It is of interest to decide if an analytical separation of the metal ions can be

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

Explanation / Answer

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%

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