The K s p of M g O H 2 is 1 × 10 – 12 . 0.01 M M g O H 2 will precipitate at the limiting p H -

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Question: Answered & Verified by Expert

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

p H

-

ChemistryIonic EquilibriumNEET

Options:

A 3
B 9
C 5
D 8

Solution:

1817 Upvotes Verified Answer

The correct answer is: 9

[M g^{2 +}] {[O H^{-}]}^{2} = 10^{- 12}

0.01 \times {[O H^{-}]}^{2} = 10^{- 12}

[{O H}^{-}] = 10^{- 5} M

∴ [H^{+}] = 10^{- 9} M

and

p H = - log 10^{- 9} = 9

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