Before You Begin

Question #44094

Difficulty: 8/10 | 🟢 Free

The Question

Consider a sparingly soluble salt \ce{MA} with a solubility product $K_{sp} = 1.0 \times 10^{-6}$ at $298 \text{ K}$ . The anion \ce{A^-} is the conjugate base of a weak acid \ce{HA} with an ionization constant $K_a = 1.0 \times 10^{-9}$ . Calculate the solubility of \ce{MA} in a $0.10 \text{ M}$ aqueous solution of \ce{MCl}.

(Assume \ce{MCl} is completely dissociated, no other complexation or precipitation reactions occur, and ignore activity coefficients.)

Options

A.
$2.0 \times 10^{-5} \text{ M}$
B.
$1.0 \times 10^{-5} \text{ M}$
C.
$1.0 \times 10^{-3} \text{ M}$
D.
$1.6 \times 10^{-5} \text{ M}$

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

Salt Hydrolysis Logic|Common Ion Effect on Solubility|Ionic Product of Water (Kw)

Why This Question Is Tricky

Common Mistakes

This problem tests the 'Common Ion Effect coupled with Salt Hydrolysis'.

Students typically assume that the solubility $s$ is simply $K_{sp}/[\ce{M^+}]$ , ignoring that the anion undergoes significant hydrolysis.

The trap is that the common ion fixes the equilibrium concentration of the unhydrolyzed anion $[\ce{A^-}]$ , not the total solubility.

To find the true solubility, one must calculate the additional \ce{MA} that dissolves to replenish the \ce{A^-} lost to hydrolysis, requiring the simultaneous solution of the $K_{sp}$ and $K_h$ equations.

Quick Facts About This Question

Question Information

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