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Question: If the solubility product \(K _ { s p }\) of a sparingly soluble salt \(M X _ { 2 }\) at \(1.0 \...

If the solubility product KspK _ { s p } of a sparingly soluble salt MX2M X _ { 2 } at 1.0×10−111.0 \times 10 ^ { - 11 }, the solubility of the salt in mole litre–1 at this temperature will be.

A

2.46×10142.46 \times 10 ^ { 14 }

B

1.36×10−41.36 \times 10 ^ { - 4 }

C

2.60×10−72.60 \times 10 ^ { - 7 }

D

1.20×10−101.20 \times 10 ^ { - 10 }

Answer

1.36×10−41.36 \times 10 ^ { - 4 }

Explanation

Solution

Ksp=4S3K _ { s p } = 4 S ^ { 3 }

S=Ksp43=1×10−1143=1.35×10−4S = \sqrt [ 3 ] { \frac { K _ { s p } } { 4 } } = \sqrt [ 3 ] { \frac { 1 \times 10 ^ { - 11 } } { 4 } } = 1.35 \times 10 ^ { - 4 }