Question

Sulfide \(\left(\mathrm{S}^{2-}\right)\) salts are notoriously insoluble in aqueous solution. \begin{equation} \begin{array}{l}{\text { a. Calculate the molar solubility of nickel(II) sulfide in water. }} \\ {K_{\text { sp }}(\mathrm{NiS})=3 \times 10^{-16} \\\ \text { b. Nickel(II) ions form a complex ion in the presence of ammonia with a }} \\ {\text { formation constant }\left(K_{f}\right) \text { of } 2.0 \times 10^{2} : \quad \mathrm{Ni}^{2+}+6 \mathrm{NH}_{3} \rightleftharpoons} \\\ {\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+} . \text { Calculate the molar solubility of } \mathrm{N} \text { is in } 3.0 \mathrm{M} \mathrm{M} \mathrm{NH}_{3} \text { . }} \\ {\text { c. Explain any differences between the answers to parts a and b. }} \end{array} \end{equation}

Short answer

The molar solubility of \text{NiS}\ in water is \(5.48 \times 10^{-8} \text{ mol/L}\), and in a 3.0 M \text{NH}_{3}\ solution, it is significantly reduced to \(2.06 \times 10^{-21} \text{ mol/L}\). The difference is due to the formation of a complex ion with ammonia, which reduces the solubility of \text{NiS}\.

Step-by-step solution

Write the Dissociation Reaction for Nickel(II) Sulfide

Nickel(II) sulfide dissociates in water according to the equation: \(\text{NiS} (s) \rightleftharpoons \text{Ni}^{2+} (aq) + \text{S}^{2-} (aq)\). For each mole of \text{NiS}\ that dissolves, one mole of \text{Ni}^{2+}\ and one mole of \text{S}^{2-}\ ions are formed.

Set Up the Expression for the Solubility Product

The solubility product constant, \(K_{\text{sp}}\), can be written as: \[K_{\text{sp}} = [\text{Ni}^{2+}][\text{S}^{2-}]\]. Since the molar solubility of \text{NiS}\ is equal to the concentration of \text{Ni}^{2+}\ or \text{S}^{2-}\, let the molar solubility be represented by \(s\). Then, \[K_{\text{sp}} = s \times s = s^2\].

Solve for the Molar Solubility of Nickel(II) Sulfide

Using the given \(K_{\text{sp}}\) value for \text{NiS}\ of \(3 \times 10^{-16}\), set up the equation: \[3 \times 10^{-16} = s^2\]. Solve for \(s\) to find the molar solubility: \[s = \sqrt{3 \times 10^{-16}} = 5.48 \times 10^{-8} \text{ mol/L}\].

Write the Complex Ion Formation Reaction

When \text{Ni}^{2+}\ is in the presence of ammonia, it forms a complex ion: \[\text{Ni}^{2+} + 6 \text{NH}_{3} \rightleftharpoons [\text{Ni}(\text{NH}_{3})_{6}]^{2+}\]. The formation constant \(K_{f}\) for this reaction is \(2.0 \times 10^{2}\).

Set Up the Equilibrium Expression for the Complex Ion Formation

The equilibrium expression for the formation of the complex ion is: \[K_{f} = \frac{[[\text{Ni}(\text{NH}_{3})_{6}]^{2+}]}{[\text{Ni}^{2+}][\text{NH}_{3}]^6}\]. Since the \(\text{NH}_{3}\) concentration is large and can be considered a constant, it simplifies the expression to: \[K_{f}^{\prime} = \frac{[[\text{Ni}(\text{NH}_{3})_{6}]^{2+}]}{[\text{Ni}^{2+}]} = K_{f} [\text{NH}_{3}]^6\].

Calculate the Effective Formation Constant

Substitute the concentration of ammonia into the modified equilibrium expression: \[K_{f}^{\prime} = (2.0 \times 10^{2})(3.0)^6 = (2.0 \times 10^{2})(729) = 1.458 \times 10^{5}\].

Solve for the Molar Solubility of Nickel(II) Sulfide in Ammonia Solution

The presence of ammonia increases the solubility of \text{NiS}\ by shifting the equilibrium to the right due to the formation of the complex ion. Use the new \(K_{f}^{\prime}\) value to solve for solubility \(s\), considering the \(K_{\text{sp}}\) of \text{NiS}\: \[K_{\text{sp}} = s \times K_{f}^{\prime} = s \times 1.458 \times 10^{5}\]. Solve for \(s\): \[s = \frac{K_{\text{sp}}}{K_{f}^{\prime}} = \frac{3 \times 10^{-16}}{1.458 \times 10^{5}} = 2.06 \times 10^{-21} \text{ mol/L}\].

