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

Before we proceed, we need to calculate the initial concentrations of \(\mathrm{Cd}^{2+}\) and \(\mathrm{NH}_{3}\) after mixing the solutions. Initial concentration of \(\mathrm{Cd}^{2+}\): $$ \frac{1.0 \mathrm{mL}\cdot 1.0 M}{(1.0+1.0) \mathrm{mL}} = \frac{1.0}{2.000} M = 0.5 M $$ Initial concentration of \(\mathrm{NH}_{3}\): $$ \frac{(1.0\mathrm{L})(5.0 M)}{1.001\mathrm{L}} \approx 5.0 M $$

We now use the complex formation constant to determine the concentration of the complex ion \(\mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\). Let x be the concentration of \(\mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\). From the chemical equation and equilibrium constant expression: $$ K = \frac{[\mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4}^{2+}]}{[\mathrm{Cd}^{2+}][\mathrm{NH}_{3}]^4} = 1.0 \times 10^{7} $$ Substitute the concentrations found in step 1: $$ 1.0 \times 10^{7} = \frac{x}{(0.5-x)(5.0-4x)^4} $$ Since K is large, we assume that x is close to 0.5 M. Therefore, we can simplify and solve for x using these assumptions: $$ 1.0 \times 10^{7} \approx \frac{x}{(0.5)(5.0)^4} $$ $$ x \approx 0.5 M - \frac{1.0 \times 10^{-7}}{625} $$ The concentration of the complex \(\mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) is approximately 0.5 M.

Now, let's use the solubility product constant \(K_{sp}\) to see if a precipitate of \(\mathrm{Cd}(\mathrm{OH})_{2}\) will form. We know the reaction \(\mathrm{Cd}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Cd}^{2+}(a q)+2\mathrm{OH}^{-}(a q)\) with \(K_{sp} = 5.9 \times 10^{-15}\). To determine if a precipitate will form, we need to find the concentration of \(\mathrm{OH}^{-}\) at equilibrium. Let y represent the concentration of \(\mathrm{OH}^{-}\). $$ K_{sp} = [\mathrm{Cd}^{2+}][\mathrm{OH}^{-}]^2 = 5.9 \times 10^{-15} $$ Since the concentration of \(\mathrm{Cd}^{2+}\) is very low due to the complex formation, we will use the remaining concentration of \(\mathrm{Cd}^{2+}\) from step 2: $$ 5.9 \times 10^{-15} = \left(\frac{1.0 \times 10^{-7}}{625}\right)(y)^2 $$ Solving for y, the concentration of \(\mathrm{OH}^{-}\) is: $$ y \approx \sqrt{\frac{5.9 \times 10^{-15}\times 625}{1.0 \times 10^{-7}}} \approx 3.0 \times 10^{-4}\,\mathrm{M} $$ Since the concentration of \(\mathrm{OH}^{-}\) is significantly lower than the initial concentration of \(\mathrm{NH}_{3}\) and \(\mathrm{OH}^{-}\) is not consumed in the formation of the complex ion, very little \(\mathrm{Cd}(\mathrm{OH})_{2}\) will precipitate out of the solution. Therefore, a precipitate of \(\mathrm{Cd}(\mathrm{OH})_{2}\) will not form under these conditions.