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Question

A sparingly soluble compound M(OH)x has Ksp=4×1012 and solubility =104 M . Then the value of x is:

A
1
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B
3
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C
2
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D
2.5
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Solution

The correct option is C 2
The equilibrium between the undissolved solid M(OH)x and the ions in a saturated solution can be represented by the equation :

M(OH)x (s)Mx+ (aq)+x OH (aq)
s xs
Here solubility is s=104

Ksp=[Mx+][OH]x
Ksp=s×(xs)x
Putting the values,
4×1012=104×(x×104)x4×108=(x×104)x

By inspection, we can find that the relation will hold good
when x=2.

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