We have a water supply of neutral pH with lead Pb2+ and magnesium Mg2+ in the wa

Question

We have a water supply of neutral pH with lead Pb2+ and magnesium Mg2+ in the water supply. The concentration of the Pb2+ is .010M and the concentration of the Mg2+ is .010M. Using sodium carbonate Na2CO3 determine if it is possible to precipitate 99.99% of the Pb2+ and leave the Mg2+ behind.

PbCO3(s) <----> Pb2+      +         CO32-          Ksp = 7.4 X 10-14

MgCO3(s) <----> Mg2+         +     CO32-          Ksp =   3.5 X 10-8

7.4 *10-4 = [1*10-6][CO32-]

100% - 99% =.01%

0.010M PB *.0001= 1*10-6

7.4*10-14/1*10-6    = [CO32-]=7.4*10-8M

Q=[.010][7.4*10-8]

KSP less than were good

Explanation / Answer

MgCO3

Ksp = [Mg2+][CO3^2-]

[CO3^2-] = 3.5 x 10^-8/0.1 = 3.5 x 10^-7 M

This is the carbonate ion concentration required to precipitate Mg2+ ion from solution

Calculate the [Pb2+] remaining in solution at this CO3^2- concentration when Mg2+ has begun precipitation

PbCO3

Ksp = [Pb2+][CO3^2-]

[Pb2+] = 7.4 x 10^-14/3.5 x 10^-7 = 2.11 x 10^-7 M

Percent [Pb2+] precipitated from solution = 100 - [(2.11 x 10^-7/0.1) x 100] = 99.999%

So it is possible to preipitate 99.999% Pb2+ from solution leaving behind Mg2+ ion.

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