This is a Ksp problem with a common ion add on.
Two equilibria. Sr(NO3)2 is a strong electrolyte and ionizes 100% in solution.
..........Sr(NO3)2 ==> Sr^2+ + 2NO3^-
I..........0.100........0........0
C.........-0.100.......0.100...0.200
E...........0..........0.100...0.200
SrF2 is a slightly soluble electrolyte and only dissolves partially. Ksp is in charge.
.........SrF2 --> Sr^2+ + 2F^-
I........solid....0.......0
C........solid....x.......2x
E........solid....x.......2x
so Ksp expression is
Ksp = (Sr^2+)(F^-)^2
Substitute as follows:
(Sr^2+) = x from the SrF2 and 0.1 for the Sr(NO3)2 salt = x+0.1
(F^-) = 2x
and solve for x.
Note: It isn't necessary to solve a full quadratic equation since x + 0.1 for all practical purposes is 0.1.
calculate the solubility of SrF2 (ksp=7.9E-10) in 0.100 M Sr(NO3)2 ?
The answer is 4.44E-5 M how do you get that?
I Have no clue?
4 answers
what is the final answer?
i still need help.
Follow up on your later post but the answer really is 4.44E-5M. You just didn't set it up right. Probably didn't square F^- OR didn't include the 2 which squared is 4.