............BaCO3 ==> Ba^2+ + CO3^2-I
I............solid.....0.......0
C............solid.....x.......x
E............solid.....x.......x
Ksp =(Na^2+)(CO3^2-)
.......Na2CO3 --> 2Na^+ + CO3^2-
I.......0.05M......0.......0
C......-0.05......0.1.....0.05
E.........0.......0.1.....0.05
Substitute ito the Ksp equation.
(Ba^2+) = x
(CO3^2-) = x + 0.05
Solve for x = (BaCO3) in mols/L = M
That gives you mols/L. Convert to mols/2.5 L, then convert to grams. grams = mols x molar mass.
what mass of barium carbonate will dissolve in a 2.50L solutions of 0.0500 molar sodium carbonate
1 answer