Question

Arrange the following solids in order of decreasing solubility, CaF2, K sp=4.0 × 10-11; Ag2CO3, K sp=8.1 × 10-12; Ba3(PO4)2, K sp=6.0 × 10-39; and Ag3(PO4)2, K sp=1.0 × 10-31. View Available Hint(s) Arrange the following solids in order of decreasing solubility, CaF2, K sp=4.0 × 10-11; Ag2CO3, K sp=8.1 × 10-12; Ba3(PO4)2, K sp=6.0 × 10-39; and Ag3(PO4)2, K sp=1.0 × 10-31. Ba3(PO4)2 > Ag3(PO4)2 > CaF2 > Ag2CO3 CaF2 > Ag2CO3 > Ag3(PO4)2 > Ba3(PO4)2 Ba3(PO4)2 > CaF2 > Ag3(PO4)2 > Ag2CO3 Ag2CO3 > Ba3(PO4)2 > Ag3(PO4)2 > CaF2

Answer 1

CaF2 > Ag2CO3 > Ag3(PO4)2 > Ba3(PO4)2

Step-by-step explanation:

Ksp which is solubility product konstant shows equilibrium between a solids and its respective ions in a solution. And the lower it is the less soluble the ion compound will be. And for CaF2 we have the highest konstant and for Ba3(PO4)2 we have it the lowest.