Question

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 \(5.0\) \(M\) \(\mathrm{NH}_{3} ?\) $$\begin{array}{r} \mathrm{Cd}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) \\ K=1.0 \times 10^{7} \\ \mathrm{Cd}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Cd}^{2+}(a q)+2 \mathrm{OH}^{-}(a q) \\ K_{\mathrm{sp}}=5.9 \times 10^{-15} \end{array}$$

Short answer

A precipitate of \(\mathrm{Cd}(\mathrm{OH})_{2}\) will not form under these conditions, as the concentration of \(\mathrm{OH}^{-}\) at equilibrium is significantly lower than the initial concentration of \(\mathrm{NH}_{3}\) and very little \(\mathrm{Cd}(\mathrm{OH})_{2}\) will precipitate out of the solution due to the formation of the complex ion \(\mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\).

Step-by-step solution

Calculate the concentrations of reactants

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 \(\mathrm{NH}_{3}\): $$ \frac{(1.0\mathrm{L})(5.0 M)}{1.001\mathrm{L}} \approx 5.0 M $$

Calculate the concentration of the complex

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.

Determine if a precipitate will form

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.

Key concepts

Solubility Product Constant (Ksp)

The Solubility Product Constant, or \(K_{sp}\), is a special type of equilibrium constant that applies to the dissolution of sparingly soluble compounds. In a chemical reaction where a solid ionic compound dissolves and dissociates into its ions, \(K_{sp}\) helps determine how much of the solid will dissolve in the solution, forming a saturated solution.
The expression for \(K_{sp}\) is formed by taking the product of the equilibrium concentrations of the ionic species, each raised to the power of their stoichiometric coefficients. For the dissolution of cadmium hydroxide, \(\text{Cd(OH)}_{2(s)} \rightleftharpoons \text{Cd}^{2+}_{(aq)} + 2 \text{OH}^-_{(aq)}\), the \(K_{sp}\) expression is \(K_{sp} = [\text{Cd}^{2+}][\text{OH}^-]^2\).
Understanding \(K_{sp}\) is vital for predicting whether a precipitate will form when different solutions are mixed. If the product of the ion concentrations in the solution exceeds \(K_{sp}\), more ions will recombine to form the precipitate.

Complex Ion Formation

Complex ion formation occurs when a central metal ion, like \(\text{Cd}^{2+}\), binds with surrounding molecules or ions, such as ammonia (\(\text{NH}_3\)), to form a complex ion. This process can significantly affect solubility because it effectively removes free metal ions from the solution.
The formation constant, \(K\), quantifies the stability of the complex ion. It is given by the ratio of the complex ion concentration to the product of the concentrations of the unbound species. For \(\text{Cd(NH}_3)_4^{2+}\), the equation is \[K = \frac{[\text{Cd(NH}_3)_4^{2+}]}{[\text{Cd}^{2+}][\text{NH}_3]^4} = 1.0 \times 10^7\]
Complex ion formation reduces the concentration of free \(\text{Cd}^{2+}\) ions available for forming a precipitate. In the presented solution, the large \(K\) value suggests that \(\text{Cd}^{2+}\) predominantly exists as a part of the complex, minimizing the likelihood of \(\text{Cd(OH)}_2\) precipitation.

Equilibrium Concentrations

When dealing with equilibrium reactions, it's crucial to calculate the equilibrium concentrations of different species involved. This allows us to understand the exact conditions in the system and predict reactions, such as precipitation.
For example, in a reaction involving cadmium nitrate and ammonia, calculating the concentrations of \(\text{Cd}^{2+}\) and \(\text{NH}_3\) after mixing, and subsequently the concentration of the complex \(\text{Cd(NH}_3)_4^{2+}\), helps us assess the system's balance.
Steps for calculation include:

• Using initial volumes and molarity to find initial concentrations.
• Applying formation constants like \(K\) to estimate complex ion concentrations.
• Assessing solubility using \(K_{sp}\) and remaining free ion concentrations, helping determine if further reactions will occur, such as precipitation.

This detailed understanding of equilibrium concentrations assists in predicting and manipulating chemical reactions, tailored to specific environments.