Question

Calculate the molar solubility of fFe(OH)₂ in pure water. A) 1.5 x 10⁻⁵ M

B) 1.0 x 10⁻⁹ M
C) 2.0 x 10⁻³ M
D) 5.0 x 10⁻⁷ M

Answer 1

The molar solubility of Fe(OH)₂ in pure water is calculated using the Ksp expression and is found to be approximately 1.06 × 10⁻¶ M, which does not exactly match the provided options.

Step-by-step explanation:

To calculate the molar solubility of Fe(OH)₂ in pure water, we use the solubility product constant (Ksp) for Fe(OH)₂, which can be expressed in the equation for the dissociation of the solid into its ions in solution:

Fe(OH)₂ (s) → Fe²⁺ (aq) + 2OH⁻ (aq)

The Ksp expression for this equilibrium is:

Ksp = [Fe²⁺][OH⁻]²

From the provided information, the Ksp for Fe(OH)₂ is 4.87 × 10⁻¹⁷. Let 's' represent the solubility of Fe(OH)₂ in moles per liter. At equilibrium, we would have [Fe²⁺] = 's' and [OH⁻] = 2s because for every mole of Fe(OH)₂ that dissolves, 1 mole of Fe²⁺ and 2 moles of OH⁻ are produced.

Substituting into the Ksp expression yields:

Ksp = s(2s)² = 4s³

Solving for 's' gives us the molar solubility:

s = (Ksp/4)¹/³ = (4.87 × 10⁻¹⁷ / 4)¹/³

s = (1.2175 × 10⁻¹⁷)¹/³

s = 1.06 × 10⁻¶ M

This value is not an exact match for one of the options provided in the question, indicating there might be a mistake in the information given or in the question options themselves. However, the closest answer to our calculated value is B) 1.0 × 10⁻¹ M, assuming a typo might have been made in the exponent of the options.