Calculate the \(\mathrm{pH}\) of each of the following solutions. a. \(0.10 \mathrm{MCH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\) b. \(0.050 M \mathrm{NaCN}\)

In the given exercise, we have two different solutions: a. \(0.10 M \mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\) is an ammonium salt, which dissociates in water to form \(\mathrm{CH}_{3}\mathrm{NH}_{3}^{+}\) ions and \(\mathrm{Cl}^-\) ions. The \(\mathrm{CH}_{3}\mathrm{NH}_{3}^{+}\) ion acts as an acid, donating a proton to water to form \(\mathrm{CH}_{3}\mathrm{NH}_{2}\) and \(\mathrm{H_3O^+}\) ions. b. \(0.050 M \mathrm{NaCN}\) is a salt of a weak acid \(\mathrm{HCN}\) and strong base \(\mathrm{NaOH}\). It dissociates in water to form \(\mathrm{Na}^{+}\) and \(\mathrm{CN}^{-}\) ions. \(\mathrm{CN}^-\) will act as a weak base, taking a proton from water, generating \(\mathrm{HCN}\) and \(\mathrm{OH}^-\) ions. Now that we have identified the nature of each compound in the given solutions, let's proceed to calculate the concentration of \(\mathrm{H^{+}}\) ions in each case.

a. For \(0.10 M \mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\) solution: As we know, the reaction of \(\mathrm{CH}_{3}\mathrm{NH}_{3}^{+}\) with water is: \[\mathrm{CH}_{3} \mathrm{NH}_{3}^{+} + \mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{CH}_{3} \mathrm{NH}_{2} + \mathrm{H}_{3} \mathrm{O}^{+}\] The \(K_{a}\) expression for this equation is: \[K_{a} = \frac{[\mathrm{H_3O}^+][\mathrm{CH_3NH_2}]}{[\mathrm{CH_3NH_3}^+]}\] Since the initial concentration of the acid is given, we can use the ICE table to determine the final concentrations of each species: Initial: \([\mathrm{CH_3NH_3^+}] = 0.10 M\), \([\mathrm{H_3O^+}] = 0\), \([\mathrm{CH_3NH_2}] = 0\) Change: \([\mathrm{CH_3NH_3^+}] = -x\), \([\mathrm{H_3O^+}] = x\), \([\mathrm{CH_3NH_2}] = x\) Equilibrium: \([\mathrm{CH_3NH_3^+}] = 0.10-x\), \([\mathrm{H_3O^+}] = x\), \([\mathrm{CH_3NH_2}] = x\) Now, substitute these equilibrium concentrations into the \(K_a\) expression and solve for \(x\) to find \([\mathrm{H_3O^+}]\). Then, use the definition of pH to calculate the pH of the solution: \[\mathrm{pH} = -\log_{10}[\mathrm{H_3O}^+]\] b. For \(0.050 M \mathrm{NaCN}\) solution: Since \(\mathrm{CN}^-\) will act as a weak base, we need to consider its reaction with water: \[\mathrm{CN}^- + \mathrm{H_2O} \rightleftharpoons \mathrm{HCN} + \mathrm{OH}^-\] The \(K_b\) expression for this equation is: \[K_b = \frac{[\mathrm{HCN}][\mathrm{OH^-}]}{[\mathrm{CN^-}]}\] Similar to step a, we can use an ICE table to determine the equilibrium concentrations of each species. Initial: \([\mathrm{CN^-}] = 0.050 M\), \([\mathrm{HCN}] = 0\), \([\mathrm{OH^-}] = 0\) Change: \([\mathrm{CN^-}] = -x\), \([\mathrm{HCN}] = x\), \([\mathrm{OH^-}] = x\) Equilibrium: \([\mathrm{CN^-}] = 0.050-x\), \([\mathrm{HCN}] = x\), \([\mathrm{OH^-}] = x\) Substitute these equilibrium concentrations into the \(K_b\) expression and solve for \(x\) to find \([\mathrm{OH^-}]\). Then, calculate \([\mathrm{H_3O^+}]\) using the relation: \[[\mathrm{H_3O^+}] [\mathrm{OH^-}] = K_w\] Finally, calculate the pH using the definition of pH: \[\mathrm{pH} = -\log_{10}[\mathrm{H_3O}^+]\]