Question

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

Short answer

The initial concentration of \(\mathrm{Cd^{2+}}\) is \(C_{\mathrm{Cd^{2+}}} = 0.00100 \mathrm{M}\). Using the equilibrium constant for the formation of \(\mathrm{Cd(NH_3)_4^{2+}}\), we find that \([\mathrm{Cd(NH_3)_4^{2+}}]=4.95 \times 10^{-5} \mathrm{M}\). Subtracting this from the initial concentration of \(\mathrm{Cd^{2+}}\), we get \([Cd^{2+}]_{eq} = 9.51 \times 10^{-4} \mathrm{M}\). The concentration of \(\mathrm{OH^-}\) ions, considering the basic property of \(\mathrm{NH_3}\), is calculated to be \([\mathrm{OH^-}] = 5.16 \times 10^{-4} \mathrm{M}\). Now, calculating the ion product \(Q = [\mathrm{Cd^{2+}}]_{eq} \times [\mathrm{OH^-}]^2 = 2.54 \times 10^{-15}\). Comparing this value to \(K_{sp}\) of \(\mathrm{Cd(OH)_2}\), we find that \(Q < K_{sp}\). Therefore, no precipitate of \(\mathrm{Cd(OH)_2}\) will form in this solution.

Step-by-step solution

Determine the initial concentration of species

Initially, the reaction between \(\mathrm{Cd^{2+}}\) and \(\mathrm{NH_3}\) will tend to form the \(\mathrm{Cd(NH_3)_4^{2+}}\) complex. So first, we need to determine the initial concentration of \(\mathrm{Cd^{2+}}\) in the solution. The volume of the solution is \(1.0 \mathrm{~mL} + 1.0 \mathrm{~L} = 1001 \mathrm{~mL}\). So, the initial concentration of \(\mathrm{Cd^{2+}}\) can be calculated as follows: \[C_{\mathrm{Cd^{2+}}} = \frac{1.0 \mathrm{~mL} \times 1.0 \mathrm{M}}{1001 \mathrm{~mL}}\]

Calculate the concentration of \(\mathrm{Cd(NH_3)_4^{2+}}\)

We know that the equilibrium constant for the formation of \(\mathrm{Cd(NH_3)_4^{2+}}\) complex is \(K=1.0 \times 10^7\). We can use this value to calculate the concentration of \(\mathrm{Cd(NH_3)_4^{2+}}\) at equilibrium: \[[\mathrm{Cd(NH_3)_4^{2+}}] = \frac{K \times [Cd^{2+}] \times [NH_3]^4}{1}\] Plug in the values and evaluate the expression to find the equilibrium concentration of \(\mathrm{Cd(NH_3)_4^{2+}}\).

Calculate the equilibrium concentration of \(\mathrm{Cd^{2+}}\) and \(\mathrm{OH^-}\)

Now, we need to calculate the concentration of \(\mathrm{Cd^{2+}}\) ions at equilibrium. As \(\mathrm{Cd^{2+}}\) ions are consumed in the formation of the complex, their concentration at equilibrium will decrease: \[[Cd^{2+}]_{eq} = [Cd^{2+}]_{initial} - [\mathrm{Cd(NH_3)_4^{2+}}]\] Next, let's find the concentration of \(\mathrm{OH^-}\). Since \(\mathrm{NH_3}\) acts as a base, it can react with water to produce \(\mathrm{OH^-}\): \[\mathrm{NH_3}(aq) + H_2O(l) \rightleftharpoons \mathrm{NH_4^+}(aq) + \mathrm{OH^-}(aq)\] Using the concentration of \(\mathrm{NH_3}\), we can determine the equilibrium concentration of \(\mathrm{OH^-}\).

Calculate the ion product Q

Now that we have the equilibrium concentrations of \(\mathrm{Cd^{2+}}\) and \(\mathrm{OH^-}\), we can calculate the ion product Q: \[Q = [\mathrm{Cd^{2+}}]_{eq} \times [\mathrm{OH^-}]^2\]

Compare Q with \(K_{sp}\)

Finally, we will compare the value of Q to the solubility product constant \(K_{sp}\) of \(\mathrm{Cd(OH)_2}\). If \(Q > K_{sp}\), a precipitate of \(\mathrm{Cd(OH)_2}\) will form. If \(Q \le K_{sp}\), there will be no precipitate. Given the equilibrium concentrations from the previous steps, compare the value of \(Q\) to the \(K_{sp}=5.9 \times 10^{-15}\) and determine if a precipitate will form in the solution.

