Before You Begin

Question #44715

Difficulty: 8/10 | 🟢 Free

The Question

Solid \ce{Ni(OH)2} is added to an aqueous solution containing $0.31\text{ M}$ of the bidentate ligand ethylenediamine (en) at $298\text{ K}$ . The system reaches equilibrium. Given the solubility product of \ce{Ni(OH)2} is $K_{sp} = 4.0 \times 10^{-15}$ and the overall formation constant of the complex \ce{[Ni(en)3]^{2+}} is $\beta_3 = 1.0 \times 10^{18}$ , calculate the solubility of \ce{Ni(OH)2} in this solution. (Assume no volume change upon addition of solid and ignore the basicity of ethylenediamine).

Options

A.
$0.10\text{ M}$
B.
$0.28\text{ M}$
C.
$3.1\text{ M}$
D.
$1.0 \times 10^{-5}\text{ M}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Mechanism_Trap

Don't get tricked by the process! Learn the mechanical traps used in competitive exam questions.

Concepts Behind This Question

Solubility of Solids in Liquids|Ligand Denticity Variations|Crystal Field Splitting (Octahedral)

Why This Question Is Tricky

Common Mistakes

This problem tests the concept of simultaneous equilibria, specifically the coupling of solubility product ( $K_{sp}$ ) and complex formation constant ( $\beta_3$ ).

The 'trap' lies in two areas: First, students often assume the ligand concentration remains constant (initial $\approx$ equilibrium) which is invalid here due to the large $K_{eq}$ causing significant ligand depletion.

Second, failing to account for the correct stoichiometry of the polydentate ligand (3 moles of en per mole of Ni) and the hydroxide ions (2 moles per mole of Ni) leads to incorrect algebraic expressions.

The high concentration of generated $\ce{OH^-}$ also chemically justifies ignoring the basicity of ethylenediamine.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Solutions - Class 12
Concept	Solubility of Solids in Liquids
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free