posted by guest .
How much water at 25 degrees celsius is needed to dissolve 10 grams of baruim phosphate?
Ba3(PO4)2 ==> 3Ba^+2 + 2PO4^-3
Set up Ksp chart, find solubility from Ksp which will be in moles/L. Convert that to grams to give you g/L, then determine the amount of water necessaary to dissolve 10 g.
the Ksp of barium phosphate is 1.3996*10^-8 how much water would you need to to dissolve this amount of baruim phospahte?
the molar solubilty of baruim phosphate is 8.42335*10^-6 g/L
what formula would i use to figure out the water?
Don't you know how to calculate the solubility of Ba3(PO4)2?
Ba3(PO4)2 ==> 3Ba^+3 + 2PO4^-3
Ksp = (Ba^+2)^3(PO4^-3)^2
If you let S = solubility of Ba3(PO4)2, then (Ba^+2) = 3S and (PO4^-3) = 2S.
Substitute 3S and 2S in the Ksp expression to obtain
(3S)^3(2S)^2 = Ksp.
Solve for S which tells you the solubility of the compound in moles/L.
Convert that to grams/L THEN you can work on how much water would be necessary to dissolve 10 g. I didn't work it out but I expect a bunch. By the way, post your work if you get stuck. This much is a freebie for you.