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Standard XII

Chemistry

Molarity

Mg OH 2 is pr...

Question

$M g (O H)_{2}$ is precipitated when $N a O H$ is added to a solution of $M g^{2 +}$ . If the final concentration of $M g^{2 +}$ is $10^{- 10}$ M, the concentration of $O H^{-}$ (M) in the solution is:
$[Solubility product for M g (O H)_{2} = 5.6 \times 10^{- 12}]$

0.056

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0.12

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0.24

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0.025

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Solution

The correct option is C 0.24
$M g (O H)_{2} ⇌ M g^{2 +} + 2 O H^{-}$
$K_{s p} = [M g^{2 +}] [O H^{-}]^{2} 5.6 \times 10^{- 12} = [10^{- 10}] [O H^{-}]^{2} [O H^{-}] = \sqrt{5.6 \times 10^{- 2}} = 0.24 M$

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Similar questions

M g (O H)_{2}

is precipitated when NaOH is added to a solution of

M g^{2 +}

. If the final concentration of

M g^{2 +}

10^{- 10}

M, the concentration of

O H^{-} (M)

in the solution is [Solubility product for

M g (O H)_{2} = 5.6 \times 10^{- 12}

Q. The solubility product constant Ksp of

M g (O H)_{2}

9.0 \times 10^{- 12}

. If a solution is

0.010

M with respect to

M g^{2 +}

ion, what is the maximum hydroxide ion concentration which could be present without causing the precipitation of

M g (O H)_{2}

Q. Solubility product of

M g {(O H)}_{2}

1 \times 10^{- 11}

. At what pH, precipitation of

M g {(O H)}_{2}

will begin from 0.1 M

{M g}^{2 +}

solution ;-

Q. At

25^{o} C

, the solubility product of

M g (O H)_{2}

1.0 \times 10^{- 11}

. At which

p H

, will

M g^{2 +}

ions start precipitating in the form of

M g (O H)_{2}

from a solution of 0.001 M

M g^{2 +}

ions?

Q. Solubility product of

M g (O H)_{2}

9 \times 10^{- 12}

. If concentration of

M g^{+ 2}

in saturated solution is

0.01

M, then concentration of

O H^{-}

is?

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Mg(OH)2 is precipitated when NaOH is added to a solution of Mg2+. If the final concentration of Mg2+ is 10–10 M, the concentration of OH– (M) in the solution is: [Solubility product for Mg(OH)2=5.6×10–12]

The correct option is C 0.24Mg(OH)2⇌Mg2++2OH− Ksp=[Mg2+][OH–]25.6×10–12=[10–10][OH–]2[OH–]=√5.6×10−2=0.24 M

$M g (O H)_{2}$ is precipitated when $N a O H$ is added to a solution of $M g^{2 +}$ . If the final concentration of $M g^{2 +}$ is $10^{- 10}$ M, the concentration of $O H^{-}$ (M) in the solution is:
$[Solubility product for M g (O H)_{2} = 5.6 \times 10^{- 12}]$

The correct option is C 0.24
$M g (O H)_{2} ⇌ M g^{2 +} + 2 O H^{-}$
$K_{s p} = [M g^{2 +}] [O H^{-}]^{2} 5.6 \times 10^{- 12} = [10^{- 10}] [O H^{-}]^{2} [O H^{-}] = \sqrt{5.6 \times 10^{- 2}} = 0.24 M$