Oscar A.
asked • 03/22/20The solubility of silver(I)phosphate at a given temperature is 2 g/L. Calculate the Ksp at this temperature. After you get your answer, take the negative log and enter that (so it's like you're taking the pKsp)
The answer is 7.85
Samuel F. answered • 03/22/20
Chemical Engineer with 5+ Years of Tutoring Experience
Hello again Oscar!
First, we need to convert the solubility for mol/L. The molecular weight of this compound is ( g/mol. So the solubility is:
2/418.58 = 4.78 .10-3 mol/L
The equilibrium is the following:
Ag3PO4 ↔ 3Ag+ + PO43-
The Ksp = [Ag+]3*[PO43-] = (3S)3*(S) = 27S4 ; where S is the solubility in mol/L
Ksp = 1.4 . 10-8
-log Ksp = 7.85
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Samuel F.
The "concentration" of silver phosphate would be the molar density, that can be understood as the density divided by the molar weight. Neither change when the compound is dissolved. In more advanced levels of Chemistry, we use the "activity" instead of the concentration, which is a concept that might be worth taking a look.03/22/20