How much Fe++ could be present in water containing 2 x 1e-02 M HCO3-
(bicarbonate) without causing the precipitation of FeCO3 (Ksp=10^(-10.7))
Hint: carbonate eq
Ka2=4.7e-11

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Clarification of Question by hohum-ga on 24 Jul 2006 02:04 PDT

Pleeease answer soon!! thnxs :)

The following could be helpful to you:
Fe   ----------> Fe++ + 2e- (Anodic reaction) 

H2CO3 + e- ----> HCO3- + H (Cathodic reaction) 

Fe + 2H2CO3 ---> Fe++ + 2 HCO3- + H2

The build up of the bicarbonate ion can lead to an increase in the pH
of the solution till conditions promoting precipitation of iron
carbonate are reached, leading to reaction given below:

Fe + 2HCO3- ---> FeCO3+ H2O+CO2 


Regards,
livioflores-ga

In the first place we have the balanced equation:
Fe++ + HCO3- <-> FeCO3(s)   (equation 1)
Then we must look at the solubility of the bicarbonate in the water:
2HCO3- + 2H2O <-> 2CO3-- + 2H3O+      (equation 2)

We know the solubility constant of equation 1, and we know the Ka of
equation 2. We also know the concentration of HCO3-, therefore we can
work backwards and find the minimum amount of Fe++ needed to
precipitate FeCO3. Thus, anything below this concentration will be
your answer.

Ka2 = [CO3--]^2 / [HCO3-]^2 = 4.7e-11 = [CO3--]^2 / [2x10^-2 M]

Thus [CO3--] = sqrt( [4.7x10^-11] x [2x10^-2 M] ) = 9.70x10^-7 M

Now, using the Ksp we get the equation: Ksp = [Fe++][CO3--] / 1  (1
because pure compounds are counted as 1 in this form)

Ksp = 10^-10.7 = [Fe++]x(9.70x10^-7 M)
[Fe++] = (10^-10.7) / (9.70x10^-7 M) = 2.06x10^-5 M. 

Therefore, if your concentration of Fe++ is less than this value, the
FeCO3 precipitate will not form. If it is above, then you will have a
precipitate formed.

Hope this helps.

michalski-ga