Before You Begin

Question #44086

Difficulty: 8/10 | 🟢 Free

The Question

Solid $\ce{CaHPO4}$ is equilibrated with a buffer solution at $298\text{ K}$ maintained at $\text{pH} = 5.0$ . The solution initially contains $0.10\text{ M } \ce{CaCl2}$ . Given the solubility product of $\ce{CaHPO4}$ is $K_{sp} = 1.0 \times 10^{-7}$ and the successive acid dissociation constants for $\ce{H3PO4}$ are $K_{a1} = 10^{-3}$ , $K_{a2} = 10^{-7}$ , and $K_{a3} = 10^{-12}$ . Assuming infinite buffer capacity and ideal behavior, what is the molar solubility of $\ce{CaHPO4}$ in this solution?

Options

A.
$1.02 \times 10^{-4}\text{ mol/L}$
B.
$1.00 \times 10^{-4}\text{ mol/L}$
C.
$1.01 \times 10^{-4}\text{ mol/L}$
D.
$1.00 \times 10^{-6}\text{ mol/L}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Exception_Exposure

Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.

Concepts Behind This Question

Common Ion Effect on Solubility|Amphiprotic Ion pH|Buffer Capacity and pH Range

Why This Question Is Tricky

Common Mistakes

This problem tests the intersection of the Common Ion Effect and Amphiprotic Ion Hydrolysis.

Students often mechanically apply the common ion effect ( $S = K_{sp}/C$ ) without considering that the anion ( $\ce{HPO4^{2-}}$ ) acts as a base at lower pH values.

The 'hidden variable' is the secondary chemical sink: the conversion of $\ce{HPO4^{2-}}$ to $\ce{H2PO4^-}$ and $\ce{H3PO4}$ , which forces over 100 times more solid to dissolve than predicted by the common ion effect alone.

Distractors target the omission of minor species in the mass balance equation.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Equilibrium - Class 11 Part 1
Concept	Common Ion Effect on Solubility
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free