[TaggerFeld]

Hi everyone. I am trying to solve this problem. I keep having trouble with it. I have written down the solubility reactions and Ksp, but still unable to figure this out. The original question is:"How many Fe++ can be present in H2O or water containing 2 x 1e-02 M of HCO3- w/out directly causing the precipitation of FeCO3 which has a Ksp value of =10^(-10.7)? Thanks!

[Borek]

You first step should be calculation of pOH of HCO₃^- solution.

http://www.chembuddy.com/?left=pH-calculation&right=pH-amphiprotic-salt

[TaggerFeld]

I had thought something similar to this and tried, but did not work. My reasoning:

First write the solubility reaction:
                   FeCO3 <=> Fe(2+) + CO3(2-)
initial (M)      0                 0            2*e-2
Equilibrium    x                 x            2*e-2 + x

The relationship for Ksp is:
Ksp=[Fe(2+)]*[CO3(2-)]=x(2e-2+x)=10**-10.7=2e-11.
or x^2 + 2xe-2 -2e-11=0
The solution to this equation is x=10^-9 (M)

So the maximum concentration of Fe2+ in order to avoid precipitation is 10^-9 (M), but it is wrong.

[Borek]

Oops, sorry - my first take was that you should check if Fe(OH)₂ doesn't precipitate from the solution (that's why pOH), but it asks specifically about carbonate...

So you have to find out what is the concentration of CO₃^2- in the HCO₃^- solution. Still acid/base equilibrium question. Then plug this CO₃^2- into Kso to calculate Fe²⁺ concentration. Probably no need for ICE table, but you will have to compare concentrations to see how much carbonates were used by precipitation.