The \(\mathrm{pH}\) of a \(0.016-M\) aqueous solution of \(p\) -toluidine \(\left(\mathrm{CH}_{3} \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{NH}_{2}\right)\) is 8.60 . Calculate \(K_{\mathrm{b}}\) .

Recall the relationship between pH, pOH, and the ion product constant for water, Kw: \(pH + pOH = 14\) Given the pH of the p-toluidine solution, we can find the pOH as follows: \(pOH = 14 - pH = 14 - 8.60 = 5.40\)

Now that we have the pOH, we can calculate the concentration of OH- ions using the following formula: \[OH^{-} = 10^{-pOH}\] Plugging in the found pOH value: \[OH^{-} = 10^{-5.40} = 3.98 \times 10^{-6}\,M\]

The Kb expression for p-toluidine \((CH_{3}C_{6}H_{4}NH_{2})\) is as follows: \[K_{b} = \frac{[OH^{-}][CH_{3}C_{6}H_{4}NH_{3}^{+}]}{[CH_{3}C_{6}H_{4}NH_{2}]}\] At equilibrium, the concentration of p-toluidine, [\(CH_{3}C_{6}H_{4}NH_{2}\)], is equal to its initial concentration minus the concentration of OH- ions produced. Since p-toluidine is a weak base and does not dissociate completely, we can assume that the change in its concentration will be small, so it can be approximated as follows: \[[CH_{3}C_{6}H_{4}NH_{2}] \approx 0.016\,M - [OH^{-}]\] The concentration of p-toluidine cation, [\(CH_{3}C_{6}H_{4}NH_{3}^{+}\)], is equal to the concentration of OH- ions produced: \[[CH_{3}C_{6}H_{4}NH_{3}^{+}] = [OH^{-}]\] Now we can substitute these expressions into the Kb expression: \[K_{b} = \frac{[OH^{-}][OH^{-}]}{(0.016\,M - [OH^{-}])}\]

We can now solve for Kb using the concentration of OH- ions calculated in Step 2: \[K_{b} = \frac{(3.98 \times 10^{-6})^2}{(0.016 - 3.98 \times 10^{-6})}\] Simplifying the expression and solving for Kb: \[K_{b} \approx 9.96 \times 10^{-10}\] So, the base dissociation constant, Kb, for p-toluidine is approximately \(9.96 \times 10^{-10}\).