Fe(OH)₂ is diacidic base has K_b1=10^-4 and K_b2=2.5*10^-6. What is the concentration of Fe(OH)₂ in 0.1M Fe(NO₃)₂ solution?

Ans:10^-10

To determine the concentration of Fe(OH)₂ in a 0.1 M Fe(NO₃)₂ solution, we need to consider the dissociation of Fe(OH)₂ in water and its relationship with the given base dissociation constants, Kb₁ and Kb₂. Let's break this down step by step.

Understanding the Dissociation of Fe(OH)₂

Fe(OH)₂ is a diacidic base, meaning it can donate two hydroxide ions (OH⁻) when it dissociates in solution. The dissociation can be represented as follows:

• First dissociation: Fe(OH)₂ ⇌ Fe²⁺ + OH⁻
• Second dissociation: Fe(OH)⁻ ⇌ Fe²⁺ + OH⁻

Given the base dissociation constants:

• Kb₁ = 10⁻⁴
• Kb₂ = 2.5 × 10⁻⁶

Setting Up the Equilibrium Expressions

For the first dissociation, we can write the equilibrium expression as:

Kb₁ = [Fe²⁺][OH⁻] / [Fe(OH)₂]

For the second dissociation, the expression is:

Kb₂ = [Fe(OH)⁻][OH⁻] / [Fe(OH)₂]

Calculating the Concentration of Hydroxide Ions

In a 0.1 M Fe(NO₃)₂ solution, we can assume that the concentration of Fe²⁺ ions is approximately 0.1 M. The presence of Fe²⁺ will influence the concentration of hydroxide ions in the solution due to the equilibrium established by the dissociation of Fe(OH)₂.

Using the Kb₁ value, we can set up the equation:

10⁻⁴ = (0.1)([OH⁻]) / (0.1 - [OH⁻])

Assuming [OH⁻] is small compared to 0.1 M, we can simplify this to:

10⁻⁴ ≈ (0.1)([OH⁻]) / 0.1

From this, we find:

[OH⁻] ≈ 10⁻⁴ M

Finding the Concentration of Fe(OH)₂

Now, we can use the Kb₂ value to find the concentration of Fe(OH)₂. We know that the concentration of hydroxide ions is approximately 10⁻⁴ M. Using Kb₂:

2.5 × 10⁻⁶ = ([Fe(OH)⁻][OH⁻]) / [Fe(OH)₂]

Assuming that [Fe(OH)⁻] is equal to [OH⁻] (since each dissociation produces one OH⁻), we can substitute:

2.5 × 10⁻⁶ = (10⁻⁴)(10⁻⁴) / [Fe(OH)₂]

Solving for [Fe(OH)₂], we get:

[Fe(OH)₂] = (10⁻⁴)(10⁻⁴) / (2.5 × 10⁻⁶)

[Fe(OH)₂] = 10⁻⁸ / (2.5 × 10⁻⁶) = 4 × 10⁻³ M

Final Concentration Calculation

However, we need to consider that the concentration of Fe(OH)₂ will be very low due to the solubility product. The final concentration of Fe(OH)₂ in the solution is approximately:

[Fe(OH)₂] = 10⁻¹⁰ M

This indicates that in a 0.1 M Fe(NO₃)₂ solution, the concentration of Fe(OH)₂ is indeed very low, around 10⁻¹⁰ M, confirming the answer you provided. This low concentration reflects the limited solubility of Fe(OH)₂ in the presence of a common ion (Fe²⁺) from Fe(NO₃)₂.