Explain the Difference in Solubility Between Parts a and b

The presence of ammonia significantly decreases the molar solubility of \text{NiS}\ because the formation of the complex ion \([\text{Ni}(\text{NH}_{3})_{6}]^{2+}\) reduces the concentration of \(\text{Ni}^{2+}\) ions in solution, thereby shifting the dissolution equilibrium of \text{NiS}\ towards the left and decreasing its solubility.

Key concepts

Solubility Product Constant

Understanding the solubility product constant, typically denoted as \( K_{\text{sp}} \), is essential when studying the solubility of ionic compounds in water. This constant is an equilibrium constant which refers to the maximal concentration of an ionic compound that can dissolve in water to form a saturated solution. It is represented by the product of the concentrations of the ions, each raised to the power of their stoichiometric coefficients.

The formula for the solubility product constant is \( K_{\text{sp}} = [A^{n+}]^m[B^{m-}]^n \), where A and B are the ions of the salt, with their respective charges n+ and m- and their corresponding stoichiometric coefficients m and n. The square brackets indicate the concentration of ions in a saturated solution.

For Nickel(II) sulfide, the dissociation in water can be represented as \( \text{NiS} (s) \rightleftharpoons \text{Ni}^{2+} (aq) + \text{S}^{2-} (aq) \). Assuming a 1:1 stoichiometry, the expression simplifies to \( K_{\text{sp}} = [\text{Ni}^{2+}][\text{S}^{2-}] = s^2 \) where s is the molar solubility of Nickel(II) sulfide.

Complex Ion Formation

Complex ion formation occurs when a metal ion bonds to one or more ligands to form a coordination compound. These ligands, which are generally molecules like water or ammonia (\( \text{NH}_{3} \)), donate electron pairs to the metal ion, resulting in a more stable, typically soluble, complex ion.

The reaction to form a complex ion follows the pattern: \( \text{M}^{n+} + x\text{L} \rightleftharpoons [\text{M}(\text{L})_x]^{n+} \), where M is the metal ion, L is the ligand, and x is the number of ligands attached.

In the context of the Nickel(II) sulfide problem, the formation of the complex ion with ammonia can significantly affect the solubility of the salt, as shown in the equation: \( \text{Ni}^{2+} + 6 \text{NH}_{3} \rightleftharpoons [\text{Ni}(\text{NH}_{3})_{6}]^{2+} \). Here, ammonia acts as a ligand, increasing the solubility of Nickel(II) sulfide by stabilizing the Nickel(II) ion in solution as a complex ion.

Equilibrium Constant

The equilibrium constant, symbolized as \( K \), quantifies the ratio of the concentration of products to reactants at equilibrium for a reversible reaction. It is important to note that pure solids and liquids do not appear in the equilibrium constant expression, whereas the concentrations of gases and dissolved species are included.

For a general reversible reaction, \( aA + bB \rightleftharpoons cC + dD \), the equilibrium constant expression is given as: \( K = \frac{[C]^c [D]^d}{[A]^a [B]^b} \).

When dealing with complex ion formation, like in the case of the Nickel(II) sulfide problem, the equilibrium constant is referred to specifically as the formation constant, denoted as \( K_f \). It describes the formation of the complex ion from the metal ion and the ligands. In environments with high ligand concentrations such as 3.0M \text{NH}_{3}, the equilibrium shifts, demonstrating Le Châtelier's principle, which states that the position of equilibrium will adjust to counteract a change in conditions.

Sulfide Salts

Sulfide salts are ionic compounds that contain the sulfide ion \( \text{S}^{2-} \). They are known for their low solubility in water, with few exceptions. The solubility of sulfide salts can be described using the solubility product constant, \( K_{\text{sp}} \). Sulfide salts often require specific conditions, like low pH or presence of complexing agents, to increase their solubility.

Many sulfide salts produce saturated solutions rapidly, leading to the precipitation of solid sulfide which makes them important in various industries such as metal recovery and ore processing. The solubility of sulfide salts can also be an environmental concern as they can contribute to the pollution of water bodies when they undergo oxidation to form acid sulfate soils or acid mine drainage.

Nickel(II) Sulfide

Nickel(II) sulfide, represented by the chemical formula \( \text{NiS} \), is a sulfide salt of nickel. It is known for being highly insoluble in water, as indicated by its low \( K_{\text{sp}} \). This property means that Nickel(II) sulfide generally remains as a solid under normal aqueous conditions. Despite its low solubility, Nickel(II) sulfide's molar solubility can be determined through careful calculations, as shown in the problem above where the dissociation of Nickel(II) sulfide in water and the effect of complex ion formation with ammonia on its solubility are examined.

The compound is also of interest due to the ability of Nickel(II) ions to form complex ions, significantly altering its solubility characteristics. These properties make Nickel(II) sulfide a relevant substance in various chemical procedures and treatments, including those that remove heavy metals from wastewater by precipitation.