The K sp of Mg ( OH ) 2 is 1 × 1 0 - 12 . A 0.01 M MgC l 2 solution will precipitate at what limiting pH value?

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Question: Answered & Verified by Expert

The

K_{sp}

of Mg

(OH)_{2}

is

1 \times 1 0^{- 12} .

A 0.01 M

MgC l_{2}

solution will precipitate at what limiting pH value?

ChemistryIonic EquilibriumJEE Main

Solution:

1971 Upvotes Verified Answer

The correct answer is: 9

K_{sp} = [M g^{2 +}] {[O H^{-}]}^{2}

[O H^{-}] = \sqrt{\frac{1 \times 1 0^{- 12}}{0.01}}

[O H^{-}] = 1 0^{- 5}

pOH = 5

pH = 9

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