Duplicate Question
The question on this page has been marked as a duplicate question.
Original Question
You have a 250ml solution of 0.13M calcium iodate. How many grams of potassium iodate need to be added to the solution to begin...Asked by anonymous
You have a 250ml solution of 0.13M calcium iodate. How many grams of potassium iodate need to be added to the solution to begin to precipitate calcium iodate? The solubility product constant for calcium iodate is6.47 *10^-6.
B) you have a 250ml solution of 0.13 M calcium iodate. How many total grams need to be added to the solution until it is saturated in potassium iodate? The ksp for potassium Iodate is 0.183.
pls help me with this problem
B) you have a 250ml solution of 0.13 M calcium iodate. How many total grams need to be added to the solution until it is saturated in potassium iodate? The ksp for potassium Iodate is 0.183.
pls help me with this problem
Answers
Answered by
DrBob222
Neither problem makes any sense to me. With Ksp = 6.47E-6, the solubility of Ca(IO3)2 is 0.0117 in a saturated solution. Therefore, you can't have a 0.13M solution. You don't need to add any KIO3; Ca(IO3)2 is already pptd.