Asked by Please Help Me, I need to learn.
If a solution containing 0.015 M each of bromide, chloride, iodide, and thiocyanate, is treated with Cu+, in what order will the anions precipitate?
Answers
Answered by
DrBob222
Take CuBr as an example: The Ksp value is 6.3E-9 but make sure you use the Ksp values in your notes or your text.
................CuBr ===> Cu^+ + Br^-
I................solid...........0...........0
C..............solid............x............x
E..............solid.............x............x
Ksp = 6.3E-9 = (Cu^+)(Br^-)
(Cu^+) = x
(Br^-) = 0.15 M from the problem.
Substitute
6.3E-9 = (x)(0.15)..
Then x = (Cu^+) = 6.3E-9/0.15 = ?
Do the same for all of the salts. The one with the smallest value for Cu will ppt first followed in order by the others. Post your work if you get stuck.
................CuBr ===> Cu^+ + Br^-
I................solid...........0...........0
C..............solid............x............x
E..............solid.............x............x
Ksp = 6.3E-9 = (Cu^+)(Br^-)
(Cu^+) = x
(Br^-) = 0.15 M from the problem.
Substitute
6.3E-9 = (x)(0.15)..
Then x = (Cu^+) = 6.3E-9/0.15 = ?
Do the same for all of the salts. The one with the smallest value for Cu will ppt first followed in order by the others. Post your work if you get stuck.
Answered by
Please Help Me, I need to learn.
Thank you so much for your help.