Question

M (OH)x has Ksp=4x10-12 and solubility is 10-4M hence x is

Answer 1

Solubility product for any reaction is equal to the product of concentration of the products.

For given reaction Ksp=[M][OH]

$M(OH)^{x}$ (s)  ↔ $M^{x+}$ (aq) + x. $OH^{-}$(aq)

$K_{sp}$ = (S) . $(xS)^{2}$

 4 X  $10^{-12}$ = $(10)^{-4}$ $(10^{-4x})^{x}$

 4 X $10^{-12}$ = $x^{2}$.$10^{-4x}$.  $10^{-4}$

 4 X $10^{-12}$ = $x^{2}$. $(10^{-4})^{x+1}$  

 $2^{2}$ x $(10^{-4})^{3}$   = $x^{x}$. $(10^{-4})^{x+1}$  

Comparing both left and right sides of above equation we get x = 2