Asked by matt
How many milligrams of magnesium palmitate will precipitate from 965ml of this solution when it is cooled from 50C to 25C? For Mg(C16H21O2)2, Ksp= 4.8x10(-12) at 50C and 3.3x10(-12) at 25C.
Answers
Answered by
DrBob222
Mg(pal)2 ==> Mg^2+ + 2pal^-
....x.........x.......2x
Ksp = (Mg^2+)(pal)^2/[Mg(pal)2]
Substitute into Ksp expression and use Ksp @ 50C to determine the solubility at 50 C. x = solubility in moles/L. Convert to grams/L, then to grams/965 mL.
Repeat the above procedure using Ksp @ 25 C and determine grams/965 mL.
Then subtract grams/965mL of the two to determine the grams that will ppt and convert to mg.
....x.........x.......2x
Ksp = (Mg^2+)(pal)^2/[Mg(pal)2]
Substitute into Ksp expression and use Ksp @ 50C to determine the solubility at 50 C. x = solubility in moles/L. Convert to grams/L, then to grams/965 mL.
Repeat the above procedure using Ksp @ 25 C and determine grams/965 mL.
Then subtract grams/965mL of the two to determine the grams that will ppt and convert to mg.
Answered by
matt
thank you!
Answered by
student
Ksp = (Mg^2+)(pal)^2/[Mg(pal)2]- expression is incorrect. mg(pal)2 is not part of the ksp expression as the reactant is in the solid state. only Ksp = [Mg^2+][pal]^2 is relevant.
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