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Chapter 16: Problem 53

Codeine, \(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{O}_{3} \mathrm{N},\) is an opiate, has analgesic and antidiarrheal properties, and is widely used. In water, codeine is a weak base. A handbook gives \(\mathrm{p} K_{\mathrm{a}}=6.05\) for protonated codeine, \(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{O}_{3} \mathrm{NH}^{+} .\) Write the reaction for \(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{O}_{3} \mathrm{NH}^{+}\) and calculate \(\mathrm{p} K_{\mathrm{b}}\) for codeine.

Short Answer

Expert verified
The \(pK_b\) for codeine is approximately 7.95.

Step by step solution

01

Write the reaction for protonated codeine

The protonated codeine (\(\mathrm{C}_{18}\mathrm{H}_{21}\mathrm{O}_{3}\mathrm{NH}^{+}\)) works as a weak acid and can donate a proton to water in the solution, thus becoming codeine (\(\mathrm{C}_{18}\mathrm{H}_{21}\mathrm{O}_{3}\mathrm{N}\)). \n\nThe reaction would be: \(\mathrm{C}_{18}\mathrm{H}_{21}\mathrm{O}_{3}\mathrm{NH}^{+} + H_2O \leftrightarrow \mathrm{C}_{18}\mathrm{H}_{21}\mathrm{O}_{3}\mathrm{N} + H_3O^{+}\)
02

Formulate the definition of \(pK_b\)

The \(pK_b\) value represents the negative base-10 logarithm of the base dissociation constant, \(K_b\). Codeine, in water, behaves as a weak base, so it can accept a proton from water. However, in this problem we're not directly given \(K_b\), but we're given \(pK_a\) for the protonated version of codeine.
03

Calculate \(pK_b\)

Using the provided \(pK_a\), we can use the relation \(pK_w = pK_a + pK_b\). \n\nAt room temperature, the \(pK_w\) of water is approximately 14. So we can use this to calculate \(pK_b\) as follows: \(\n\n pK_b = pK_w - pK_a \)\n\nSubstitute the given and known values: \(\n\n pK_b = 14 - 6.05 = 7.95\)

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Most popular questions from this chapter

Of the following, the amphiprotic ion is (a) \(\mathrm{HCO}_{3}^{-}\) (b) \(\mathrm{CO}_{3}^{2-} ;\) (c) \(\mathrm{NH}_{4}^{+} ;\) (d) \(\mathrm{CH}_{3} \mathrm{NH}_{3}^{+} ;\) (e) \(\mathrm{ClO}_{4}^{-}\).The \(\mathrm{pH}\) in \(0.10 \mathrm{M} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}(\mathrm{aq})\) must be (a) equal to \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) in \(0.10 \mathrm{M} \mathrm{HNO}_{2}(\mathrm{aq}) ;\) (b) less than the \(\mathrm{pH}\) in \(0.10 \mathrm{M} \mathrm{HI}(\mathrm{aq}) ;\) (c) greater than the \(\mathrm{pH}\) in \(0.10 \mathrm{M} \mathrm{HBr}(\mathrm{aq}) ;\) (d) equal to \(1.0.\)

The following four equilibria lie to the right: \(\mathrm{N}_{2} \mathrm{H}_{5}^{+}+\) \(\mathrm{CH}_{3} \mathrm{NH}_{2} \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}+\mathrm{CH}_{3} \mathrm{NH}_{3}^{+} ; \mathrm{H}_{2} \mathrm{SO}_{3}+\mathrm{F}^{-} \longrightarrow\) \(\mathrm{HSO}_{3}^{-}+\mathrm{HF} ; \mathrm{CH}_{3} \mathrm{NH}_{3}^{+}+\mathrm{OH}^{-} \longrightarrow \mathrm{CH}_{3} \mathrm{NH}_{2}+\) \(\mathrm{H}_{2} \mathrm{O} ;\) and \(\mathrm{HF}+\mathrm{N}_{2} \mathrm{H}_{4} \longrightarrow \mathrm{F}^{-}+\mathrm{N}_{2} \mathrm{H}_{5}^{+}\) (a) Rank all the acids involved in order of decreasing acid strength. (b) Rank all the bases involved in order of decreasing base strength. (c) State whether each of the following two equilibria lies primarily to the right or to the left: (i) \(\mathrm{HF}+\mathrm{OH}^{-} \longrightarrow \mathrm{F}^{-}+\mathrm{H}_{2} \mathrm{O} ;\) (ii) \(\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}+\) \(\mathrm{HSO}_{3}^{-} \longrightarrow \mathrm{CH}_{3} \mathrm{NH}_{2}+\mathrm{H}_{2} \mathrm{SO}_{3}\).

The antimalarial drug quinine, \(\mathrm{C}_{20} \mathrm{H}_{24} \mathrm{O}_{2} \mathrm{N}_{2},\) is a diprotic base with a water solubility of \(1.00 \mathrm{g} / 1900 \mathrm{mL}\) of solution. (a) Write equations for the ionization equilibria corresponding to \(\mathrm{p} K_{\mathrm{b}_{1}}=6.0\) and \(\mathrm{p} K_{\mathrm{b}_{2}}=9.8\) (b) What is the \(\mathrm{pH}\) of saturated aqueous quinine?

Suppose you wanted to produce an aqueous solution of \(\mathrm{pH}=8.65\) by dissolving one of the following salts in water. Which salt would you use, and at what molarity? (a) \(\mathrm{NH}_{4} \mathrm{Cl} ;\) (b) \(\mathrm{KHSO}_{4} ;\) (c) \(\mathrm{KNO}_{2}\); (d) \(\mathrm{NaNO}_{3}\).

Calculate \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) for each solution:(a) \(0.00165 \mathrm{M} \mathrm{HNO}_{3} ;\) (b) \(0.0087 \mathrm{M} \mathrm{KOH} ;\) (c) \(0.00213 \mathrm{M}\) \(\operatorname{Sr}(\mathrm{OH})_{2} ;(\mathrm{d}) 5.8 \times 10^{-4} \mathrm{M} \mathrm{HI}\)
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