Solubility of $M {g (O H)}_{2}$ $Mg(OH)_2$ is $1.6$ $1.6$ x $10^{- 4}$ $10^-4$ $mol/L$ $"mol/L"$ at $298$ $298$ $K$ $K$ . What is its solubility product?

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Solubility of $M {g (O H)}_{2}$ $Mg(OH)_2$ is $1.6$ $1.6$ x $10^{- 4}$ $10^-4$ $mol/L$ $"mol/L"$ at $298$ $298$ $K$ $K$ . What is its solubility product?

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Answer 1 · 2016-01-31T00:20:34

$1.6 \cdot 10^{- 11}$ $1.6 * 10^(-11)$

Explanation:

Magnesium hydroxide, $Mg {(OH)}_{2}$ $"Mg"("OH")_2$ , is considered Insoluble in aqueous solution, which implies that the ions it forms will be in equilibrium with the solid.

Magnesium hydroxide dissociates only partially to form magnesium cations, ${Mg}^{2 +}$ $"Mg"^(2+)$ , and hydroxide anions, ${OH}^{-}$ $"OH"^(-)$

$Mg {(OH)}_{2(s]} ⇌ {Mg}_{(aq]}^{2 +} + 2 {OH}_{(aq]}^{-}$ $"Mg"("OH")_text(2(s]) rightleftharpoons "Mg"_text((aq])^(2+) + color(red)(2)"OH"_text((aq])^(-)$

For an insoluble compound, its molar solubility tells you how many moles of the compound can be dissolved per liter of aqueous solution before reaching saturation.

In your case, a molar solubility of

$s = 1.6 \cdot 10^{- 4}$ $s = 1.6 * 10^(-4)$

means that you can only dissolve $1.6 \cdot 10^{- 4}$ $1.6 * 10^(-4)$ moles of magnesium in a liter of water at that temperature.

Take a look at the dissociation equilibrium. Notice that every mole of magnesium hydroxide that dissociates produces $1$ $1$ mole of magnesium cations and $2$ $color(red)(2)$ moles of hydroxide anions.

This tells you that if you successfully dissolve $1.6 \cdot 10^{- 4}$ $1.6 * 10^(-4)$ moles of magnesium hydroxide in one liter of solution, number of moles of each ion will be

$n_{M g^{2 +}}^{2 +} = 1 \times 1.6 \cdot 10^{- 4} = 1.6 \cdot 10^{- 4} {moles Mg}^{2 +}$ $n_(Mg^(2+)) = 1 xx 1.6 * 10^(-4) = 1.6 * 10^(-4)"moles Mg"^(2+)$

and

$n_{O H^{-}}^{-} = 2 \times 1.6 \cdot 10^{- 4} = 3.2 \cdot 10^{- 4} moles$ $n_(OH^(-)) = color(red)(2) xx 1.6 * 10^(-4) = 3.2 * 10^(-4)"moles"$

Since we're working with one liter of solution, you can cay that

$[{Mg}^{2 +}] = 1.6 \cdot 10^{- 4} M$ $["Mg"^(2+)] = 1.6 * 10^(-4)"M"$

$[{OH}^{-}] = 3.2 \cdot 10^{- 4} M$ $["OH"^(-)] = 3.2 * 10^(-4)"M"$

By definition, the solubility product constant, $K_{s p}$ $K_(sp)$ , will be equal to

$K_{s p} = [{Mg}^{2 +}] \cdot {[{OH}^{-}]}^{2}$ $K_(sp) = ["Mg"^(2+)] * ["OH"^(-)]^color(red)(2)$

Plug in these values to get

$K_{s p} = 1.6 \cdot 10^{- 4} \cdot {(3.2 \cdot 10^{- 4})}^{2}$ $K_(sp) = 1.6 * 10^(-4) * (3.2 * 10^(-4))^color(red)(2)$

$K_{s p} = 1.6 \cdot 10^{- 11} \to$ $K_(sp) = color(green)(1.6 * 10^(-11)) ->$ rounded to two sig figs

The listed value for magnesium hydroxide's solubility product is $1.6 \cdot 10^{- 11}$ $1.6 * 10^(-11)$ , so this is an excellent result.

http://www.wiredchemist.com/chemistry/data/solubility-product-constants

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Solubility of $M {g (O H)}_{2}$ $Mg(OH)_2$ is $1.6$ $1.6$ x $10^{- 4}$ $10^-4$ $mol/L$ $"mol/L"$ at $298$ $298$ $K$ $K$ . What is its solubility product?

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Explanation:

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Explanation:

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Solubility of $M {g (O H)}_{2}$ $Mg(OH)_2$ is $1.6$ $1.6$ x $10^{- 4}$ $10^-4$ $mol/L$ $"mol/L"$ at $298$ $298$ $K$ $K$ . What is its solubility product?