Question

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

• A. $4$
• B. $1$
• C. $3$
• D. $2$

Answer 1

Answer: (D)

$2$

$M(OH{)}_x\rightleftharpoons M^{x+}+xO{H}^{-}$

$S$ $xS$

$K_{sp}=\left[M^{x+}\right]\left[O{H}^{-}\right]=\left(S\right)(xS{)}^x=4\times {10}^{-9}$

Solubility is ${10}^{-3}M$.

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

It is possible only when $x=2$.