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

Oct 12, 2016

For a $1 \cdot L$ volume of $0.01 \cdot m o l \cdot {L}^{-} 1$ $M g {\left(O H\right)}_{2}$, $0.28 \cdot g$ salt are required.

#### Explanation:

$\text{Molarity}$ $=$ $\text{Moles"/"Volume}$ $=$ $\frac{0.01 \cdot m o l}{1.0 \cdot L}$.

And thus we need to dissolve $0.01 \cdot m o l \times 58.32 \cdot g \cdot m o {l}^{-} 1 \times \frac{1}{2}$ $=$ $0.28 \cdot g$ in a $1 \cdot L$ volume. Why did I include the $\frac{1}{2}$?

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

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