The solubility of CuBr is 2×10−4molL−1. at 25oC. The Ksp value for CuBr is:
A
4×10−8mol2L−2
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B
4×10−11mol2L−1
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C
4×10−4mol2L−2
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D
4×10−15mol2L−2
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Solution
The correct option is A4×10−8mol2L−2 Given, s=2×10−4molL−1 the dissociation of CuBr is given by CuBr(s)⇌Cu+(aq)+Br−(aq)t=0100ateqb1−sss solubility product: Ksp=[Cu+][Br−]=s×s s2=(2×10−4)2=4×10−8mol2L−2