Problem 27 - Chemistry

Question

Approximately 0.14 g nickel(II) hydroxide, Ni(OH) \(_{2}(s),\) dissolves per liter of water at \(20^{\circ} \mathrm{C}\) . Calculate \(K_{\text { sp }}\) for \(\mathrm{Ni}(\mathrm{OH})_{2}(s)\) at this temperature.

Step-by-Step Solution

Answer

The solubility product constant, \(K_{sp}\), for Ni(OH)₂(s) at 20°C is approximately \(4.37 * 10^{-8}\).

Step 1: Write the balanced chemical equation for the dissolution of Ni(OH)₂(s) in water

The balanced chemical equation for the dissolution of nickel(II) hydroxide in water is: Ni(OH)₂(s) ⇌ Ni²⁺(aq) + 2OH⁻(aq)

Step 2: Write the equilibrium expression for this reaction

The equilibrium expression, Ksp, for the dissolution of Ni(OH)₂(s) in water is given by: Ksp = [Ni²⁺] * [OH⁻]² where [Ni²⁺] and [OH⁻] represent the equilibrium concentrations of the Ni²⁺ and OH⁻ ions, respectively.

Step 3: Calculate molar solubility, s, of Ni(OH)₂ in water

We are given that approximately 0.14 g of Ni(OH)₂ dissolves per liter of water at 20°C. To convert this mass to moles, we need the molar mass of Ni(OH)₂: Molar mass of Ni(OH)₂ = 58.69 g/mol (Ni) + 2 * (15.999 g/mol + 1.0079 g/mol) = 92.708 g/mol for Ni(OH)₂ Now, we can find the molar solubility, s: s = (0.14 g Ni(OH)₂) / (92.708 g/mol Ni(OH)₂) = 1.51 * 10⁻³ mol/L

Step 4: Determine the equilibrium concentrations of Ni²⁺ and OH⁻ ions in the saturated solution

From the balanced chemical equation, we can see that for every mole of Ni(OH)₂ that dissolves, one mole of Ni²⁺ and two moles of OH⁻ ions are produced. Thus, we can use the molar solubility, s, to determine the equilibrium concentrations of Ni²⁺ and OH⁻ ions in the saturated solution: [Ni²⁺] = 1.51 * 10⁻³ mol/L [OH⁻] = 2 * 1.51 * 10⁻³ mol/L = 3.02 * 10⁻³ mol/L

Step 5: Calculate Ksp for Ni(OH)₂(s)

Finally, we can use the equilibrium concentrations of Ni²⁺ and OH⁻ ions in the saturated solution to calculate Ksp for Ni(OH)₂(s): Ksp = [Ni²⁺] * [OH⁻]² Ksp = (1.51 * 10⁻³)(3.02 * 10⁻³)² Ksp = 4.37 * 10⁻⁸ So, the solubility product constant, Ksp, for Ni(OH)₂(s) at 20°C is approximately 4.37 * 10⁻⁸.

Key Concepts

Dissolution

Dissolution is the process where a solid substance dissolves in a solvent, forming a solution. For nickel(II) hydroxide, Ni(OH)₂, this involves mixing with water. The solid compound splits into ions. In the classroom or lab, you'd often see it written as:

• Ni(OH)₂(s) ⇌ Ni²⁺(aq) + 2OH⁻(aq)

During dissolution, the solid's particles disperse and interact with the solvent, in this case, water. The compounds break into smaller particles or ions and evenly mix into the liquid. Having a balanced chemical equation helps understand the composition and the number of ions that result from the process.

Chemical Equilibrium

In dissolution, chemical equilibrium occurs when the rate of the solid dissolving equals the rate of ions recombining into the solid. At this point, the concentrations of ions in solution remain constant. The equilibrium is depicted by consistent forward and reverse reactions, represented by the double arrow (⇌) in chemical equations.

• This means the dissolution of Ni(OH)₂ into Ni²⁺ and OH⁻ ions and their precipitation back to solid form occur at the same rate.

This dynamic balance is crucial for understanding how much of the solid can dissolve and what concentrations ions achieve in a solution at equilibrium.

Molar Solubility

Molar solubility represents the number of moles of a substance that can dissolve per liter of solution at a specific temperature. For Ni(OH)₂, it's calculated from the mass that dissolves in water. Knowing that 0.14 g dissolves in a liter, you'd convert this mass to moles using its molar mass.

• Molar solubility of Ni(OH)₂: 1.51 x 10⁻³ mol/L

Molar solubility helps predict how much of a substance can dissolve in a given volume of solvent before reaching saturation. Above this point, the substance may not dissolve further without changing conditions.

Equilibrium Concentrations

Equilibrium concentrations indicate the concentrations of ions in a saturated solution at equilibrium. To find them, you relate molar solubility to ion production.

• From Ni(OH)₂: [Ni²⁺] equals the molar solubility, which is 1.51 x 10⁻³ mol/L.
• Since each mole of Ni(OH)₂ produces two moles of OH⁻, [OH⁻] is 2 * 1.51 x 10⁻³ mol/L = 3.02 x 10⁻³ mol/L.

These equilibrium concentrations are vital for calculations involving the solubility product constant, Ksp, which quantifies the solubility of a compound.