Part A Learning Goal To learn how to calculate the solubility from Ksp and vice

Question

Part A Learning Goal To learn how to calculate the solubility from Ksp and vice versa. A saturated solution of magnesium fuoride, MgF2, was prepared by dissolving solid MgF2 in water. The concentration of Mg2+ ion in the solution was found to be 1.18x103 M.Calculate Ksp for MgF2 Express your answer numerically. Consider the following equilibrium between a solid salt and its dissolved form (ions) in a saturated solution: CaF2 (s)Ca2+(aq)2F (aq) View Available Hint(s) At equilibriurm, the ion concentrations remain constant because the rate of dissolution of solid CaF2 equals the rate of the ion crystallization. The equilibrium constant for the dissolution reaction is Ksp is called the solubility product and can be determined experimentally by measuring the solubility, which is the amount of compound that dissolves per unit volume of saturated solution. Submit Part B The value of Kw, for silver sulfate, Ag SO4, is 1.20 x 10-5. Calculate the solubility of Ag2SO4 in grams per lter. Express your answer numerically in grams per liter View Available Hint(s) solubility g/L

Explanation / Answer

part A)

concentration of Mg+2 = 1.18 x 10^-3 M

concentration of F- 2.36 x 10^-3

MgF2   --------------> Mg+2 +    2 F-

Ksp = [Mg+2][F-]^2

      = (1.18 x 10^-3) (2.36 x 10^-3)^2

Ksp = 6.57 x 10^-9

part B)

Ag2SO4 --------------> 2 Ag+   +   SO42-

                                       2 S             S

Ksp = (2S)^2 x S

1.20 x 10^-5 = 4 S^3

S = 0.0144 M

solubility = 4.50 g / L

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