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A solution is prepared by mixing 150.0 mL of 0.200 M Mg2+ and 250.0 mL of...

A solution is prepared by mixing 150.0 mL of 0.200 M Mg2+ and 250.0 mL of 0.100 M F-. A precipitate forms. Calculate the concentrations of Mg2+ and F- at equilibrium. Ksp = 6.4 × 10-9

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Expert Solution

150mL of 0.2M magnesium ion solution will contain (0.2/1000)x150 = 0.03moles of the metal ions. Similarly, 250mL of 0.1M fluoride ions contain 0.025moles of fluoride ions. The expression for Ksp is given as Ksp = [Mg2+][2F-]2. Now, it is given that Ksp = 6.4x10-9 implying that 6.4x10-9 = [x][2x]2 where x is the quantity of magnesium fluoride dissolved. Solving for x, we get 4x3 = Ksp and x = cube root of (6.4x10-9)/4 = 1.1696x10-3 giving the concentrations of magnesium to be 1.1696x10-3M and that of fluoride to be twice as much, which amounts to 2.3392x10-3M.

Thus, the concentrations of magnesium cations at equilibrium will be 1.1696x10-3M and that of fluoride will be 2.3392x10-3M.

NOTE: Since the concentration of ions put into the solution is quite high compared to its solubility and the need for quantity of precipitate in grams is not asked, this calculation can be done even without finding the no.of moles of the anion.

queen_honey_blossom answered 2 years ago
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