A) 0.0047 g/L
B) 0.047 g/L
C) 0.0087 g/L
D) 0.087 g/L
Correct Answer: C
Solution :
\[Mg{{(OH)}_{2}}M{{g}^{2+}}+2O{{H}^{-}}\] \[{{K}_{sp}}=[M{{g}^{2+}}]{{[O{{H}^{-}}]}^{2}}\] \[1.4\times {{10}^{-11}}=(s){{(2s)}^{2}}\] \[1.4\times {{10}^{-11}}=4{{s}^{3}}\] \[s=1.5\times {{10}^{-4}}mol/L\] (\[\because \]Mol. wt. of\[Mg{{(OH)}_{2}}=52\]) \[=58\times 1.5\times {{10}^{-4}}\] \[=0.0087\text{ }g/L\]
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