Question

The solubility products of $Al(OH{)}_3$ and $Zn(OH{)}_2$ are $8.5\times {10}^{-23}$ and $1.8\times {10}^{-14}$ respectively. If $N{H}_{4}OH$ is added to a solution containing $A{l}^{3+}$ and $Z{n}^{2+}$ ions, then substance precipitated first is:

• A. $Al(OH{)}_3$
• B. $Zn(OH{)}_2$
• C. both together
• D. none of these

Answer 1

The correct option is A $Al(OH{)}_3$

Given

Solubility product of

$Al(OH{)}_3=8.5\ast {10}^{-23}$

$Zn(OH{)}_3=1.8\ast {10}^{-14}$

Solution

Solution product of

$Al(OH{)}_3=A{l}^{3+}+3(OH{)}^{-}$

$K_{sp}=s\ast (3s{)}^3$

$8.5\ast {10}^{-23}=27s^4$

$s=(\frac{8.5\ast {10}^{-23}}{27}{)}^{\frac{1}{4}}$

$s=1.33\ast {10}^{-6}$

Solution product of

$Zn(OH{)}_2=Z{n}^{2+}+2(OH{)}^{-}$

$K_{sp}=s\ast (2s{)}^2$

$1.8\ast {10}^{-14}=4s^3$

$s=(\frac{1.8\ast {10}^{-14}}{4}{)}^{\frac{1}{3}}$

$s=16.5\ast {10}^{-6}$

As we can see Solubility product of $Al(OH{)}_3$ is less than $Zn(OH{)}_2$ therefore $Al(OH{)}_3$ is less soluble and will precipitate first.

The correct option is A