Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: meenu on November 12, 2008, 01:59:35 AM
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You have a saturated solution of Mg ( C16H31O2)2 at 50 C
How many milligrams of Magnesiuum palmitate Mg ( C16H31O2)2 will precipitate from 965 ml of this solution when it is cooled at 25 C .
Ksp at 50 C = 4.8 x 10^ -12
Ksp at 25 C = 3.3 x 10^ -12
answer options (mg) 0.1 , 1000 , 100 , 10
what im doing is
Ksp = [Mg+] [2 P]^2 P stands for ( C16H31O2 - )
finding the concentration of ions in mols per litre at 50 c and at 25 C
and then subtracting . After that converting concentration in milligrams.
molar mass = 534
???
im getting
0.05674 gram / litre - 0.0500 gram / litr
0.00674 gram / litre
for 0.965 litre solution we have 0.00650 grams or 6.5 mg
but not getting the correct answer ??? ???
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finding the concentration of ions in mols per litre at 50 c and at 25 C
and then subtracting . After that converting concentration in milligrams.
Show how you do this part for 50 deg C.
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Mg(P)2 :rarrow: Mg + + 2 P-
:larrow:
s mol s mol 2s mol
Ksp = [Mg] [2 P]2
let [Mg]= s mol L-1
[ P] = 2s mol L-1
4.8 x 10-12 = (s) (2s) 2
4.8 x 10-12 = 4 s3
s 3= 4.8 x 10-12) / 4
s = 1.06 x 10-4 mol L-1
s = 1.06 x 10-4 x 534 grams L-1
s =0.05674 grams L-1
same thing i did for concentration at 25 C
but after subtracting two concentration im not getting correct value of the precipitate ??? ???
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Your approach and answer both look correct, no idea why there is no correct answer between these given.