The solubility of is affected by the presence of , which provides a common ion ( ). The solubility product constant ( ) for is: -
Step 1: Define solubility
Let the solubility of in the solution be (in ). Dissolving contributes:
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Step 2: Account for common ion from
The solution contains , which dissociates completely:
Thus, the total is: -
Step 3: Apply
Substitute into the expression: -
Step 4: Simplify using common ion effect
Since is sparingly soluble, its solubility is very small ( ). Therefore, we can approximate: So: Solve for : -
Step 5: Verify approximation
If , then: The approximation holds since is much smaller than . Let’s confirm exactly: This matches the given , confirming the solution.
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Step 6: Check options
The calculated solubility matches option (C).