The Question is from neet-chemistry chemical-equilibrium solubility-product, and it was created by the author Mr. S M HUSSAIN

Question

A sparingly soluble salt MX$_2$ has a solubility product K$_{sp}$ = 4 x 10$^{-12}$. In a solution containing 0.01 M NaX, the molar solubility of MX$_2$ is:

Mr. S M HUSSAIN · Accepted Answer

## CORE INFORMATION
✓ **Answer**: A
## CONCEPT FRAMEWORK
	**Primary Concept**: Solubility Product and Common Ion Effect
	**Related Topics**: Equilibrium, Ionic Compounds, Solubility
	**Prerequisites**: Understanding of K$_{sp}$, molar solubility, and the common ion effect.
## SOLUTION PATHWAY
**Given Information**:
	K$_{sp}$(MX$_2$) = 4 x 10$^{-12}$
	[NaX] = 0.01 M, which implies [X$^-$] = 0.01 M due to complete dissociation of NaX.
**Method & Steps**:
	1. Write the solubility equilibrium expression for MX$_2$:
		MX$_2$(s) $ightleftharpoons$ M$^{2+}$(aq) + 2X$^-$ (aq)
		Reasoning: This represents the dissolution of MX$_2$ in solution.
	2. Express K$_{sp}$ in terms of molar solubility (s):
		Let 's' be the molar solubility of MX$_2$. Then, [M$^{2+}$] = s and [X$^-$] = 2s + 0.01 (due to the common ion effect from NaX).
		K$_{sp}$ = [M$^{2+}$][X$^-$]$^2$ = s(2s + 0.01)$^2$
		Working: Since K$_{sp}$ is very small and [X$^-$] from NaX is relatively high, we can approximate 2s + 0.01 $\approx$ 0.01.
		Key Point: This approximation simplifies the calculation and is valid when the solubility of the salt is significantly reduced by the common ion effect.
	3. Solve for 's':
		4 x 10$^{-12}$ = s(0.01)$^2$
		Result: s = (4 x 10$^{-12}$) / (1 x 10$^{-4}$) = 4 x 10$^{-8}$ M
		Verification: The calculated solubility is much smaller than 0.01 M, justifying the approximation made in step 2.
## OPTION ANALYSIS
[A] ✓: 4 x 10$^{-8}$ M (Correct, as calculated above)
[B] :x: 2 x 10$^{-6}$ M (Incorrect, this value would be obtained if the common ion effect was ignored)
[C] :x: 1 x 10$^{-5}$ M (Incorrect, this is even further off from the correct value)
[D] :x: 4 x 10$^{-10}$ M (Incorrect, this value doesn't account for the squared term of [X$^-$] in the K$_{sp}$ expression)
## LEARNING AIDS
	**Common Mistakes**:
		1. Ignoring the common ion effect: This would lead to a much higher (and incorrect) calculated solubility.
			How to avoid: Always consider the concentration of common ions already present in the solution.
		2. Forgetting to square the [X$^-$] term in the K$_{sp}$ expression: This would lead to an incorrect calculation.
			How to avoid: Carefully write the K$_{sp}$ expression based on the balanced dissolution equation.
	**Quick Tips**:
		1. When K$_{sp}$ is very small and a common ion is present, the approximation 2s + [common ion] $\approx$ [common ion] is usually valid.
		2. Remember that the common ion effect always *decreases* the solubility of a sparingly soluble salt.

Mr. S M HUSSAIN · Answer

The correct Option is "4 x 10$^{-8}$ M"

Mr. S M HUSSAIN · Answer

2 x 10$^{-6}$ M

Mr. S M HUSSAIN · Answer

1 x 10$^{-5}$ M

Mr. S M HUSSAIN · Answer

4 x 10$^{-10}$ M

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