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JEE Main 2022 · Chemistry · Ionic Equilibrium · medium

The solubility product of Mg(OH)2 at 298 K is 4×10^-12 (log 2=0.30). The value of 10× (pH of its saturated aqueous solution) is

Numerical answer

The solubility product of Mg(OH)2\mathrm{Mg(OH)_2} at 298 K298\ \mathrm K is 4×10−124\times10^{-12} (log⁡2=0.30\log2=0.30). The value of 10×10\times (pH of its saturated aqueous solution) is
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Answer: 103

Let solubility be ss: Ksp=[Mg2+][OH−]2=s(2s)2=4s3=4×10−12K_{sp}=[\mathrm{Mg^{2+}}][\mathrm{OH^-}]^2=s(2s)^2=4s^3=4\times10^{-12}, so s=10−4 Ms=10^{-4}\ \mathrm M and [OH−]=2s=2×10−4 M[\mathrm{OH^-}]=2s=2\times10^{-4}\ \mathrm M. pOH=4−log⁡2=3.70\mathrm{pOH}=4-\log2=3.70, so pH=14−3.70=10.30\mathrm{pH}=14-3.70=10.30. Therefore 10×pH=10310\times\mathrm{pH}=103.
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