Question

Mg2+ + 2(OH-) -> Mg(OH)2

(note that this is a equilibrium reaction so it should have both forward and backward equal signs)

I need to find the concentration of [Mg2+] and [OH-] with the given concentration of [Mg(OH)2] of 1.5 x 10^-12 mol^3l^3

Answer 1

Answer on Question #61305 - Chemistry - Inorganic Chemistry

$$\mathrm{Mg}^{2+} + 2\mathrm{OH}^{-} \leftrightarrow \mathrm{Mg(OH)}_{2}$$

The solution constant $K_{s}$:

$$K_{s} = \left[\mathrm{Mg}^{2+}\right] \times \left[2\mathrm{OH}^{-}\right]^{2} = s \times (2s)^{2} = 4s^{3} = 1.5 \times 10^{-12} \, (\mathrm{mol}^{3} \, \mathrm{l}^{3})$$ $$s = \sqrt[3]{1.5/4} \times 10^{-4} = 0.7211 \times 10^{-4} = 7.21 \times 10^{-5} \, (\mathrm{mol} \cdot \mathrm{l})$$

The concentration of $\left[\mathrm{Mg}^{2+}\right] = 7.21 \times 10^{-5} \, \mathrm{mol} \cdot \mathrm{l}$, and the concentration of $\left[\mathrm{OH}^{-}\right] = (1.5 \times 10^{-12}) / (7.21 \times 10^{-5}) = 2.08 \times 10^{-8} \, \mathrm{mol} \cdot \mathrm{l}$.

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