Aniline, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2},\) is a weak base with \(K_{\mathrm{b}}=3.8 \times 10^{-10}\) (a) Write out the reaction that occurs when aniline reacts with water. (b) (i) What is the concentration of each species at equilibrium in a solution of 0.25\(M \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2} ?\) (ii) What is the pH value for the solution in (i)?

Aniline is a base, so when it is added to water, it accepts a proton (H+) from water, becoming an ammonium ion. The water molecule becomes a hydroxide ion. The reaction can be written as\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \leftrightharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}(aq) + \mathrm{OH}^{-}(aq)\)

We use the equilibrium constant \(K_b\) to create an ICE table and solve for the concentrations. 'I' stands for Initial, 'C' for Change, and 'E' for Equilibrium. - For aniline, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\): - I: 0.25M - C: -x M - E: 0.25 - x M- For the ammonium ion, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\): - I: 0M - C: +x M - E: x M- For the hydroxide ion, \(\mathrm{OH}^{-}\): - I: 0M - C: +x M - E: x MWe plug the 'E' values into the \(K_b\) expression,\(K_b = \frac{[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}][\mathrm{OH}^{-}]}{[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}]} = \frac{(x)(x)}{(0.25-x)} = 3.8 x 10^{-10}\)Solving the \(K_b\) equation will give the value of x, which can be used to find the concentrations.

The pH is calculated using the concentration of the \(\mathrm{OH}^-\) ions at equilibrium using the formula: \(pOH = -\log [\mathrm{OH}^-]\)Then calculate the pH from the pOH using the formula:\(pH = 14 - pOH\)