100 mL of an aqueous solution is 0.200 M in Ba(NO 3 ) 2 and 0.0500 M in Ca(NO 3
ID: 1029492 • Letter: 1
Question
100 mL of an aqueous solution is 0.200 M in Ba(NO3)2 and 0.0500 M in Ca(NO3)2. One wishes to use selective precipitation to separate the cations. To do so one adds the soluble salt
Na2SO4 (s) to the solution.
(a) Calculate the final SO42- concentration that will effect the best separation of cations.
(b) Calculate the mass of Na2SO4 required to be added to the solution to achieve this final concentration of SO42-.
(c) Calculate the fraction of the original Ba2+ ions that reamin in solution at this SO42- concentration.
Note: The Ksp was not given in the problem.
Explanation / Answer
The Ksp values are required (obtained from the internet).
Ksp (BaSO4) = 1.08*10-10
Ksp (CaSO4) = 4.93*10-5
(a) Both BaSO4 and CaSO4 are 1:1 (cation:anion) salts; hence, we can simply compare the Ksp values to predict which salt precipitates out first. Since CaSO4 has a higher solubility product, hence, CaSO4 is more soluble. Consequently, BaSO4 will precipitate out the solution first followed by CaSO4.
The separation is affected by adding solid Na2SO4 to the solution containing both Ba2+ and Ca2+ ions. When the concentration of SO42- reaches the value such that the Ksp of CaSO4 is reached, CaSO4 can no longer remain in the solution and begins to precipitate out. At all SO42- concentrations lower than this value, CaSO4 remains in solution while BaSO4 precipitates out. Therefore, in order to determine the concentration of SO42- required for best separation of the cations, we work with the Ksp of CaSO4.
Ksp (CaSO4) = [Ca2+][SO42-]
=====> 4.93*10-5 = (0.0500)*[SO42-]
=====> [SO42-] = (4.93*10-5)/(0.0500) = 9.86*10-4 M (ans).
(b) Moles of SO42- corresponding to 9.86*10-4 M SO42- = (100 mL)*(1 L/1000 mL)*(9.86*10-4 M) = 9.86*10-5 mole.
Molar mass of Na2SO4 = (2*22.989 + 1*32.065 + 4*15.999) g/mol = 142.039 g/mol.
Mass of solid Na2SO4 required = (9.86*10-5 mole)*(142.039 g/mol)*(1000 mg/1 g) = 140.050 mg (ans).
(c) Determine [Ba2+] when [SO42-] = 9.86*10-4 M. Use the Ksp for BaSO4.
Ksp (BaSO4) = [Ba2+][SO42-]
=====> 1.08*10-10 = [Ba2+]*(9.86*10-4)
=====> [Ba2+] = (1.08*10-10)/(9.86*10-4) = 1.09*10-7 M.
Fraction of Ba2+ that stays in solution = (1.09*10-7 M)/(0.200 M) = 5.45*10-7 (ans).
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