The Ksp of MgOH2 is 1- 10-12- 0-01 m MgOH2 will precipitate at the limiting pH equal to-

The solubility product expression is: #160; #160; K_SP = [Mg^2+][OH^ ]^2 Rearrange the above expression: [OH^ ] = √(K_sp/[Mg^2+]) Substitute values in ...

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Byju's Answer

Standard XII

Chemistry

Ostwald Dilution Law

The Ksp of ...

Question

The $K_{s p}$ of $M g (O H)_{2}$ is $1 \times 10^{- 12}$ . $0.01 m$ $M g (O H)_{2}$ will precipitate at the limiting pH equal to:

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Solution

The solubility product expression is:

$K_{S P} = [M g^{2 +}] [O H^{-}]^{2}$

Rearrange the above expression:

$[O H^{-}] = \sqrt{\frac{K_{s p}}{[M g^{2 +}]}}$

Substitute values in the above expression and calculate hydroxide ion concentration-

$[O H^{-}] = \sqrt{\frac{1 \times 10^{- 12}}{0.01}}$

$[O H^{-}] = 1 \times 10^{- 5}$ M

Calculate pOH from the hydroxide ion concentration:

$p O H = - l o g [O H^{-}] = - l o g (1 \times 10^{- 5})$

$p O H = 5$

Calculate pH from pOH-

$p H = 14 - p O H$

$p H = 14 - 5$

$p H = 9$

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Similar questions

Q. The

K_{s p}

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

1 \times 10^{- 12}

0.10 M

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

will precipitate at the limiting pH:

K_{s p}

M g (O H)_{2}

1 \times 10^{- 12}, 0.01 M M g C l_{2}

will be precipitating at the limiting pH :

Q. Calculate pH at which

M g (O H)_{2}

begin to precipitate form a solution containing 0.1 M

M g^{2 +}

ions.

K_{s p}

M g (O H)_{2} = 1 \times 10^{- 11}

Q. What is the maximum pH of a solution

0.10 M

{M g}^{2 +}

from which

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

will nor precipitate? Given,

K_{s p} [M g {(O H)}_{2}] = 1.2 \times 10^{- 11}

Q. Calculate the concentration (in

m o l L^{- 1}

) of hydroxide ion necessary to precipitate

M g (O H)_{2}

from

0.1 M

solution of

M g S O_{4}

[K_{s p}

M g (O H)_{2} = 5.5 \times 10^{- 12}]

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The Ksp of Mg(OH)2 is 1×10−12. 0.01 m Mg(OH)2 will precipitate at the limiting pH equal to:

The $K_{s p}$ of $M g (O H)_{2}$ is $1 \times 10^{- 12}$ . $0.01 m$ $M g (O H)_{2}$ will precipitate at the limiting pH equal to: