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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. \(\mathrm{A} 5.0 \times 10^{-3} \mathrm{M}\) solution of codeine has a pH of \(9.95 .\) Calculate the value of \(K_{b}\) for this substance. What is the \(\mathrm{pK}_{b}\) for this base?

Short Answer

Expert verified
The value of \(K_b\) for codeine is approximately \(1.585 \times 10^{-6}\), and the \(pK_b\) is approximately 5.80.

Step by step solution

01

Calculate the concentration of hydroxide ions (OH⁻) using the given pH value.

We know that the pH of the solution is 9.95. The relationship between pH and pOH is given by the equation: \(pH + pOH = 14\) We can find the pOH value as follows: \(pOH = 14 - pH = 14 - 9.95 = 4.05\) Now, we can calculate the concentration of hydroxide ions, [OH⁻], using the formula: \(pOH = -\log{[OH^-]}\) Rearrange the formula to solve for [OH⁻]: \([OH^-] = 10^{-pOH} = 10^{-4.05} \approx 8.913 \times 10^{-5} M\)
02

Calculate the concentration of codeine ions (C18H21NO3+) in the solution.

Given that the initial concentration of codeine is \(5.0 \times 10^{-3} M\), we can set up an expression for the equilibrium concentrations of codeine and its ions: Initial: \( C_{18} H_{21} NO_{3} \rightarrow 5.0 \times 10^{-3} - x\) \( OH^- \rightarrow x\) \(C_{18} H_{21} NO_{3}^+ \rightarrow x\) We know that [OH⁻] = x, and we have already calculated the concentration of hydroxide ions (OH⁻) in Step 1. Therefore, x is approximately equal to \(8.913 \times 10^{-5} M\).
03

Calculate the value of Kb using the equilibrium concentrations.

With the concentrations calculated in Step 2, we can now find Kb. The reaction for codeine with water in the equilibrium can be represented as follows: \(C_{18} H_{21} NO_{3}(aq) + H_2O(l) \rightleftharpoons C_{18} H_{21} NO_{3}^+(aq) + OH^-(aq)\) The formula for the equilibrium constant Kb is given by: \(K_b = \frac{[C_{18} H_{21} NO_{3}^+][OH^-]}{[C_{18} H_{21} NO_{3}]}\) Insert the equilibrium concentrations from Step 2 and calculate Kb: \(K_b = \frac{(8.913 \times 10^{-5})(8.913 \times 10^{-5})}{5.0 \times 10^{-3} - 8.913 \times 10^{-5}} \approx 1.585 \times 10^{-6}\)
04

Calculate the pKb value for the base.

Now that we have Kb, we can find the pKb using the formula: \(pK_b = -\log{K_b}\) Insert the Kb value and calculate pKb: \(pK_b = -\log{(1.585 \times 10^{-6})} \approx 5.80\) The value of Kb for codeine is approximately \(1.585 \times 10^{-6}\), and the pKb is approximately 5.80.

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Most popular questions from this chapter

The iodate ion is reduced by sulfite according to the following reaction: $$\mathrm{IO}_{3}^{-}(a q)+3 \mathrm{SO}_{3}^{2-}(a q) \longrightarrow \mathrm{I}^{-}(a q)+3 \mathrm{SO}_{4}^{2-}(a q)$$ The rate of this reaction is found to be first order in \(\mathrm{IO}_{3}^{-}\) , first order in \(\mathrm{SO}_{3}^{2-}\) , and first order in \(\mathrm{H}^{+}\) . (a) Write the rate law for the reaction. (b) By what factor will the rate of the reaction change if the pH is lowered from 5.00 to 3.50\(?\) Does the reaction proceed more quickly or more slowly at the lower pH? (c) By using the concepts discussed in Section 14.6, ex-plain how the reaction can be pH-dependent even though H' does not appear in the overall reaction.

An unknown salt is either NaF, NaCl, or NaOCl. When 0.050 mol of the salt is dissolved in water to form 0.500 L of solution, the pH of the solution is 8.08. What is the identity of the salt?

A 0.100\(M\) solution of chloroacetic acid \(\left(\mathrm{ClCH}_{2} \mathrm{COOH}\right)\) is 11.0\(\%\) ionized. Using this information, calculate \(\left[\mathrm{ClCH}_{2} \mathrm{COO}^{-}\right],\left[\mathrm{H}^{+}\right],\left[\mathrm{ClCH}_{2} \mathrm{COOH}\right],\) and \(K_{a}\) for chloroacetic acid.

At \(50^{\circ} \mathrm{C},\) the ion-product constant for \(\mathrm{H}_{2} \mathrm{O}\) has the value \(K_{w}=5.48 \times 10^{-14}\) . (a) What is the pH of pure water at \(50^{\circ} \mathrm{C} ?\) (b) Based on the change in \(K_{w}\) with temperature, predict whether \(\Delta H\) is positive, negative, or zero for the autoionization reaction of water: $$2 \mathrm{H}_{2} \mathrm{O}(l) \Longrightarrow \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q)$$

Label each of the following as being a strong acid, a weak acid, or a species with negligible acidity. In each case write the formula of its conjugate base, and indicate whether the conjugate base is a strong base, a weak base, or a species with negligible basicity: \((\mathbf{a}) \mathrm{HCOOH}\), \((\mathbf{b})\mathrm{H}_{2},(\mathrm{c}) \mathrm{CH}_{4},(\mathbf{d}) \mathrm{HF},(\mathbf{e}) \mathrm{NH}_{4}^{+}\)
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