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?
what is direct and derived measurement??
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
Well, let's find out the final concentration of barium ions, shall we? Ba2+ ions come from BaCl2, so we need to calculate the moles of Ba2+ in 100 mL of 0.1 M BaCl2.
100 mL is 0.1 L. So, moles of Ba2+ = concentration (M) × volume (L) = 0.1 M × 0.1 L = 0.01 moles.
Now, we mix this with 100 mL of 0.05 M H2SO4. Since the reaction between BaCl2 and H2SO4 is 1:1, the moles of Ba2+ are still 0.01.
Finally, we need to find the final volume of the solution. We mixed 100 mL of BaCl2 with 100 mL of H2SO4, so the final volume is 200 mL or 0.2 L.
So, the final concentration of barium ions (Ba2+) is moles/volume = 0.01 moles / 0.2 L = 0.05 M.
Therefore, the final concentration of barium ions (Ba2+) in the solution is 0.05 M. But hey, if you happen to find any clowns swimming in the solution, let me know—I'll try to keep them entertained!
To determine the final concentration of barium ions (Ba2+) in the solution, we can use the concept of stoichiometry and the volume of the solution.
First, let's identify the balanced chemical equation for the reaction between BaCl2 (barium chloride) and H2SO4 (sulfuric acid):
BaCl2 + H2SO4 -> BaSO4 + 2HCl
From the balanced equation, we can see that 1 mole of BaCl2 reacts with 1 mole of H2SO4 to form 1 mole of BaSO4 (barium sulfate) and 2 moles of HCl (hydrochloric acid).
Next, let's calculate the moles of BaCl2 and H2SO4 in the solution using their respective molarities (M) and volumes.
For BaCl2:
Moles = Molarity x Volume
Moles of BaCl2 = 0.1 M x 0.1 L (100 mL converted to liters) = 0.01 moles
For H2SO4:
Moles = Molarity x Volume
Moles of H2SO4 = 0.05 M x 0.1 L (100 mL converted to liters) = 0.005 moles
Since the stoichiometric ratio between BaCl2 and Ba2+ is 1:1, the number of moles of Ba2+ formed will be the same as the moles of BaCl2. Therefore, the final concentration of Ba2+ in the solution will be:
Final concentration = Moles / Volume of solution
Final concentration of Ba2+ = 0.01 moles / 0.2 L (sum of the volumes of BaCl2 and H2SO4) = 0.05 M
Therefore, the final concentration of barium ions (Ba2+) in the solution is 0.05 M.
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.