contestada

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

Respuesta :

ted7jet

Answer:

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

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.

Otras preguntas