In: Chemistry
It is of interest to decide if an analytical separation of the metal ions can be effected by selective precipitation of carbonates from a solution that is 9.88×10-2 M in Fe2+ and 0.106 M in Ni2+.
NiCO3 | Ksp = 6.60×10-9 |
FeCO3 | Ksp = 3.50×10-11 |
To analyze this problem, answer the following questions.
(1) What carbonate concentration is needed to precipitate 99.9% of
the metal that forms the least soluble carbonate?
M
(2) When 99.9% of the least soluble carbonate has precipitated,
will all of the metal that forms the more soluble carbonate still
remain in solution?
(3) What is the upper limit on the carbonate ion concentration if
the more soluble compound is not to precipitate? M
(4) If the [CO32-] is at
this upper limit, what percentage of the metal that forms the least
soluble carbonate remains in solution? %
Leas tsoluble = FeCO3, sicne it has lower Ksp
Q1
find CO3-2 required to precipitate hte least soluble
Ksp = [Fe+2][CO3-2]
99.99% of Fe+2 = (100-99.99) = 0.01 % left
0.01/100*(9.88*10^-2) = 0.00000988 M of Fe+2 left in solution
3.5*10^-11 = (0.00000988)([CO3-2]
[CO3-2] = (3.5*10^-11)/(0.00000988) = 0.000003542 M = 3.542*10^-6 M
Q2
find the other metal at this point
Ksp = [Ni+2][CO3-2]
(6.6*10^-9) = [Ni2+] ( 3.542*10^-6)
[Ni2+] = (6.6*10^-9) / (3.542*10^-6) = 0.0018633
therefore, not all the Nickel has precipitated
Q3
upper limit so the soluble compound will not precipitate
Ksp =[Ni2+][CO3-2]
(6.6*10^-9) / (0.106) = [CO3-2]
[CO3-] = 6.2264*10^-8 M then Ni2+ wont form preciptiate
Q4
if CO3-2 is at this value, find metal of leaast soluble
Ksp = [Fe+2][CO3-2]
3.5*10^-11 = [Fe+2] ( 6.2264*10^-8)
[Fe+2] = (3.5*10^-11) / ( 6.2264*10^-8) = 0.0005621 M
% remains = (0.0005621)/(9.88*10^-2)*100 =0.568 %