Key concepts

Precipitation Reaction

In chemistry, precipitation reactions occur when two solutions containing soluble compounds are mixed, leading to the formation of an insoluble product, or precipitate. This occurs because the product formed has such low solubility in water that it separates out of solution as a solid. In the given exercise, the reaction we are investigating is whether the compound \(\mathrm{Cd(OH)_2}\) will precipitate out of solution.

• The reactants in the solution are \(\mathrm{Cd^{2+}}\) and \(\mathrm{OH^-}\) ions.
• When these ions combine, they can form \(\mathrm{Cd(OH)_2(s)}\), a solid that indicates precipitation.

To predict precipitation, we compare the calculated ion product \(Q\) against the known solubility product constant \(K_{sp}\). The solubility product constant represents the product of ionic concentrations in a saturated solution at equilibrium. If the ion product \(Q\), calculated from the concentrations of ions in the solution, is greater than \(K_{sp}\), a precipitate will form. Although a precipitate has not yet formed, the potential for precipitation exists, and the solution is said to be supersaturated under these conditions.

Complex Ion Formation

Complex ion formation involves the creation of a charged species consisting of a central metal ion bonded to one or more molecules or ions. In our exercise, we examine the formation of the complex ion \(\mathrm{Cd(NH_3)_4^{2+}}\), where \(\mathrm{Cd^{2+}}\) ions in solution bind with ammonia molecules \((\mathrm{NH_3})\).

• The stability of this complex ion is quantified by its formation constant, \(K=1.0 \times 10^7\).
• This large \(K\) value suggests a strong tendency for \(\mathrm{Cd^{2+}}\) ions to form complexes with ammonia.

The complex ion formation reduces the concentration of free \(\mathrm{Cd^{2+}}\) ions, impacting its interaction with hydroxide ions \((\mathrm{OH^-})\) to potentially form a precipitate. Understanding these dynamics is crucial in predicting both the occurrence and prevention of precipitation, as the bound \(\mathrm{Cd^{2+}}\) ions are less available to form \(\mathrm{Cd(OH)_2}\).

Ion Product (Q)

The ion product \(Q\) in a solution is a calculated value that helps predict whether a precipitate will form. In our scenario, it is determined by the concentrations of the ions that would form the solid, specifically \(\mathrm{Cd^{2+}}\) and \(\mathrm{OH^-}\) ions:\[ Q = [\mathrm{Cd^{2+}}] \times [\mathrm{OH^-}]^2 \]

• \(Q\) varies with the conditions of the solution, unlike \(K_{sp}\) which is a constant value at a given temperature.
• \(Q\) is useful for comparing against \(K_{sp}\) to determine state of the system; specifically, whether the system is in equilibrium, saturated, or supersaturated.

When \(Q\) is greater than \(K_{sp}\), the solution is said to be supersaturated, and a precipitate will likely form. If \(Q\) is less than \(K_{sp}\), the solution is unsaturated and no precipitate forms. When \(Q\) equals \(K_{sp}\), the system is in equilibrium, meaning the solution is saturated and any further addition of ions would lead to precipitation. Understanding \(Q\) is key to predicting and controlling precipitation reactions.

Solubility Product Constant (Ksp)

The solubility product constant \(K_{sp}\) is a crucial equilibrium constant that quantifies the solubility of an ionic compound. Specifically, it defines the maximum product of the ion concentrations in a solution at equilibrium for a slightly soluble salt.

• For \(\mathrm{Cd(OH)_2}\), \(K_{sp}=5.9 \times 10^{-15}\).
• This low \(K_{sp}\) value indicates that \(\mathrm{Cd(OH)_2}\) is only slightly soluble under normal conditions.

When solving precipitation problems, comparing the ion product \(Q\) to \(K_{sp}\) helps decide the state of the solution. If \(Q\) exceeds \(K_{sp}\), it suggests that the concentrations of ions are higher than what can remain dissolved, leading to precipitation. \(K_{sp}\) is vital to understanding solubility and plays a key role in reactions where the solubility is impacted by changes in conditions, such as the pH of the solution or the presence of complexing agents like \(\mathrm{NH_3}\). Keeping this in mind allows chemists to manipulate conditions to favor or prevent precipitate formation.