What is the finaly concentration of barium ions, Ba2+, in soulution with 100 ml of .1 M BaCl2 is mixed with 100ml of .05M H2SO4?

Answer 1

BaCl2 + H2SO4 ==> BaSO4 + 2HCl

BaCl2 = 0.1 M x 0.1 L = 0.01 mols
H2SO4 = 0.05 M x 0.1 L = 0.005 mols.

Therefore, all of the H2SO4 will react, 1/2 of the BaCl2 will react.
Mols BaCl2 at end of reaction = 0.01 - 0.005 = 0.005 mols.

Mols H2SO4 at end of reaction = 0.005 - 0.005 = 0

Mols BaSO4 formed = 0.005.
Mols HCl formed = 2 x 0.005 = 0.01.

So you have two sources for Ba^+2. One is the mols Ba^+2 from BaCl2 and the other is any Ba^+2 from BaSO4. But BaSO4 is quite insoluble and the amount of Ba^+2 contributed by BaSO4 is quite small and we can neglect it. So Ba^+2 from BaCl2 is simply mols/L = 0.005/total volume. total volume is 100 mL + 100 mL or 200 mL = 0.2 L. You do the math.

Answer 2

what is direct and derived measurement??

Answer 3

I'm not sure of your question. We have direct measurements in the SI system and derived units. Here is a link.
http://physics.nist.gov/cuu/Units/units.html