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Topic: solubility pproduct - ammount of precipitate  (Read 5110 times)

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Offline meenu

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solubility pproduct - ammount of precipitate
« on: November 12, 2008, 01:59:35 AM »
 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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Re: solubility pproduct - ammount of precipitate
« Reply #1 on: November 12, 2008, 02:53:26 AM »
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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Offline meenu

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Re: solubility pproduct - ammount of precipitate
« Reply #2 on: November 12, 2008, 05:20:09 AM »
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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Re: solubility pproduct - ammount of precipitate
« Reply #3 on: November 12, 2008, 09:32:38 AM »
Your approach and answer both look correct, no idea why there is no correct answer between these given.
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