- Joined
- Feb 11, 2009
- Messages
- 230
- Reaction score
- 0
I had this problem in my Kaplan study guide:
How does the addition of a small amount of ZnF2 to a saturated solution of PbF2 affect this solution?
A. ZnF2 does not dissolve and PbF2 remains in solution
B. ZnF2 dissolves and PbF2 precipitates.
C. ZnF2 dissolves PbF2 remains in solution
D. ZnF2 does not dissolve PbF2 precipitates
The answer is A, but in my ExamKrackers book it states:
A common ion added to a saturated solution will shift the equilibrium increasing precipitate. A common ion added to a solution that is not saturated will NOT shift the equilibrium, because in an unsaturated solution, there is no equilibrium to shift.
So, why isn't the answer B? wouldn't the ZnF2 add F ions to the solution, thus forcing PbF2 to precipitate?
The passage even states that ZnF2 is more soluble than PbF2.
How does the addition of a small amount of ZnF2 to a saturated solution of PbF2 affect this solution?
A. ZnF2 does not dissolve and PbF2 remains in solution
B. ZnF2 dissolves and PbF2 precipitates.
C. ZnF2 dissolves PbF2 remains in solution
D. ZnF2 does not dissolve PbF2 precipitates
The answer is A, but in my ExamKrackers book it states:
A common ion added to a saturated solution will shift the equilibrium increasing precipitate. A common ion added to a solution that is not saturated will NOT shift the equilibrium, because in an unsaturated solution, there is no equilibrium to shift.
So, why isn't the answer B? wouldn't the ZnF2 add F ions to the solution, thus forcing PbF2 to precipitate?
The passage even states that ZnF2 is more soluble than PbF2.
Last edited: