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Chapter 14: Problem 104

Codeine \(\left(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{3}\right)\) 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 \times 10^{-3}-M\( solution of codeine is 9.59 , calculate \)K_{\mathrm{b}}$ .

Short Answer

Expert verified
The base dissociation constant, Kb, for codeine is approximately 8.91 × 10^-9.

Step by step solution

01

Calculate pOH

To find the pOH, we will use the formula: pOH = 14 - pH From the given information, pH = 9.59. So, pOH = 14 - 9.59 = 4.41.
02

Calculate [OH-] concentration

Now, let's find the concentration of hydroxide ions [OH-] using the formula: [OH-] = 10^(-pOH) [OH-] = 10^(-4.41) = 3.89 × 10^-5 M.
03

Set up the reaction

Since codeine acts as a base, it will react with water to produce hydroxide ions and the conjugate acid: C18H21NO3 + H2O ⇌ C18H21NO3H+ + OH-
04

Write the Kb expression

Now, we will set up the Kb expression for the given reaction: Kb = \(\frac{[C_{18}H_{21}NO_{3}H^{+}][OH^{-}]}{[C_{18}H_{21}NO_{3}]}\)
05

Find the change in concentration

As codeine reacts with water, there will be a change in the concentrations of the species involved. Initially, we have: [C18H21NO3] = 1.7 × 10^-3 M [C18H21NO3H+] = 0 [OH-] = 3.89 × 10^-5 M (from Step 2) After the reaction reaches equilibrium, we will have: [C18H21NO3] = 1.7 × 10^-3 - x M [C18H21NO3H+] = x M [OH-] = 3.89 × 10^-5 + x M
06

Use the Kb expression to find Kb

Now, we will plug the equilibrium concentrations into the Kb expression: Kb = \(\frac{x(3.89 \times 10^{-5}+x)}{1.7 \times 10^{-3}-x}\) Since x is much smaller than 3.89 × 10^-5 M, we can approximate x as negligible. Therefore, the expression becomes: Kb ≈ \(\frac{x(3.89 \times 10^{-5})}{1.7 \times 10^{-3}}\) Now, we use the fact that the hydroxide ion concentration ([OH-]) is equal to the concentration of the conjugate acid (C18H21NO3H+) at equilibrium: x = 3.89 × 10^-5 M Finally, we can solve for Kb: Kb ≈ \(\frac{(3.89 \times 10^{-5})(3.89 \times 10^{-5})}{1.7 \times 10^{-3}}\) = 8.91 × 10^-9 So, the base dissociation constant, Kb, for codeine is approximately 8.91 × 10^-9.

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Most popular questions from this chapter

A solution of formic acid (HCOOH, \(K_{\mathrm{a}}=1.8 \times 10^{-4} )\) has a \(\mathrm{pH}\) of 2.70 . Calculate the initial concentration of formic acid in this solution.

A codeine-containing cough syrup lists codeine sulfate as a major ingredient instead of codeine. The Merck Index gives $\mathrm{C}_{36} \mathrm{H}_{44} \mathrm{N}_{2} \mathrm{O}_{10} \mathrm{S}$ as the formula for codeine sulfate. Describe the composition of codeine sulfate. (See Exercise \(104 . )\) Why is codeine sulfate used instead of codeine?

Calculate the pH of each of the following solutions. a. 0.12\(M \mathrm{KNO}_{2}\) b. 0.45 \(\mathrm{MNaOCl}\) c. 0.40 \(\mathrm{M} \mathrm{N} \mathrm{a} \mathrm{OCl}\)

Using your results from Exercise \(133,\) place the species in each of the following groups in order of increasing base strength. a. \(\mathrm{OH}^{-}, \mathrm{SH}^{-}, \mathrm{SeH}^{-}\) b. \(\mathrm{NH}_{3}, \mathrm{PH}_{3}\) c. \(\mathrm{NH}_{3}, \mathrm{HONH}_{2}\)

Write the reaction and the corresponding \(K_{\mathrm{b}}\) equilibrium expression for each of the following substances acting as bases in water. a. \(\mathrm{NH}_{3}\) b. \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\)
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