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

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

\operatorname{Mor}(a q)+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \operatorname{HMor}^{+}(a q)+\mathrm{OH}^{-}(a q)

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

STEP 2 Substituting into the equilibrium equation

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

gives

\frac{x^{2}}{0.0075-x}=1.6 \times 10^{-6}

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

0.0075-x \simeq 0.0075

and you can write the equilibrium-constant equation as

\frac{x^{2}}{0.0075} \simeq 1.6 \times 10^{-6}

Hence,

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

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

\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]=1.0 \times 10^{-14}

Substituting into this equation gives

\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] \times 1.1 \times 10^{-4}=1.0 \times 10^{-14}

Hence,

\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\frac{1.0 \times 10^{-14}}{1.1 \times 10^{-4}}=9.1 \times 10^{-11}

The \mathrm{pH} of the solution is

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

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

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