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Find the solubility of Cr(oh)3 when it suspended in 0.01M of Mg(oh)2

To determine the solubility of chromium(III) hydroxide, Cr(OH)3, when it is suspended in a 0.01 M solution of magnesium hydroxide, Mg(OH)2, we need to consider the common ion effect and the solubility product constant (Ksp) of Cr(OH)3. Let’s break this down step by step.

Understanding the Basics

Chromium(III) hydroxide is a sparingly soluble compound, and its solubility can be represented by the following equilibrium reaction:

Cr(OH)3 (s) ⇌ Cr³⁺ (aq) + 3 OH⁻ (aq)

The solubility product constant (Ksp) for Cr(OH)3 is given by:

Ksp = [Cr³⁺][OH⁻]³

Common Ion Effect

When Cr(OH)3 is added to a solution containing Mg(OH)2, the presence of hydroxide ions (OH⁻) from Mg(OH)2 affects the solubility of Cr(OH)3. The common ion effect states that the solubility of a salt decreases in a solution that already contains one of its constituent ions. In this case, the hydroxide ions from Mg(OH)2 will shift the equilibrium to the left, reducing the solubility of Cr(OH)3.

Calculating Hydroxide Ion Concentration

First, we need to calculate the concentration of hydroxide ions in the 0.01 M Mg(OH)2 solution. Since Mg(OH)2 dissociates into Mg²⁺ and 2 OH⁻ ions, the concentration of OH⁻ ions will be:

[OH⁻] = 2 × [Mg(OH)2] = 2 × 0.01 M = 0.02 M

Setting Up the Ksp Expression

Now, we can substitute this concentration into the Ksp expression for Cr(OH)3. Let 's' be the solubility of Cr(OH)3 in this solution:

Ksp = [Cr³⁺][OH⁻]³ = s × (0.02)³

Assuming the concentration of Cr³⁺ ions is equal to 's' (the solubility of Cr(OH)3), we can rewrite the Ksp expression:

Ksp = s × (0.02)³

Finding Ksp for Cr(OH)3

The Ksp value for Cr(OH)3 is approximately 6.7 × 10⁻³³ at 25°C. Now we can set up the equation:

6.7 × 10⁻³³ = s × (0.02)³

Calculating (0.02)³ gives us:

(0.02)³ = 0.000008 = 8 × 10⁻⁶

Solving for Solubility

Now we can substitute this value back into the equation:

6.7 × 10⁻³³ = s × 8 × 10⁻⁶

To find 's', we rearrange the equation:

s = (6.7 × 10⁻³³) / (8 × 10⁻⁶)

Calculating this gives:

s ≈ 8.375 × 10⁻²⁸ M

Final Thoughts

The solubility of Cr(OH)3 in a 0.01 M Mg(OH)2 solution is approximately 8.375 × 10⁻²⁸ M. This illustrates how the presence of a common ion, in this case, hydroxide ions, significantly reduces the solubility of a sparingly soluble salt. Understanding these concepts is crucial in fields like analytical chemistry and environmental science, where solubility plays a key role in various chemical processes.