At 250C, the solubility product of MgOH2 is 4.0×10−12 . Then calculate pH? (Take log2=0.30 )

MgOH2→ Mg+2+2OH− s s 2sksp=4s3=4×10−12, ∴s=10−4OH−=2s=2×10−4 pOH=4−log102=4−0.3=3.7pH=14−pOH=14−3.7=10.3

At 250C, the solubility product of MgOH2 is 4.0×10−12 . Then calculate pH? (Take log2=0.30 )

MgOH2→ Mg+2+2OH− s s 2sksp=4s3=4×10−12, ∴s=10−4OH−=2s=2×10−4 pOH=4−log102=4−0.3=3.7pH=14−pOH=14−3.7=10.3

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At $25^{0} C$ , the solubility product of $Mg {(OH)}_{2}$ is $4 .0 \times 10^{- 12}$ . Then calculate pH? (Take $\log 2 = 0.30$ )

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4.7

10.3

9.3

3.7

answer is D.

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Detailed Solution

$\begin{array}{l} Mg {(OH)}_{2} \overset{}{\to} {Mg}^{+ 2} + 2 {OH}^{-} \\ s s 2 s \end{array}$
$k_{sp} = 4 s^{3} = 4 \times 10^{- 12}, ∴ s = 10^{- 4}$
$[{OH}^{-}] = 2 s = 2 \times 10^{- 4}$ ${pOH=4−log}_{10}^{2} = 4 - 0 .3 = 3 .7$
$pH = 14 - pOH = 14 - 3 .7 = 10 .3$