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Solutions

Question
CBSEENCH12010989

An aqueous solution contains an unknown concentration of Ba2+. When 50 mL of a 1 M solution of Na2SO4 is added, BaSO4 just begins to precipitate. The final volume is 500 mL. The solubility product of BaSO4 is 1 x 10–10. What is the original concentration of Ba2+?

  • 1.0 x 10–10 M

  • 5 x 10–9 M

  • 2x10–9 M

  • 1.1 x 10–9M

Solution

D.

1.1 x 10–9M

1 M Na2SO450 ml  +  Ba+2 450 ml Final Solution500 ml   Concentration of SO4-2 in Ba+2 solutionM1V1 = M2V21  x 50 = M2 x 500M2 = 110for just precipitationI.P = Ksp[Ba+2] [SO42-] = Ksp(BaSO4)[Ba+2] x110 = 10-10[Ba+2] = 10-9 M in 500 ml Solution 

For the calculation of [Ba+2] in original solution (450 ml)

M1 x 450 = 10-9 x 500M1 = 500450 x 10-9 = 1.11 x 10-9 M[M1 = Molarity of Ba+2 in original solution (450 ml)]