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What is the pH of a saturated solution of Zn(OH)₂? (Given Ksp = 1.8 × 10⁻¹⁴)

User Visual
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Final answer:

The pH of a saturated solution of Zn(OH)₂, given the Ksp value of 1.8 × 10⁻¹⁴, is calculated as 9.52 after establishing the relationship between solubility, ion concentration, and pH.

Step-by-step explanation:

To determine the pH of a saturated solution of Zn(OH)₂ given the solubility product constant, Ksp, is 1.8 × 10⁻¹⁴, we need to establish the relationship between solubility, ion concentration, and pH.

First, we write the dissociation of Zn(OH)₂ in water:
Zn(OH)₂(s) ⇌ Zn²⁺(aq) + 2OH⁻(aq)

Let 's' be the molar solubility of Zn(OH)₂, then the concentrations of Zn²⁺ and OH⁻ at equilibrium will be 's' and '2s', respectively. The Ksp expression for Zn(OH)₂ is:
Ksp = [Zn²⁺][OH⁻]² = s(2s)² = 4s³

Substituting the given Ksp value, we get:
1.8 × 10⁻ⁱ⁴ = 4s³ ⇒ s = √[1.8 × 10⁻ⁱ⁴ / 4] = 1.65 × 10⁻⁵ M

Now that we have 's', we can find the concentration of OH⁻:
[OH⁻] = 2s = 3.3 × 10⁻⁵ M
Using the formula pOH = -log[OH⁻], we find the pOH of the solution. The pOH value is then used to find the pH, as pH + pOH = 14 at 25°C. Solving:

pOH = -log(3.3 × 10⁻⁵) ≈ 4.48
pH = 14 - pOH ≈ 14 - 4.48 = 9.52

Therefore, the pH of the saturated solution of Zn(OH)₂ is approximately 9.52.

User Blazs
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