Question

The solubility product $Mg(OH{)}_2$ in water at ${25}^{\circ }C$ is $2.56\times {10}^{-13}(mol/lt{)}^3$ while that of $AI(OH{)}^3$is $4.32\times {10}^{-34}(mol/lt{)}^4$. lf $s_{1}and{s}_2$ are the solubilities of $Mg(OH{)}_2$ and $AI(OH{)}^3$ in water in moillt at ${25}^{\circ }C$, what is the ratio. $s_1/s_2$ ?

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

Answer 1

Answer: (B)

$2\times {10}^4$

$K_{sp}=\left(Mg\right)(OH{)}^2=(s\left)\right(2s{)}^2=4s^3{s}_1=\sqrt[3]{32.56X{10}^{-13}4}=0.84X{10}^{-4}{K}_{sp}=\left(Al\right){\left(OH\right)}^3=\left(s\right)(3s{)}^3=27s^4{s}_2=\sqrt[4]{4\frac{4.32X{10}^{-34}}{27}}=0.67X{10}^{-8}\frac{s_1}{s_2}=1.25X{10}^4\simeq 2X{10}^4$