Solubility of #Mg(OH)_2# is #1.6# x #10^-4# #"mol/L"# at #298# #K#. What is its solubility product?

1 Answer
Jan 31, 2016

Answer:

#1.6 * 10^(-11)#

Explanation:

Magnesium hydroxide, #"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+)#, and hydroxide anions, #"OH"^(-)#

#"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 * 10^(-4)#

means that you can only dissolve #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# mole of magnesium cations and #color(red)(2)# moles of hydroxide anions.

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

#n_(Mg^(2+)) = 1 xx 1.6 * 10^(-4) = 1.6 * 10^(-4)"moles Mg"^(2+)#

and

#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 * 10^(-4)"M"#

#["OH"^(-)] = 3.2 * 10^(-4)"M"#

By definition, the solubility product constant, #K_(sp)#, will be equal to

#K_(sp) = ["Mg"^(2+)] * ["OH"^(-)]^color(red)(2)#

Plug in these values to get

#K_(sp) = 1.6 * 10^(-4) * (3.2 * 10^(-4))^color(red)(2)#

#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 * 10^(-11)#, so this is an excellent result.

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