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For the reaction MgF2(s) <=> Mg2+(aq) + 2 F-(aq), at a given temperature, K = 7.4...

For the reaction MgF2(s) <=> Mg2+(aq) + 2 F-(aq), at a given temperature, K = 7.4 x 10-11. What MASS (you’ll need to calculate molarity!) of MgF2 could be added to 2L of water, such that ALL the MgF2 dissolves? How soluble is MgF2 in water (qualitative answer…ie, VERY, NOT VERY, etc).

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Answer #1

at equilibirum [Mg2+] = S , [F-] = 2S where S is solubility of MgF2

K = [Mg2+] [F-]^2

7.4 x 10^ -11 = (S) (2S)^2

7.4 x 10^ -11 = 4S^3

S = 2.64 x 10^ - 4 M

volume of solution = 2L

Moles of MgF2 needed = Molarity of soluble MgF2 x solution volume in L

= 2.64 x 10^ -4 M x 2L = 0.000529 mol

Mass of MgF2 = moles of MgF2 x molar mass of MgF2

= 0.000529 mol x 62.3 g/mol

= 0.033 g

Thus 0.033 grams of MgF2 when dissolved in 2L water then all MgF2 dissolves

MgF2 is Not Very soluble in water

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