Question

If $K_{sp}$ of ${M\left(OH\right)}_3$ is $1\times {10}^{-12}$ then 0.001M. $M^{3+}$ is precipitated at a $pH>9$.

If true enter 1, else enter 0. ________.

Answer 1

The statement, "If $K_{sp}$ of ${M\left(OH\right)}_3$ is $1\times {10}^{-12}$ then 0.001M. $M^{3+}$ is precipitated at a $pH>9$." is true.
The precipitation of ${M\left(OH\right)}_3$ will occur when, its ionic product exceeds solubility product.
$\left[M^{3+}\right][{OH}^{-}{]}^3>K_{sp}$
The concentration of metal ion is 0.001 M and the value of the solubility product is $1\times {10}^{-12}$ .
Hence, precipitation will occur when ${\left[{OH}^{-}\right]}^3>1\times {10}^{-9}$ or $\left[{OH}^{-}\right]>1\times {10}^{-3}$
Thus, for precipitation to occur, $pOH$ should be less than 3 and $pH$ should be more than 9.