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How would you calculate the molar solubility of Mn(OH)2 (Ksp = 1.6e-13) in pure water?

1 Answer

anor277 · Stefan V.

Nov 11, 2015

#Mn(OH)_2(s) rightleftharpoons Mn^(2+) + 2OH^-#

#K_(sp) = 1.6xx10^(-13)#

Explanation:

#K_(sp) = 1.6xx10^(-13)=[Mn^(2+)][OH^-]^2#

If we call the solubility of manganese(II) hydroxide #S#, then by the stoichiometry of the reaction:

#K_(sp)=S(2S)^2=4S^3#.

So #S = root(3)((1.6 xx 10^(-13))/4)#

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