Question

A saturated solution of nickel(II) sulfide contains approximately \(3.6 \times 10^{-4} \mathrm{g}\) of dissolved NiS per liter at \(20^{\circ} \mathrm{C}\). Calculate the solubility product \(K_{\mathrm{sp}}\) for NiS at \(20^{\circ} \mathrm{C}\).

Short answer

The solubility product, \(K_{\mathrm{sp}}\), for NiS at \(20^{\circ}\mathrm{C}\) is approximately \(1.58 \times 10^{-11}\).

Step-by-step solution

Calculate the number of moles of dissolved NiS

First, we need to convert the given solubility from grams to moles. To do this, we will use the molar mass of NiS which is (58.69 + 32.07) g/mol. Number of moles = \(\cfrac{3.6 \times 10^{-4}\mathrm{g}}{(58.69+32.07) \mathrm{g/mol}}= \cfrac{3.6 \times 10^{-4}\mathrm{g}}{90.76 \mathrm{g/mol}}\) Number of moles = 3.97 x 10^{-6} moles/L

Write the dissociation reaction and the solubility product expression

The dissociation reaction for nickel(II) sulfide (NiS) in water can be represented as: NiS (s) ⇌ Ni²⁺ (aq) + S²⁻ (aq) The solubility product expression, Ksp, is the product of the concentrations of the dissolved ions: Ksp = [Ni²⁺][S²⁻] Since the number of moles of Ni²⁺ and S²⁻ produced is equal, their concentrations will also be equal.

Plug in the concentrations and solve for Ksp

Now, we will plug in the concentrations of Ni²⁺ and S²⁻ from the dissociation of NiS, both of which are equal to 3.97 x 10^{-6} moles/L. Ksp = (3.97 x 10^{-6})(3.97 x 10^{-6}) Ksp = \(1.58 \times 10^{-11}\) The solubility product, Ksp, for NiS at 20 degrees Celsius is approximately \(1.58 \times 10^{-11}\).

Key concepts

Nickel(II) Sulfide

Nickel(II) sulfide, abbreviated as NiS, is an important compound in the study of solubility and chemical equilibria. It is an insoluble salt that plays a crucial role in many industrial processes, especially in metallurgy. NiS has a unique position because it is often used in electroplating processes to apply a nickel finish to surfaces. Despite its low solubility, it is essential to understand how NiS behaves in aqueous environments.

Understanding the solubility of nickel(II) sulfide is vital for predicting whether it will precipitate or stay dissolved under certain conditions. This knowledge is particularly useful in chemical engineering and environmental science where nickel compounds impact both product formation and pollution control strategies.

Dissolution Process

The dissolution process of nickel(II) sulfide involves a solid substance (NiS) interacting with a solvent, typically water, leading to the formation of ions. When NiS is mixed with water, some of it will dissolve into free ions, specifically nickel ions (\(\text{Ni}^{2+}\)) and sulfide ions (\(\text{S}^{2-}\)).

The equation for this process is:- NiS (s) ⇌ Ni²⁺ (aq) + S²⁻ (aq)This reaction illustrates that for each mole of NiS that dissolves, one mole of Ni²⁺ and one mole of S²⁻ are produced. The dissolution of NiS is a reversible process and will reach equilibrium, where the rate of dissolution equals the rate at which the ions recombine to form the solid NiS.

Ions Concentration

The concentration of ions in a solution tells us how much of the compound has dissolved. In the case of nickel(II) sulfide, when it dissolves, it produces equal concentrations of nickel ions and sulfide ions in the solution.

Since the dissolution reaction results in one mole of Ni²⁺ and one mole of S²⁻ per mole of NiS dissolved, their concentrations at equilibrium remain the same. For example, if the solubility of NiS is calculated as approximately \(3.97 \times 10^{-6}\) moles per liter, then the concentrations of both Ni²⁺ and S²⁻ will be the same value.

Knowing the concentration of these ions helps us calculate the solubility product (Ksp), which reflects how much of a compound can dissolve in water. It is a crucial measure in chemistry, influencing everything from laboratory experiments to the formation of mineral deposits.

Chemical Equilibrium

Chemical equilibrium occurs when the rate of the forward reaction (dissolution) equals the rate of the reverse reaction (precipitation) in a closed system. In the case of nickel(II) sulfide dissolving in water, equilibrium is established between the solid NiS and the dissolved ions, Ni²⁺ and S²⁻.

At this state, the concentration of dissolved ions remains stable over time, even though the ions are continuously dissolving and recombining. The chemical equilibrium for the dissolution of NiS can be expressed in terms of the solubility product constant, Ksp, which is given as: - Ksp = [Ni²⁺][S²⁻] This expression indicates that the product of the concentrations of the ions is constant at a given temperature when the system is at equilibrium. Ksp is a pivotal concept that helps predict whether a precipitate will form when solutions containing these ions are mixed. A higher Ksp value means greater solubility of the compound in water.