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Common Ion Effect

Solubility pr...

Question

Solubility product of $A l (O H)_{3}$ and $Z n (O H)_{2}$ are $8.5 \times 10^{- 23}$ and $1.8 \times 10^{- 14}$ respectively. If both $A l^{3 +}$ and $Z n^{2 +}$ ions are present in a solution, which one will precipitate first on addition of $N H_{4} O H$ ?

A

A l (O H)_{3}

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B

Z n (O H)_{2}

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C

M g (O H)_{2}

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D

C a (O H)_{2}

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Solution

The correct option is A $A l (O H)_{3}$
The value of solubility product for $A l (O H)_{3}$ is $8.5 \times 10^{- 23}$ .

It is much lower than the value of the solubility product for $Z n (O H)_{2}$ which is $1.8 \times 10^{- 14}$ .
Hence, when both $A l^{3 +}$ and $Z n^{2 +}$ ions are present in the solution and as $N H_{4} O H$ is added, $A l (O H)_{3}$ will precipitate first as the solubility product of $A l (O H)_{3}$ is quite low.

Hence, its ionic product will exceed its solubility product at a very low concentration of $O H^{-}$ ions.

Option A is correct.

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Similar questions

Q. The solubility products of

A l (O H)_{3}

Z n (O H)_{2}

8.5 \times 10^{- 23}

1.8 \times 10^{- 14}

respectively. If

N H_{4} O H

is added to a solution containing

A l^{3 +}

Z n^{2 +}

ions, then substance precipitated first is:

Q. The solubility products of

A l {(O H)}_{3}

Z n {(O H)}_{2}

8.5 \times 10^{- 23}

1.8 \times 10^{- 14}

at room temperature. If the solution contains

{A l}^{3 +}

{Z n}^{2 +}

ions, the ion first precipitated by adding

{N H}_{4} O H

Solubility products of $A l (O H)_{3}$ and $Z n (O H)_{2}$ are $2.7 \times 10^{- 23} a n d 3.2 \times 10^{- 14}$ respectively. If to a solution of 0.1 M each of $A l^{3 +} a n d Z n^{2 +} i o n s, N H_{4} O H$ is added in increasing amounts which of the following will be precipitated first ?

Solubility products of $A l (O H)_{3}$ and $Z n (O H)_{2}$ are $2.7 \times 10^{- 23} a n d 3.2 \times 10^{- 14}$ respectively. If to a solution of 0.1 M each of $A l^{3 +} a n d Z n^{2 +} i o n s, N H_{4} O H$ is added in increasing amounts which of the following will be precipitated first ?

K_{s p}

{A l (O H)}_{3}

{Z n (O H)}_{2}

\times

10^{- 23}

\times

10^{- 14}

respectively. If

{N H}_{4} O H

is added in a solution of

{A l}^{+ 3}

{Z n}^{+ 2}

which will precipitate earlier:

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