Question

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

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is B 2

$M(OH{)}_x$ will ionise as
$M(OH{)}_x\rightleftharpoons M^{+x}+xO{H}^{-}$
${10}^{-4}x\times {10}^{-4}$

$\therefore K_{sp}=\left[M^{+x}\right][O{H}^{-}{]}^x$


putting the values,
$\therefore \left({10}^{-4}\right)(x\times {10}^{-4}{)}^x=4\times {10}^{-12}$
By inspection, we can find that the relation will hold good,
when $x=2$.