Ksp of MgOH2 is 1- 10-12- 0-01 M MgCl2 will be precipitating at the limiting pOH

The correct option is D 5 According to given data, #160; [OH^ ] needed for precipitation of Mg(OH)_2=( 10 ^ 12 10 ^ 2)^1/2 =( 10 ^ 10)^1/2M ...

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Standard XII

Chemistry

Common Ion Effect

Ksp of MgOH2...

Question

$K_{s p} o f M g (O H)_{2}$ is $1 \times 10^{- 12}, 0.01 M M g C l_{2}$ will be precipitating at the limiting $p O H$

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Solution

The correct option is D $5$
According to given data, $[O H^{-}]$ needed for precipitation of
$M g (O H)_{2} = (\frac{10^{- 12}}{10^{- 2}})^{1 / 2} = (10^{- 10})^{1 / 2} M = 10^{- 5} M$
$p O H = - l o g [O H^{-}] = 5$

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Question 4

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KspofMg(OH)2 is 1×10−12,0.01MMgCl2 will be precipitating at the limiting pOH

The correct option is D 5According to given data, [OH−] needed for precipitation ofMg(OH)2=(10−1210−2)1/2=(10−10)1/2M=10−5MpOH=−log[OH−]=5

$K_{s p} o f M g (O H)_{2}$ is $1 \times 10^{- 12}, 0.01 M M g C l_{2}$ will be precipitating at the limiting $p O H$

The correct option is D $5$
According to given data, $[O H^{-}]$ needed for precipitation of
$M g (O H)_{2} = (\frac{10^{- 12}}{10^{- 2}})^{1 / 2} = (10^{- 10})^{1 / 2} M = 10^{- 5} M$
$p O H = - l o g [O H^{-}] = 5$