If the molar solubility (inmolL−1) of a sparingly soluble salt AB3 is s. The corresponding solubility product is Ksp. The solubility (s) can be written in terms of Ksp by the relation:
A
s=(128Ksp)14
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B
s=(Ksp27)14
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C
s=(Ksp256)15
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D
s=(Ksp3)14
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Solution
The correct option is Bs=(Ksp27)14 Let "s" molL−1 be the solubility of the AB3 AB3(s)⇌A3+(aq)+3B−(aq) t=teqc−ss3s
so, Ksp=(s)1(3s)3Ksp=27s4
Thus the expression for solubility becomes, s4=(Ksp27) ⇒s=(Ksp27)14