A student attempts to dissolve 0.25 grams of MgF2 in 0.20 L of water. With this

Question

A student attempts to dissolve 0.25 grams of MgF2 in 0.20 L of water. With this saturated solution, it was determined that [F-] = 2.4 x10^-3 M
(a) write the dissociation equation for the ionization of MgF2
(b) Given above [F-] , what is [Mg2+] and what is the molar solubility for MgF2?
(c) What is the value of Ksp for MgF2?
(d) Doe it appear that all of the solid MgF2 has dissolved into solution? Explain your reasoning?
(e) As a comparison, this student attempts the same procedures for BaF2, whose Ksp is 1.8x10^-6. Calculate the molar solubility in mole/L
(f) Describe whether 0.25 g of BaF2 would completely dissolve in 0.20L of water, and then comment on the overall solubility of BaF2 versus MgF2 under the same conditions.
(g) A 0.50 L solution contains a mixture of both Mg2+ and Ba2+ of equivalent 0.20 M concentrations. If NaF solution was added, which precipitate would appear first, and why?

Explanation / Answer

       S = [Mg2+] = [F-]/2 = 1.2x10-3 M

c.   Ksp = [Mg2+][F-]2 = Sx(2S)2 = 4S3 = 6.9x10-9

d. 0.25g MgF2/62.3g/mol/0.20L = 2x10-2 M that is > S

The dissolution was incomplete.

e. Ksp = 4S3          S = (Ksp/4)1/3 = 7.7x10-3 M          

f. 0.25 g of BaF2 will completely dissolve in 0.200 mL:

0.25 g of BaF2 / 175.34g/mol /0.200 L = 7.13 mol/L < S

g. MgF2 is less soluble (see S or Ksp values) and will precipitate first.

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