Question

Sulfide \(\left(\mathrm{S}^{2-}\right)\) salts are notoriously insoluble in aqueous solution. a. Calculate the molar solubility of nickel(II) sulfide in water.$$K_{\mathrm{sp}}(\mathrm{NiS})=3 \times 10^{-16}$$ b. Nickel(II) ions form a complex ion in the presence of ammonia with a formation constant \(\left(K_{f}\right)\) of \(2.0 \times 10^{8}\) :\(\mathrm{Ni}^{2+}+6\mathrm{NH}_{3}\rightleftharpoons\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\). Calculatethe molar solu-bility of NiS in \(3.0 \mathrm{M} \mathrm{NH}_{3}\) c. Explain any differences between the answers to parts a and b.

Short answer

a. Molar solubility of NiS in water is \(\frac{\times\sqrt{K_{\mathrm{sp}}}}{2} = \times\sqrt{3 \times 10^{-16}} = 5.48 \times 10^{-8}\text{ M}\). b. Solubility in ammonia is greater due to complex ion formation. c. Complex formation with ammonia increases the solubility compared to pure water.

Step-by-step solution

Write the equilibrium expression for the dissolution of nickel(II) sulfide

Write down the chemical equation for the dissolution of NiS in water and the corresponding equilibrium expression for its solubility product constant, \(K_{\mathrm{sp}}\): \(\mathrm{NiS(s)} \rightleftharpoons \mathrm{Ni}^{2+}(aq) + \mathrm{S}^{2-}(aq)\). The solubility product is \(K_{\mathrm{sp}} = [\mathrm{Ni}^{2+}][\mathrm{S}^{2-}]\).

Calculate the molar solubility of nickel(II) sulfide

Assume that the molar solubility of NiS is \(s\). This means that for every mole of NiS that dissolves, we get \(s\) moles of \(\mathrm{Ni}^{2+}\) and \(s\) moles of \(\mathrm{S}^{2-}\). Therefore, \(K_{\mathrm{sp}} = s^2\). Solve for \(s\) using \(K_{\mathrm{sp}} = 3 \times 10^{-16}\) to find the molar solubility in pure water.

Write down the equation for the nickel-ammonia complex

Write the equation for the formation of the nickel-ammonia complex ion and its corresponding formation constant, \(K_{f}\): \(\mathrm{Ni}^{2+} + 6\mathrm{NH}_3 \rightleftharpoons [\mathrm{Ni}(\mathrm{NH}_3)_6]^{2+}\). The equilibrium expression is \(K_{f} = \frac{[[\mathrm{Ni}(\mathrm{NH}_3)_6]^{2+}]}{[\mathrm{Ni}^{2+}][\mathrm{NH}_3]^6}\).

Calculate the molar solubility of NiS in the presence of ammonia

Since \(\mathrm{NH}_3\) is in large excess (3.0 M), its concentration doesn't change significantly. The solubility equilibrium of NiS is affected by the formation of the nickel-ammonia complex, which decreases the concentration of \(\mathrm{Ni}^{2+}\). Set up the equation \(K_{f} = \frac{s}{(3.0)^6 \times (s - \epsilon)}\), where \(s\) is the molar solubility of \(\mathrm{NiS}\) and \(\epsilon\) is the change in \(\mathrm{Ni}^{2+}\) concentration due to complex formation, which can be considered negligible due to the common ion effect. Solve for solubility \(s\) with \(K_{f} = 2.0 \times 10^{8}\).

Explain differences in solubilities

The difference in the solubilities from parts a and b can be explained by the formation of the complex ion in part b, which reduces the concentration of free \(\mathrm{Ni}^{2+}\) ions in solution and thus shifts the equilibrium to dissolve more NiS to replace the complexed nickel ions.

Key concepts

Solubility Product Constant (Ksp)

The solubility product constant, often abbreviated as Ksp, is a crucial concept in understanding the solubility of sparingly soluble compounds. It represents the maximum product of the molar concentrations of the ions of a salt in a saturated solution, each raised to the power of its stoichiometric coefficient. In simple terms, it's a constant value that helps us to determine the solubility of an ionic compound under a set of specific conditions, typically at a constant temperature.

For example, nickel(II) sulfide (NiS), a compound with very low solubility in water, can be understood through its Ksp. The chemical equation for its dissolution can be written as:NiS(s) ⇌ Ni2+(aq) + S2-(aq)
The Ksp expression for NiS would be:Ksp = [Ni2+][S2-]

Given that the solubility product constant is known for NiS, we can calculate its molar solubility, which indicates how many moles of NiS can dissolve in a liter of water to form a saturated solution.

Nickel(II) Sulfide (NiS)

Nickel(II) sulfide, NiS, is a black solid commonly found in various ores and minerals. Its chemistry is particularly interesting because of its low solubility in water, which can be quantitatively described using the solubility product constant, Ksp. This compound dissociates in water to form nickel(II) ions (Ni2+) and sulfide ions (S2-), and the equilibrium between solid NiS and its ions in solution is exclusively governed by its Ksp value.

It's important to note that NiS doesn't dissociate completely due to its limited solubility; rather, a dynamic equilibrium is established between the undissolved NiS and the ions in solution. The understanding of this sparingly soluble compound extends beyond just its dissolution as it's also part of more complex equilibria, especially when it interacts with other chemicals like ammonia.

Complex Ion Formation

Complex ion formation greatly impacts the solubility of certain ionic compounds. A complex ion is a charged species consisting of a central metal ion surrounded by molecules or ions, collectively known as ligands, which can donate a pair of electrons. These complex ions are part of larger equilibria that can alter the concentration of the free ions in solution, thus impacting overall solubility.

In the case of the NiS solubility in ammonia, ammonia acts as a ligand to form a complex ion with nickel(II) ions. The chemical equation for the formation of the nickel(II) ammonia complex is as follows:
Ni2+ + 6NH3 ⇌ [Ni(NH3)6]2+
The formation constant, Kf, describes the equilibrium between the nickel(II) ions, ammonia, and the complex ion. Since ligands like ammonia have a strong attraction to the metal ion, they effectively 'pull' the metal ions out of the solution. This decreases the metal ion concentration and subsequently influences the solubility of the original compound, NiS in this case.

Ammonia in Solubility Equilibrium

Ammonia (NH₃) plays a pivotal role in the solubility equilibrium of compounds like nickel(II) sulfide because it can profoundly increase the solubility of such compounds through complex ion formation. The addition of ammonia to a solution containing NiS changes the equilibrium condition. Ammonia binds with nickel(II) ions to form a complex ion, which reduces the concentration of free nickel ions in the solution.

This change in ion concentration leads to an increase in the solubility of NiS because the equilibrium of its dissociation is shifted to replace the nickel ions that are used up in complex formation. In systems with high concentrations of ammonia, this effect is particularly pronounced. When describing the solubility of NiS in ammonia, the Ksp of NiS and the formation constant Kf of the nickel-ammonia complex must both be considered to accurately predict the molar solubility in such solutions.