Question

It is of interest to decide if an analytical separation of the metal ions can be effected by selective precipitation of carbonates ftrom a solution that is 0.102 M im Cu? and 0.110 M in NP Kp-2.50 10-10 NiCOs Kup-6.60 10 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? When 990% of the least sol ble cat nate has precipt ated, wi all of the metal that s e more sol e carbonate sti reman (3) What is the upper limit on the carbonate ion concentration if the more soluble compound is not to precipitate? (4) Ifthe [CO,2] is at this upper limit, what percentage of the metal that forms the least solable carbonate remains in solution? solu on

Answer 1

ans)

1.

find CO3-2 required to precipitate hte least soluble

Ksp = [cu+2][CO3-2]

99.99% of cu+2 = (100-99.99) = 0.01 % left

0.01/100*(0.102) = 0.0000102 M of cu+2 left in solution

2.5*10^-10 = (0.0000102)([CO3-2]

[CO3-2] = (2.5*10^-10)/(0.0000102) = 2.45*10^-5 M

2.

find the other metal at this point

Ksp = [Ni+2][CO3-2]

(6.6*10^-9) = [Ni2+] ( 2.45*10^-5)

[Ni2+] = (6.6*10^-9) / (2.45*10^-5) = 0.000269

therefore, not all the Nickel has precipitated

3

upper limit so the soluble compound will not precipitate

Ksp =[Ni2+][CO3-2]

(6.6*10^-9) / (0.110) = [CO3-2]

[CO3-] = 6.0*10^-8 M then Ni2+ wont form preciptiate

Q4

if CO3-2 is at this value, find metal of leaast soluble

Ksp = [cu+2][CO3-2]

2.5*10^-10= [cu+2] (  6.0*10^-8)

[cu+2] = (2.5*10^-10) / (  6.0*10^-8) = 0.004166 M

% remains = (0.004166)/(0.102)*100 =4.08%