Question

The solubility product $Mg(OH{)}_2$ and $Al(OH{)}_3$ in water at ${25}^{\circ }C$ are $2.56\times {10}^{-13}$ and $4.32\times {10}^{-34}$ . If $s_1$ and $s_2$ are the solubilities of $Mg(OH{)}_2$ and $Al(OH{)}_3$ in water in mol/L 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

The correct option is B $2\times {10}^4$

$Mg(OH{)}_2\to M{g}^{2+}+2O{H}^{-}{s}_1(2S_1{)}^2{K}_{sp}ofMg(OH{)}_2=4s_1^3=2.56\times {10}^{-13}{s}_1=4\times {10}^{-5}MAl(OH{)}_3\to A{l}^{3+}+3O{H}^{-}{s}_2\left(3S_2{)}^3{K}_{sp}ofAl\right(OH{)}_3=27s_2^4=4.32\times {10}^{-34}\Rightarrow s_2^4=0.16\times {10}^{-34}\Rightarrow s_2^4=16\times {10}^{-36}\Rightarrow s_2=2\times {10}^{-9}M\Rightarrow \frac{s_1}{s_2}=\frac{4\times {10}^{-5}}{2\times {10}^{-9}}=2\times {10}^{+4}$