Question

$M(OH{)}_x$ has $K_{sp}=27\times {10}^{-12}$ and solubility in water is ${10}^{-3}M$. Calculate the value of $x$.

Answer 1

$M(OH{)}_x\rightleftharpoons M^{+x}+x\stackrel{⊝}{O}H$
$S$ $xS$
$K_{sp}=S\times (xS{)}^x=X^x\cdot S^{(1+x)}$ (Given $S={10}^{-3}$)
$27\times {10}^{-12}=x^x\cdot {10}^{-3(1+x)}\therefore x=3$
Thus hydroxide is $M(OH{)}_3$.