Solubility of #Mg(OH)_2# is #1.6# x #10^-4# #"mol/L"# at #298# #K#. What is its solubility product?
1 Answer
Explanation:
Magnesium hydroxide,
Magnesium hydroxide dissociates only partially to form magnesium cations,
#"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
Take a look at the dissociation equilibrium. Notice that every mole of magnesium hydroxide that dissociates produces
This tells you that if you successfully dissolve
#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) = ["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
http://www.wiredchemist.com/chemistry/data/solubility-product-constants