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  3. Chemistry
  4. M ( OH )x has K sp =4 × 10-12 and solubility 10-4 M . Hence, x is:

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Q. $M ( OH )_{x}$ has $K_{ sp }=4 \times 10^{-12}$ and solubility $10^{-4} M .$ Hence, $x$ is:

Equilibrium

Solution:

$M( OH ) \rightarrow \underset{S}{M^{x+}}+ \underset{xs}{x OH ^{-}}$
$\therefore (S)(x S)^{x}=K_{ sp }$
$x^{x} \,S^{x+1}=4 \times 10^{-12}$
$x^{x}\left(10^{-4}\right)^{x+1}=4 \times 10^{-12}$
$\therefore x=2$