Asked by Matt M
What is the molarity of a salt solution that is made from 31.0 grams of Ca3(P04)2 placed in a volumetric flask and filled to the 2 liter line with distilled water?
Answers
Answered by
DrBob222
I don't know if this is a REAL problem or not? If it is just a made up problem by someone who has slipped up, the answer (not real) is
mols = grams/molar mass, then
M = mols/2L = ?
However, if you recognize that Ca3(PO4)2 is an insoluble salt in H2O, then you get a REAL answer this way.
........Ca3(PO4)2 ==> 3Ca^2+ + 2PO4^3-
I........solid.........0........0
C........solid.........3x.......2x
E........solid.........3x.......2x
Ksp = (Ca^2+)^3(PO4^3-)^2
Ksp = 2.07E-33
2.07E-33 = (3x)^3(2x)^2
2.07E-33 = 27x^3*4x^2 = 108x^5
Solve for x.
x is the REAL molarity of Ca3(PO4)2 in water solution.
mols = grams/molar mass, then
M = mols/2L = ?
However, if you recognize that Ca3(PO4)2 is an insoluble salt in H2O, then you get a REAL answer this way.
........Ca3(PO4)2 ==> 3Ca^2+ + 2PO4^3-
I........solid.........0........0
C........solid.........3x.......2x
E........solid.........3x.......2x
Ksp = (Ca^2+)^3(PO4^3-)^2
Ksp = 2.07E-33
2.07E-33 = (3x)^3(2x)^2
2.07E-33 = 27x^3*4x^2 = 108x^5
Solve for x.
x is the REAL molarity of Ca3(PO4)2 in water solution.
Answered by
bob
15.5 M
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