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Byju's Answer

Standard XII

Chemistry

Redox Reactions

K sp value of...

Question

$K_{s p}$ value 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 ${N H}_{4} O H$ is added in a solution of ${A l}^{+ 3}$ and ${Z n}^{+ 2}$ which will precipitate earlier:

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

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

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Both together

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None will precipitate

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Solution

The correct option is A ${A l (O H)}_{3}$
$K_{s p}$ of $A l (O H)_{3} = 8.5 \times 10^{- 23}$
$K_{s p}$ of $Z n (O H)_{2} = 1.8 \times 10^{- 14}$
After adding of $N H_{4} O H$ in the solution of $A l^{+ 3}$ and $Z n^{+ 2}$ , $A l (O H)_{3}$ will precipitatet earlier because its solubility product value is less then $Z n (O H)_{2}$ .

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Similar questions

Q. 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

Q. The solubility products 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

N H_{4} O H

is added to a solution containing

A l^{3 +}

and

Z n^{2 +}

ions, then substance precipitated first is:

Q. The solubility products of

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

and

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

are

8.5 \times 10^{- 23}

and

1.8 \times 10^{- 14}

at room temperature. If the solution contains

{A l}^{3 +}

and

{Z n}^{2 +}

ions, the ion first precipitated by adding

{N H}_{4} O H

is:

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 ?

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Ksp value of Al(OH)3 and Zn(OH)2 are 8.5 × 10−23 and 1.8 × 10−14 respectively. If NH4OH is added in a solution of Al+3 and Zn+2 which will precipitate earlier:

The correct option is A Al(OH)3Ksp of Al(OH)3=8.5×10−23Ksp of Zn(OH)2=1.8×10−14After adding of NH4OH in the solution of Al+3 and Zn+2, Al(OH)3 will precipitatet earlier because its solubility product value is less then Zn(OH)2.

$K_{s p}$ value 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 ${N H}_{4} O H$ is added in a solution of ${A l}^{+ 3}$ and ${Z n}^{+ 2}$ which will precipitate earlier: