How do you calculate the solubility (moles/L) of Mg(OH)2 in water.The Ksp=#5.0*10^-11#?

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Answer 1 · 2017-04-23T10:32:51

We interrogate the equilibrium..........and get,

#S=2.32xx10^-4*mol*L^-1#

We interrogate the equilibrium..........

#Mg(OH)_2(s) rightleftharpoons Mg^(2+) + 2HO^(-)#

And #K_"sp"=[Mg^(2+)][""^(-)OH]^2#

And now if we call the solubility of #Mg(OH)_2# #S#, then by definition, #S=[Mg^(2+)]#, and #2S=""^(-)OH#.

And thus #K_"sp"=Sxx(2S)^2=4S^3#

And so (finally!), #S=""^(3)sqrt(K_"sp"/4)#

#S=""^(3)sqrt((5.0xx10^-11)/4)=2.32xx10^-4*mol*L^-1#

And of course we can convert this to a #"mass solubility"#, by multiplying #S# by the molar mass..........i.e.

#58.32*g*mol^-1xx2.32xx10^-4*mol*L^-1=1.35xx10^-2*g*L^-1.#

What is the #pH# of a saturated solution? This is a common follow up question.