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pH of a Solution

What will hap...

Question

What will happen if the pH of the solution of $0.001$ M $M g (N O_{3})_{2}$ solution is adjusted to $p H = 9$ ?
[Given that : $K_{s p}$ of $M g (O H)_{2} = 8.9 \times 10^{- 12}$ )].

ppt will take place

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ppt will not take place

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solution will be saturated

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none of the above

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Solution

The correct option is B ppt will not take place
At $p H = 9, p O H = 14 - p H = 5$ and $[O H^{-}] = 10^{- 5} M$

In a $0.001 M$ solution of $M g (N O_{3})_{2}$ ,
$[M g^{+ 2}] = 0.0010 M$

The ionic product for $M g (O H)_{2} = [M g] \times [O H^{-}]^{2} = 0.0010 \times (10^{- 5})^{2} = 1 \times 10^{- 13} M^{3}$
which is smaller than the $K s p$ of $8.9 \times 10^{- 12} M^{3}$

So, we don't see a ppt at this pH.

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Similar questions

[M g^{2 +}] [O H^{⊝}]^{2}

M g (O H)_{2}

M g (N O_{3})_{2}

K_{s p}

M g (O H)_{2} = 8.9 \times 10^{- 12}

25^{o} C,

1 \times 10^{- 4} M

M g (N O_{3})_{2}

M g (O H)_{2}

K_{s p} (M g (O H))_{2} = 8.9 \times 10^{- 12} M^{3} .

\sqrt{8.9} = 2.98

l o g 2.98 = 0.474

25^{o} C,

M g (O H)_{2}

10^{- 4} M

M g (N O_{3})_{2}

9.0

K_{s p} [M g (O H)_{2}]) = 10^{- 11} M^{3}

25^{\circ} C

M g (O H)_{2}

0.0001 M

M g (N O_{3})_{2}

p H

9.0

p H

K_{s p} (M g (O H)_{2}) = 10^{- 11} M^{3}

p H

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

(K_{s p} = 4 \times 10^{- 12})

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