Calculating Concentrations of Species in a Weak Base Solution Using Kb Morphine, C17H19NO3, is administered medically to relieve pain. It is a naturally occurring base, or alkaloid. What is the pH of a 0.0075 M solution of morphine at 25°C? The base-ionization constant, Kb, is 1.6 × 10^-6 at 25°C.

Question

Calculating Concentrations of Species in a Weak Base Solution Using [latex]K_{b}[/latex]

Morphine, [latex]\mathrm{C}_{17} \mathrm{H}_{19} \mathrm{NO}_{3}[/latex], is administered medically to relieve pain. It is a naturally occurring base, or alkaloid. What is the [latex]\mathrm{pH}[/latex] of a [latex]0.0075 \mathrm{M}[/latex] solution of morphine at [latex]25^{\circ} \mathrm{C}[/latex] ? The base-ionization constant, [latex]K_{b}[/latex], is [latex]1.6 imes 10^{-6}[/latex] at [latex]25^{\circ} \mathrm{C}[/latex].

PROBLEM STRATEGY

The calculation for the ionization of a weak base parallels that used with weak acids in Examples 17.2 and 17.3: you write the equation, make a table of concentrations (Step 1), set up the equilibrium-constant equation for [latex]K_{b}[/latex] (Step 2), and solve for [latex]x=[/latex] [latex]\left[\mathrm{OH}^{-} ight][/latex](Step 3). Assume the self-ionization of water can be neglected. You obtain [latex]\left[\mathrm{H}_{3} \mathrm{O}^{+} ight][/latex], and then the [latex]\mathrm{pH}[/latex], by solving [latex]K_{w}=\left[\mathrm{H}_{3} \mathrm{O}^{+} ight]\left[\mathrm{OH}^{-} ight][/latex].

Accepted Answer

Morphine, which we abbreviate as Mor, ionizes by picking up a proton from water (as does ammonia).

[latex]\operatorname{Mor}(a q)+\mathrm{H}_{2} \mathrm{O} ightleftharpoons \operatorname{HMor}^{+}(a q)+\mathrm{OH}^{-}(a q)[/latex]

STEP 1 Summarize the concentrations in a table. For the change values, note that the morphine in [latex]1 \mathrm{~L}[/latex] of solution ionizes to give [latex]x[/latex] mol HMor [latex]^{+}[/latex]and [latex]x \mathrm{~mol} \mathrm{OH}^{-}[/latex].

[latex]\begin{array}{cccc} ext { Concentration }(M) & \mathrm{Mor}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(l) & ightleftharpoons \mathrm{HMor}^{+}(\mathrm{aq})&+\mathrm{OH}^{-}(a q) \ ext { Starting } & 0.0075 & 0 & \sim 0 \ ext { Change } & -x & +x & +x \ ext { Equilibrium } & 0.0075-x & x & x \end{array} [/latex]

STEP 2 Substituting into the equilibrium equation

[latex]\frac{\left[\mathrm{HMor}^{+} ight]\left[\mathrm{OH}^{-} ight]}{[\mathrm{Mor}]}=K_{b}[/latex]

gives

[latex]\frac{x^{2}}{0.0075-x}=1.6 imes 10^{-6}[/latex]

STEP 3 If you assume that [latex]x[/latex] is small enough to neglect compared with 0.0075, you have

[latex]0.0075-x \simeq 0.0075[/latex]

and you can write the equilibrium-constant equation as

[latex]\frac{x^{2}}{0.0075} \simeq 1.6 imes 10^{-6}[/latex]

Hence,

[latex] \begin{gathered} x^{2} \simeq 1.6 imes 10^{-6} imes 0.0075=1.2 imes 10^{-8} \ x=\left[\mathrm{OH}^{-} ight] \simeq 1.1 imes 10^{-4} \end{gathered} [/latex]

You can now calculate the hydronium-ion concentration. The product of the hydronium-ion concentration and the hydroxide-ion concentration equals [latex]1.0 imes[/latex] [latex]10^{-14}[/latex] at [latex]25^{\circ} \mathrm{C}[/latex] :

[latex]\left[\mathrm{H}_{3} \mathrm{O}^{+} ight]\left[\mathrm{OH}^{-} ight]=1.0 imes 10^{-14}[/latex]

Substituting into this equation gives

[latex]\left[\mathrm{H}_{3} \mathrm{O}^{+} ight] imes 1.1 imes 10^{-4}=1.0 imes 10^{-14}[/latex]

Hence,

[latex]\left[\mathrm{H}_{3} \mathrm{O}^{+} ight]=\frac{1.0 imes 10^{-14}}{1.1 imes 10^{-4}}=9.1 imes 10^{-11}[/latex]

The [latex]\mathrm{pH}[/latex] of the solution is

[latex]\mathrm{pH}=-\log \left[\mathrm{H}_{3} \mathrm{O}^{+} ight]=-\log \left(9.1 imes 10^{-11} ight)=\mathbf{1 0 . 0 4}[/latex]

Note that you could also calculate the [latex]\mathrm{pH}[/latex] using the formula [latex]\mathrm{pH}+\mathrm{pOH}=14.00[/latex].

You would first calculate the [latex]\mathrm{pOH}[/latex]; then [latex]\mathrm{pH}=14.00-\mathrm{pOH}[/latex].

Question 17.5: Calculating Concentrations of Species in a Weak Base Solutio...

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