Chapter 15, Problem 46P

Sulfanilic acid $\left({\text{NH}}_{\text{2}}{\text{C}}_{\text{6}}{\text{H}}_{\text{4}}{\text{SO}}_{\text{3}}\text{H}\right)$ is used in manufacturing dyes. It ionizes in water according to the equilibrium equation ${\text{NH}}_{\text{2}}{\text{C}}_{\text{6}}{\text{H}}_{\text{4}}{\text{SO}}_{\text{3}}\text{H}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(\mathcal{l}\right)\underset{}{\overset{}{\rightleftarrows }}{\text{NH}}_{\text{2}}{\text{C}}_{\text{6}}{\text{H}}_{\text{4}}{\text{SO}}_3^{-}\left(aq\right)+{\text{H}}_{\text{3}}{\text{O}}^{+}\left(aq\right)K_a=5.9\times {10}^{-4}$
A buffer is prepared by dissolving 0.20 mol of sulfanilicacid and 0.13 mol of sodium sulfanilate $\left({\text{NaNH}}_{\text{2}}{\text{C}}_{\text{6}}{\text{H}}_{\text{4}}{\text{SO}}_{\text{3}}\right)$ in water and diluting to 1.00 L.

1. Compute the pH of the solution.
2. Suppose 0.040 mol of HCl is added to the buffer.Calculate the pH of the solution that results.