Question

A sparingly soluble compound $M(OH{)}_x$ has $K_{sp}=4\times {10}^{-12}$ and solubility $={10}^{-4}M$ . Then the value of $x$ is:

• A. 1
• B. 3
• C. 2
• D. 2.5

Answer 1

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 :

$M\left(OH{)}_x\right(s)\rightleftharpoons M^{x+}(aq)+xO{H}^{-}(aq)$
$sxs$
Here solubility is $s={10}^{-4}$

$\therefore K_{sp}=\left[M^{x+}\right][O{H}^{-}{]}^x$
$K_{sp}=s\times (xs{)}^x$
Putting the values,
$4\times {10}^{-12}={10}^{-4}\times (x\times {10}^{-4}{)}^{x}4\times {10}^{-8}=(x\times {10}^{-4}{)}^x$

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