Question

calculate the concentration of ions present after the reaction between 13.50mL of 2.00M BaCl2 and 2.50mL of 6.00M H2SO4

i just wanted to check my answers
[Ba^2+]=1.69M
[SO4^2-]=o.938M
[H^+]=0
[Cl^-]=1.50M

if they are wrong please explain

Answer 1

I don't understand how you obtained any of those answers.
BaCl2 + H2SO4 ==> BaSO4 + 2HCl

Here is how I see it.
You had 13.5 x 2 = 27.0 mmoles BaCl2 initially.
You had 2.50 x 6.00 M H2SO4 = 15 mmoles initially. I would construct an ICE table.
........BaCl2 + H2SO4 ==> BaSO4 + 2HCl
initial..27.0...15.0.......0........0
change...-15.0...-15.0....+15.0..+30.0
final....12.0....0.........15.0...30.0

So H^+ = 30 mmols/total volume which looks like 16 mL.
Cl is 54/16 (30 from HCl and 24 from the unused BaCl2)
Ba^+ is 12/16 = ??
H2SO4 = 0
BaSO4 is insoluble; but you can calculate the concn of Ba^+2 and SO4^- form Ksp but remember you have a common ion (Ba^+) with a concn of 12/16.