Chemical Forums

Chemistry Forums for Students => Inorganic Chemistry Forum => Topic started by: Passing on March 06, 2010, 06:24:03 PM

Title: Solubility of magnesium hydroxide
Post by: Passing on March 06, 2010, 06:24:03 PM

MgO + H₂O  ::equil:: Mg(OH)₂
In certain (small) amounts it dissolves  ::equil:: Mg²⁺ + 2OH^-

How does the solubility change by neglecting the following equilibrium?
Mg(H₂O)_x²⁺ + OH^-  ::equil:: (Mg(H₂O)_x-1(OH)]⁺ + H₂O

Title: Re: Solubility of magnesium hydroxide
Post by: Borek on March 07, 2010, 03:58:10 AM

Only equilibria I know of are

4Mg²⁺ + 4OH^- -> Mg₄(OH)₄⁴⁺ log(K) = 16.3

Mg²⁺ + OH^- -> MgOH⁺ log(K) = 2.58

Doesn't mean there no others, these I have listed in Kotrly/Sucha handbook.