Question

$M(OH{)}_x$ has $K_{sp}$ of $4\times {10}^{-9}$ and its solubility of ${10}^{-3}$M. The value of $x$ is:

Answer 1

$M(OH{)}_x\to M^{x+}+xO\overline{H}$

$K_{sp}=\left[m^{xt}\right].{\left[xO\overline{H}\right]}^x$

$4\times {10}^{-9}=\left({10}^{-3}\right)(x\times {10}^{-3}{)}^x$

$x=2$