What mass of $"magnesium hydroxide"$ is required to give a $1L$ volume of $[Mg(OH)_2]$ whose concentration is $0.01mol*L^-1$ with respect to the hydroxide?

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Answer 1 · 2016-10-12T04:03:38

For a $1*L$ volume of $0.01*mol*L^-1$ $Mg(OH)_2$ , $0.28*g$ salt are required.

$"Molarity"$ $=$ $"Moles"/"Volume"$ $=$ $(0.01*mol)/(1.0*L)$ .

And thus we need to dissolve $0.01*molxx58.32*g*mol^-1xx1/2$ $=$ $0.28*g$ in a $1*L$ volume. Why did I include the $1/2$ ?

However, $K_"sp",Mg(OH)_2=5.61×10^(−12)$ at $298K$ , which gives a solubility of $6.4xx10^-3*g*L^-1$ . The question was thus not well-proposed.

Magnesium hydroxide is thus too insoluble to provide such a concentration.

What mass of "magnesium hydroxide""magnesium hydroxide" is required to give a 1*L1*L volume of [Mg(OH)_2][Mg(OH)_2] whose concentration is 0.01*mol*L^-10.01*mol*L^-1 with respect to the hydroxide?