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Question #44080

Difficulty: 8/10 | 🟢 Free

The Question

Solid \ce{Zn(OH)2} is suspended in an aqueous solution maintained at a constant pH of 11.0 and a constant free \ce{NH3} concentration of $0.2 \text{ M}$ at $298 \text{ K}$ . Given the solubility product constant $K_{sp}$ of \ce{Zn(OH)2} is $1.0 \times 10^{-16}$ and the overall formation constant $\beta_4$ for the complex ion \ce{[Zn(NH3)4]^{2+}} is $1.0 \times 10^9$ , calculate the total molar solubility of \ce{Zn(OH)2} in this solution. (Assume no other complex ions, such as hydroxo complexes, are formed).

Options

A.
$1.6 \times 10^{-4} \text{ M}$
B.
$1.0 \times 10^{-10} \text{ M}$
C.
$3.4 \times 10^{-4} \text{ M}$
D.
$2.0 \times 10^{-2} \text{ M}$

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

Ionic Product of Water (Kw)|Common Ion Effect on Solubility|Complex Ion Oxidation States|Werner’s Theory Postulates

Why This Question Is Tricky

Common Mistakes

This problem tests the interplay between the Common Ion Effect (which suppresses solubility via high pH) and Werner's Coordination Theory (which enhances solubility via complex ion formation).

Students often isolate equilibria, calculating solubility solely from $K_{sp}$ and pH (yielding $1.0 \times 10^{-10} \text{ M}$ ), failing to recognize that the secondary complexation equilibrium acts as a massive thermodynamic sink for $\ce{Zn^{2+}}$ , driving the dissolution forward despite the high pH.

Another trap is ignoring the pH constraint and assuming dissolution in pure water with $\ce{NH3}$ , which leads to an incorrect cubic equation for solubility.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Equilibrium - Class 11 Part 1
Concept	Ionic Product of Water (Kw)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free