(K_b\). This constant is crucial for understanding how much the base ionizes and can be represented by:\[K_b = \frac{[C_{18}H_{21}NO_3^+][OH^-]}{[C_{18}H_{21}NO_3]}\]This expression indicates that at equilibrium, the product of the concentrations of the ions produced is divided by the concentration of the undissociated base.
  • The \(K_b\) value tells us about the base's strength; higher values indicate stronger bases.
  • It helps predict how the system will respond to changes (e.g., shifts in concentration).
Calculating and understanding this value is crucial for chemists when they analyze organic bases like codeine.
Organic Bases
Organic bases are an interesting and important part of chemistry. They include compounds like codeine, characterized by containing nitrogen atoms ready to accept protons (H+ ions).
- Codeine, with its structure (\(C_{18}H_{21}NO_3\)), acts as a weak organic base.
One of the key characteristics of organic bases is that they don't completely dissociate in water, unlike strong bases like NaOH. This partial dissociation is why the concept of \(K_b\), the base ionization constant, becomes significant.
In the context of organic chemistry:
  • The \(K_b\) informs us about how easily the base donates electrons to accept a proton.
  • Its measurement and comparison with other bases can influence reactions, such as in pharmaceuticals where codeine is used.
Understanding organic bases not only helps in laboratory settings but also in industrial applications, like drug formulation, where knowledge of \(K_b\) can affect drug efficacy and safety.

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Most popular questions from this chapter

Consider two solutions, solution A and solution B. [H \(\left.^{+}\right]\) in solution A is 25 times greater than that in solution \(\mathrm{B}\). What is the difference in the pH values of the two solutions?

(a) Give the conjugate base of the following Br酶nsted Lowry acids: \((\mathbf{i}) \mathrm{H}_{2} \mathrm{PO}_{4}^{-},(\mathbf{i i}) \mathrm{HBr}\). (b) Give the conjugate acid of the following Br酶nsted-Lowry bases: \((\mathbf{i}) \mathrm{CN}^{-},(\mathbf{i i}) \mathrm{HSO}_{4}^{-}\).

Ephedrine, a central nervous system stimulant, is used in nasal sprays as a decongestant. This compound is a weak organic base: $$ \mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ON}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ONH}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ A \(0.035 \mathrm{M}\) solution of ephedrine has a \(\mathrm{pH}\) of 11.33 . (a) What are the equilibrium concentrations of \(\mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ON}, \mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ONH}^{+},\) and \(\mathrm{OH}^{-} ?\) (b) Calculate \(K_{b}\) for ephedrine.

In many reactions, the addition of \(\mathrm{AlCl}_{3}\) produces the same effect as the addition of \(\mathrm{H}^{+}\). (a) Draw a Lewis structure for \(\mathrm{AlCl}_{3}\) in which no atoms carry formal charges, and determine its structure using the VSEPR method. (b) What characteristic is notable about the structure in part (a) that helps us understand the acidic character of \(\mathrm{AlCl}_{3}\) ? (c) Predict the result of the reaction between \(\mathrm{AlCl}_{3}\) and \(\mathrm{NH}_{3}\) in a solvent that does not participate as a reactant. (d) Which acid-base theory is most suitable for discussing the similarities between \(\mathrm{AlCl}_{3}\) and \(\mathrm{H}^{+}\) ?

Calculate \(\left[\mathrm{OH}^{-}\right]\) and \(\mathrm{pH}\) for each of the following strong base solutions: (a) \(0.182 \mathrm{M} \mathrm{KOH},(\mathbf{b}) 3.165 \mathrm{~g}\) of \(\mathrm{KOH}\) in 500.0 \(\mathrm{mL}\) of solution, (c) \(10.0 \mathrm{~mL}\) of \(0.0105 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\) diluted to \(500.0 \mathrm{~mL},\) (d) a solution formed by mixing \(20.0 \mathrm{~mL}\) of 0.015 \(M \mathrm{Ba}(\mathrm{OH})_{2}\) with \(40.0 \mathrm{~mL}\) of \(8.2 \times 10^{-3} \mathrm{M} \mathrm{NaOH}\).
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