Part A & B is needed. There is no total volume, this is the question straight fr

Question

Part A & B is needed. There is no total volume, this is the question straight from the paper.

A solution contains 0.25 M NaF and 0.14 M K_2SO_4. Ba(NO_3)_2 is added slowly to the solution to gradually increase the concentration of barium ions. You want to precipitate the maximum amount of sulfate (as BaSO_4), but not allow any fluoride to precipitate. What is ideal [Ba^2+] in the solution in order to accomplish this? At the [Ba^2+] you calculated for part a, what percentage of the original sulfate ion will remain in the solution?

Explanation / Answer

a) For precipitation to occur

Qsp > Ksp precepitstion will occur

Qsp < Ksp dissution continue

Qsp = Ksp solution is saturated

Where , Ksp is solubility product and Qsp is actual product of ions

Ksp for Barium sulphate = 1.08 × 10^-10

Ksp for Barium flouride = 1.84 × 10^-7

NaF is 0.25M and K2SO4 is 0.14M

Therefore, F^- concentraion is 0.25M and SO4^2- concentration is 0.14M SO4^2-

For BaF to precipitate Qsp should be > 1.84 × 10^-7.

Ksp = [Ba^2+] [ F^-] at precipitation

So , [Ba^2+] = 1.84 × 10^-7/0.25 =7.36 × 10^-7M

For BaSO4 to be precipitate Qsp of Ba^2+ and SO4^2- should be greater then its Ksp 1.08×10^-10

Ksp = [Ba^2+] [SO4^2-] at precipitation

Therefore , [ Ba^2+ ] = 1.08 × 10^-10/0.14 = 7.714 × 10^-10M

So, concentration Ba^2+ , [ Ba^2+] should be maintained between 7.714 × 10^-10M and 7.36 × 10^-7 M for precipitation of Sulphate

For maximum precipitation of BaSO4 , the concentrayion of Ba^2+ , [Ba^2+] = 7.36× 10 ^-7M

b) 1mol of Ba^2+ react with 1mol of SO4^2- to form BaSO4

Therefore, 7.76 × 10^-7M Ba^2+ react with 7.76 × 10^-7M of SO4^2-

Therefore, SO4^2- remaining is = 99.9994%

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---