A solution contains 0.050 M Ba 2+ and 0.050 MCa 2+ . A solution containing fluor

Question

A solution contains 0.050 M Ba2+ and 0.050 MCa2+. A solution containing fluoride ions is added toselectively precipitate one of the ions. At what concentration offluoride ions will a precipitate begin to form? What is theidentity of the precipitate? Ksp(BaF2) = 2.45 x 10-5 Ksp(CaF2) = 1.46 x 10-10 A solution contains 0.050 M Ba2+ and 0.050 MCa2+. A solution containing fluoride ions is added toselectively precipitate one of the ions. At what concentration offluoride ions will a precipitate begin to form? What is theidentity of the precipitate? Ksp(BaF2) = 2.45 x 10-5 Ksp(CaF2) = 1.46 x 10-10 Ksp(CaF2) = 1.46 x 10-10

Explanation / Answer

Therefore we have Ksp of BaF2 = 2.45 x10-5       We know that Ksp =[Ba+2][F-]2       We know that Ksp =[Ba+2][F-]2               2.45 x 10-5 = 0.50*][F-]2                    Thus      [F-]  = 0.007 M This is the minium concentration of the chromate ion tocause precipitation. For (CaF2                             Ksp = [F-]2[Ca+2]                        1.46 x 10-10=0.50 ][F-]2                           [F-]2   = 1.70*10-5M This is the minium concentration ti form theprecipitate. By comparing these two concentration we conclude thatthe minium concentration to form the precipitate insolution is 1.70*10-5 M                             Ksp = [F-]2[Ca+2]                        1.46 x 10-10=0.50 ][F-]2                           [F-]2   = 1.70*10-5M This is the minium concentration ti form theprecipitate. By comparing these two concentration we conclude thatthe minium concentration to form the precipitate insolution is 1.70*10-5 M

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