Solubility product of $Mg (OH)_{2}, Cd (OH)_{2}, Al (OH)_{3}$ and $Zn (OH)_{2}$ are $4 \times 10^{- 11}, 8 \times 10^{- 6}, 8.5 \times 10^{- 23}$ and $1.8 \times 10^{- 14}$ respectively. The cation, that will precipitate first as hydroxide, on adding limited quantity of ${NH}_{4} OH$ in a solution containing equimolar amount of metal cation, is :

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

${Al}^{3 +}$

${Cd}^{2 +}$

${Zn}^{2 +}$

${Mg}^{2 +}$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} {[{OH}^{-}]}_{Mg (OH)_{2}} = \sqrt{\frac{4 \times 10^{- 11}}{x}} \\ {[{OH}^{-}]}_{Cd [{OH}_{2}} = \sqrt{\frac{8 \times 10^{- 6}}{x}} \\ {[{OH}^{-}]}_{Ar (OH)_{3}} = \sqrt[3]{\frac{8.5 \times 10^{- 23}}{x}} \\ {[{OH}^{-}]}_{Zn ({OH}_{2}} = \sqrt{\frac{1.8 \times 10^{- 14}}{x}} \end{array}$

The value of the solubility product for the $Al (OH)_{3}$ is $8.5 \times 10^{- 23}$ . Its cation will precipitate first as hydroxide, on adding limited quantity of ${NH}_{4} OH$ in a solution containing equimolar amount of metal cation as its solubility product is quite low.