Calculate \(\left[\mathrm{OH}^{-}\right]\) in a solution that is (a) \(0.0062 \mathrm{M}\) \(\mathrm{Ba}(\mathrm{OH})_{2}\) and \(0.0105 \mathrm{M} \mathrm{BaCl}_{2} ;\) (b) \(0.315 \mathrm{M}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) and \(0.486 \mathrm{M} \mathrm{NH}_{3} ;\) (c) \(0.196 \mathrm{M} \mathrm{NaOH}\) and \(0.264 \mathrm{M}\) \(\mathrm{NH}_{4} \mathrm{Cl}\)

(a) In the first case, barium hydroxide (\(Ba(OH)_2\)) dissociates into barium ions and two hydroxide ions per one molecule of \(Ba(OH)_2\). Therefore, the concentration of hydroxide ions is twice the concentration of barium hydroxide, which is \(2*(0.0062M)=0.0124M\). (b) The second case involves ammonia, which reacts with water to create hydroxide ions; however, the equilibrium constant for this process is small, so we assume that the concentration of hydroxide ions is negligible relative to those introduced by bases. Therefore, the \([OH^-]\) is negligible. (c) Sodium hydroxide (\(NaOH\)) is a strong base, which completely dissociates to give one hydroxide ion per one \(NaOH\) molecule. Therefore, we get a hydroxide ion concentration equal to the sodium hydroxide concentration, which is \(0.196M\).

When a salt is introduced to the solution, it may affect the \([OH^-]\). However, in cases (a) and (c), barium chloride and ammonium chloride do not contribute to \([OH^-]\) because they do not dissociate to produce hydroxide ions. Hence, the \([OH^-]\) for cases (a) and (c) remains 0.0124M and 0.196M respectively.

To summarize, \([OH^-]\) in solution (a) is 0.0124M, in solution (b) is negligible, and in solution (c) is 0.196M.