Suppose 3.50g of zinc bromide is dissolved in 300.mL of a 65.0mM aqueous solution of potassium carbonate. Calculate the final molarity of bromide anion in the solution. You can assume the volume of the solution doesn't change when the zinc bromide is dissolved in it.

Answer 1

ZnBr2 + K2CO3 ==> ZnCO3(s + 2KBr.
Note that all of the Br2 stays in solution so just assume there is no reaction.
Final ZnBr2 (if no ZnCO3 fomeed) is 3.50 g ZnBr2/molar mass ZnBr2 = mols ZnBr2.
Then (ZnBr2) = mols ZnBr2/L solution = mols ZnBar2/0.300 L = ?
Then final (Br^-) is twice that of ZnBr2.