Questions: In the laboratory, a general chemistry student measured the pH of a 0.334 M aqueous solution of aniline, C6H5NH2 to be 9.034. Use the information she obtained to determine the Kb for this base. Kb(experiment) =

Solution Steps

The pH of the solution is given as 9.034. First, we need to find the pOH using the relationship: \[ \text{pOH} = 14 - \text{pH} \] \[ \text{pOH} = 14 - 9.034 = 4.966 \]

Next, we calculate the concentration of OH\(^-\) ions using the pOH: \[ [\text{OH}^-] = 10^{-\text{pOH}} \] \[ [\text{OH}^-] = 10^{-4.966} \approx 1.080 \times 10^{-5} \, \text{M} \]

Aniline (\(\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_2\)) is a weak base and reacts with water as follows: \[ \mathrm{C}_6\mathrm{H}_5\mathrm{NH}_2 + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{C}_6\mathrm{H}_5\mathrm{NH}_3^+ + \mathrm{OH}^- \]

The equilibrium expression for the base dissociation constant \(K_b\) is: \[ K_b = \frac{[\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_3^+][\mathrm{OH}^-]}{[\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_2]} \]

Initially, the concentration of aniline is 0.334 M. At equilibrium, let \(x\) be the concentration of \(\mathrm{OH}^-\) ions formed, which we have already calculated as \(1.080 \times 10^{-5} \, \text{M}\).

Thus, the concentration of \(\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_3^+\) is also \(1.080 \times 10^{-5} \, \text{M}\), and the concentration of \(\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_2\) is: \[ [\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_2] = 0.334 - 1.080 \times 10^{-5} \approx 0.334 \, \text{M} \]

Substitute the equilibrium concentrations into the \(K_b\) expression: \[ K_b = \frac{(1.080 \times 10^{-5})(1.080 \times 10^{-5})}{0.334} \] \[ K_b = \frac{1.1664 \times 10^{-10}}{0.334} \approx 3.491 \times 10^{-10} \]

Final Answer