here is a link to a similar problem
http://web.umr.edu/~gbert/Aj2.HTML?JAVA/Aksp2.HTMi still don't understand how to solve this question though
when you get it, please explain it to me
thank you
...just find out it show different question every time...have to paste it...is it fine??
The solubility product constant of barium fluoride in water at 25o C is
Ksp = 1.3 x 10-6 (1.3 E -6).
Calculate the solubility of this compound in water at 25oC.
BaF2(s) = Ba2+(aq) + 2 F-(aq)
solution:
BaF2(s) = Ba2+(aq) + 2 F-(aq)
Ksp = [Ba2+][F-]2 = 1.3 x 10-6
The solubility of BaF2 will be calculated as S moles/liter.
each molecule that dissolves will release 1 molecule of Ba2+ and 2 molecules of F-.
Therefore, in the saturated solution:
[Ba2+] = 1 S
[F-] = 2 S
Ksp = (1S )(2S)2 = (1122)S3 = 1.3 x 10-6
4 S3 = 1.3 x 10-6
S3 = (1.3/4) x 10-6
S3 = 3.3 x 10-7
S = (3.3 x 10-7)1/3
S = 6.9 x 10-3
refer University of Missouri-Rolla webpage
hope it may help you
good luck on your test!