Calculate the molar solubility of Ni(OH)2 in 0.10 M NaOH. The ionic product of Ni(OH)2 is 2.0 × 10–1 ?

To find the molar solubility of nickel(II) hydroxide, Ni(OH)2, in a 0.10 M sodium hydroxide (NaOH) solution, we need to consider the equilibrium established in this scenario. Nickel(II) hydroxide is a sparingly soluble compound, and its solubility can be affected by the presence of hydroxide ions from NaOH.

Understanding the Dissolution of Ni(OH)2

Nickel(II) hydroxide dissociates in water according to the following equation:

Ni(OH)2 (s) ⇌ Ni²⁺ (aq) + 2 OH⁻ (aq)

The solubility product constant (Ksp) for this reaction is given as 2.0 × 10^-1. This constant is defined as:

Ksp = [Ni²⁺][OH⁻]²

Setting Up the Equilibrium Expression

In a saturated solution of Ni(OH)2, let’s denote the molar solubility of Ni(OH)2 as 's'. Therefore, at equilibrium, the concentrations of the ions will be:

• [Ni²⁺] = s
• [OH⁻] = 0.10 M (from NaOH) + 2s (from the dissociation of Ni(OH)2)

Since the concentration of OH⁻ from NaOH is much larger than the contribution from Ni(OH)2, we can approximate [OH⁻] as 0.10 M.

Substituting into the Ksp Expression

Now we can substitute these concentrations into the Ksp expression:

Ksp = [Ni²⁺][OH⁻]² = (s)(0.10)²

Substituting the value of Ksp:

2.0 × 10^-1 = s(0.10)²

This simplifies to:

2.0 × 10^-1 = s(0.01)

Solving for Molar Solubility

Now, we can solve for 's':

s = (2.0 × 10^-1) / (0.01)

s = 20.0

Final Result

The molar solubility of Ni(OH)2 in a 0.10 M NaOH solution is 20.0 M. However, this value indicates that Ni(OH)2 is highly insoluble under these conditions, and practically, the solubility would be limited by the Ksp value. Therefore, in a real-world scenario, the actual molar solubility would be much lower than this theoretical calculation due to the saturation limits of the solution.