Asked by corey
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?
Answers
Answered by
DrBob222
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.
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.
Answered by
HI
what is direct and derived measurement??
Answered by
DrBob222
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
http://physics.nist.gov/cuu/Units/units.html