Will a precipitate of \(\mathrm{Cd}(\mathrm{OH})_{2}\) form if \(1.0 \mathrm{mL}\) of \(1.0 \mathrm{M}\) \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}\) is added to \(1.0 \mathrm{L}\) of \(5.0 \mathrm{MNH}_{3} ?\) $$\mathrm{Cd}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) \quad K=1.0 \times 10^{7}$$ $$\mathrm{Cd}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Cd}^{2+}(a q)+2 \mathrm{OH}^{-}(a q) \quad K_{\mathrm{sp}}=5.9 \times 10^{-15}$$

First, we need to calculate the moles of each species after mixing. For that, we can use the formula: moles = volume × concentration. - Moles of \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}\): \((1.0 \times 10^{-3}\,L) \times (1.0\,\mathrm{M}) = 1.0 \times 10^{-3} \, \text{moles}\) - Moles of \(\mathrm{NH}_{3}\): \((1.0\,L) \times (5.0\,\mathrm{M}) = 5.0\,\text{moles}\) Next, we need to calculate the final concentrations of each species in the mixed solution: - Final volume of the solution: \(1.0\,L + 1.0\times10^{-3}\,L = 1.001\,L\) - Concentration of \(\mathrm{Cd}^{2+}\): \(\frac{1.0\times10^{-3}\,\text{moles}}{1.001\,L} \approx 9.99\times10^{-4}\,\mathrm{M}\) - Concentration of \(\mathrm{NH}_{3}\): \(\frac{5.0\,\text{moles}}{1.001\,L} \approx 4.995\,\mathrm{M}\)

Now, we will calculate the final concentration of \(\mathrm{Cd}^{2+}\) after reacting with \(\mathrm{NH}_{3}\) to form the complex \(\mathrm{Cd}(\mathrm{NH}_{3})_{4}^{2+}\) using the reaction: $$\mathrm{Cd}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) \quad K=1.0 \times 10^{7}$$ According to the stoichiometry of the reaction, 1 mole of \(\mathrm{Cd}^{2+}\) reacts with 4 moles of \(\mathrm{NH}_{3}\). Thus, all 1.0×10^(-3) moles of \(\mathrm{Cd}^{2+}\) react with \(\frac{1}{4} \times 1.0\times10^{-3}\, \text{moles} = 2.5\times10^{-3}\, \text{moles}\) of \(\mathrm{NH}_{3}\). After the reaction, the concentration of each species becomes: - \(\mathrm{Cd}^{2+}\): \(0\,\mathrm{M}\) - \(\mathrm{NH}_{3}\): \(\frac{5.0\,\text{moles} - 2.5\times10^{-3}\,\text{moles}}{1.001\,L} \approx 4.993\,\mathrm{M}\) There will be no free \(\mathrm{Cd}^{2+}\) ions remaining in the solution since all are complexed with ammonia.

As ammonia is a weak base, it can react with water to produce \(\mathrm{OH}^-\) ions. $$\mathrm{NH}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{NH}_{4}^{+}(a q)+\mathrm{OH}^{-}(a q) \quad K_{b} = 1.8 \times 10^{-5}$$ However, since the concentration of \(\mathrm{NH}_{3}\) is very high (about 4.993 M) compared to the \(K_{b}\) value, we can assume that the \(\mathrm{OH}^-\) concentration is very small and negligible.

Now, we can calculate the reaction quotient, Q, for the solubility equilibrium of \(\mathrm{Cd}(\mathrm{OH})_{2}\) using the concentrations of the ions in the solution: $$Q = [\mathrm{Cd}^{2+}][\mathrm{OH}^{-}]^{2}$$ With no free \(\mathrm{Cd}^{2+}\) ions and an insignificant concentration of \(\mathrm{OH}^-\), Q will be close to 0. Since \(Q < K_{\mathrm{sp}}=5.9\times10^{-15}\), a precipitate of \(\mathrm{Cd}(\mathrm{OH})_{2}\) will not form under these conditions.