Question

The solubility product of a rare earth metal hydroxide $M{\left(OH\right)}_3$ at room temperature is $4.32\times {10}^{-14}$. Its solubility is:

• A. $2\times {10}^{-3}M$
• B. $2.0\times {10}^{-4}M$
• C. $2\times {10}^{-5}M$
• D. $2.0\times {10}^{-6}M$

Answer 1

The correct option is A $2\times {10}^{-3}M$

$M(OH{)}_3\rightleftharpoons M^{3+}+3O{H}^{-}$

$K_{sp}$ for $A{B}_3$ type molecules is $K_{sp}=27s^4$

$4.32\times {10}^{-14}=27s^4{s}^4=\frac{4.32\times {10}^{-14}}{27}=16\times {10}^{-12}s=2\times {10}^{-3}M$