Calculate the \(\mathrm{pH}\) of each of the following solutions \(\left(K_{a}\right.\) and \(K_{b}\) values are given in Appendix \(D\) ): (a) \(0.095 M\) propionic acid \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}\right),\) (b) \(0.100 \quad M\) hydrogen chromate ion \(\left(\mathrm{HCrO}_{4}^{-}\right),\left(\right.\) c) \(0.120 \mathrm{M}\) pyridine \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\right)\)

Write the ionization equilibrium for propionic acid: \(C_2H_5COOH \rightleftharpoons H^+ + C_2H_5COO^-\) Set up an ICE table to find the equilibrium concentrations of \(\mathrm{H}^{+}\) and \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COO}^{-}\): \[ \begin{array}{c|c c c} & [C_2H_5COOH] & [H^+] & [C_2H_5COO^-] \\ \hline I & 0.095 & 0 & 0 \\ C & -x & x & x \\ E & 0.095-x & x & x \\ \end{array} \] Use the given \(K_a\) value from Appendix D for propionic acid to find x: \(K_a = \frac{[H^+][C_2H_5COO^-]}{[C_2H_5COOH]} = \frac{x^2}{0.095-x}\) Solve for x, which represents the \(\mathrm{H}^{+}\) ion concentration, and then use the pH formula to find the pH of the solution.

Write the ionization equilibrium for hydrogen chromate ion: \(HCrO_4^- \rightleftharpoons H^+ + CrO_4^{2-}\) Follow the same ICE table and equilibrium steps as in the propionic acid calculation with the initial concentration of hydrogen chromate ion given as 0.100 M and the \(K_a\) value for hydrogen chromate from Appendix D.

Write the base ionization equilibrium for pyridine: \(C_5H_5N + H_2O \rightleftharpoons OH^- + C_5H_5NH^+\) Set up an ICE table to find the equilibrium concentrations of \(\mathrm{OH}^{-}\) and \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}\), with the initial concentration of pyridine given as 0.120 M: \[ \begin{array}{c|c c} & [OH^-] & [C_5H_5NH^+] \\ \hline I & 0 & 0 \\ C & +x & +x \\ E & x & x \\ \end{array} \] Use the given \(K_b\) value from Appendix D for pyridine to find x: \(K_b = \frac{[\mathrm{OH}^{-}][\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}]}{[\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}]} = \frac{x^2}{0.120}\) Solve for x, which represents the \(\mathrm{OH}^{-}\) ion concentration. Then, convert this to \(\mathrm{H}^{+}\) concentration by using the relation: \([\mathrm{H}^{+}][\mathrm{OH}^{-}] = K_w\), where \(K_w = 1.0 \times 10^{-14}\) Finally, use the pH formula to find the pH of the pyridine solution.