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Question #44112

Difficulty: 8/10 | 🟢 Free

The Question

A $1.0 \text{ L}$ aqueous solution at $298 \text{ K}$ contains $0.10 \text{ M}$ of a weak diprotic acid $\ce{H2X}$ ( $K_{a1} = 2.0 \times 10^{-5}$ , $K_{a2} = 5.0 \times 10^{-12}$ ) and $0.02 \text{ M}$ of $\ce{M^{2+}}$ ions. Solid $\ce{NaHX}$ is slowly added to this solution. Given the solubility product $K_{sp}(\ce{MX}) = 1.25 \times 10^{-10}$ , what is the maximum number of moles of $\ce{NaHX}$ that can be added before the precipitation of $\ce{MX}$ begins? (Assume no change in volume upon addition of solid $\ce{NaHX}$ ).

Options

A.
$0.05 \text{ mol}$
B.
$0.20 \text{ mol}$
C.
$0.025 \text{ mol}$
D.
$0.10 \text{ mol}$

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

Successive Equilibrium Constants|Common Ion Effect on Solubility

Why This Question Is Tricky

Common Mistakes

This problem tests the 'Successive Equilibrium' fallacy and Le Chatelier's Principle in a counter-intuitive way.

Students often assume that adding a common ion ( $\ce{HX-}$ ) will suppress dissociation and prevent precipitation.

However, adding $\ce{HX-}$ to $\ce{H2X}$ forms a buffer that decreases $[\ce{H+}]$ .

Because the concentration of the divalent anion $[\ce{X^{2-}}]$ is inversely proportional to the square of $[\ce{H+}]$ , this decrease in acidity causes a massive, quadratic spike in $[\ce{X^{2-}}]$ , which triggers the precipitation of the metal salt.

The trap options catch students who invert the buffer equation (Option B) or incorrectly assume $[\ce{X^{2-}}]$ depends linearly on $[\ce{H+}]$ without accounting for $K_{a1}$ (Option C).

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Equilibrium - Class 11 Part 1
Concept	Successive Equilibrium Constants
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free