Question

Suppose 5.87 g of barium acetate is dissolved in 300. mL of a 71.0 m M aqueous solution of sodium chromate.

calculate the final molarity of barium cation in the solution. You can assume the volume of the solution doesn't change when the barium acetate is dissolved in it.

Round your answer to 2 significant digits.

Answer 1

Barium acetate = Ba(C2H3O2) = Ba(Ac)2
Ba(Ac)2 + Na2CrO4 ==> BaCrO4 + 2NaAc
mols Ba(Ac)2 initially = grams/molar mass
mols Na2CrO4 initially = 0.2L x 0.071 M = ?(I assume 71.0 m M is 71 millimolar.) and that makes all of the Na2CrO4 used.
mols Ba(Ac)2 remaining = inital Ba(Ac)2 - mols Na2C4O4 = ?
Then M (Ba^2+) = mols/L soln. L soln = 0.300 + 0.071 = ?
Then remember 2 s.f.
Again, you can add what little Ba^2+ comes from the BaCrO4 ppt (use Ksp and the common ion effect for this) but I think this is insignificant and can be neglected.

Answer 2

@ DrBob222

I know my ions are the same as my moles but I am having trouble understanding what I do next. Can you explain how you broke down this problem? What steps am I missing? Thank you!