How much Fe++ could be present in water containing 0.8 x 1e-02 M HCO3-
(bicarbonate) without causing the precipitation of FeCO3 (Ksp=10**(-10.7))
To determine the maximum amount of Fe++ that can be present in water without causing the precipitation of FeCO3, we need to compare the solubility product constant (Ksp) of FeCO3 with the ion product (IP) of Fe++ and CO3--.
The Ksp expression for FeCO3 can be written as:
Ksp = [Fe++][CO3--]
Since we are given the concentration of HCO3- (bicarbonate) in the water, we can assume that the concentration of CO3-- can be calculated from the concentration of HCO3- using the equilibrium constant expression for the dissociation of bicarbonate:
[HCO3-] ⇌ [H+] + [CO3--]
The concentration of CO3-- can be calculated using the provided concentration of HCO3- (0.8 x 1e-02 M) and the pH of the solution, assuming the solution is at pH 7. For a pH of 7, the concentration of H+ can be assumed to be 1 x 10^(-7) M.
Using the equilibrium constant expression for the dissociation of bicarbonate, we have:
[HCO3-] = [H+] + [CO3--]
0.8 x 1e-02 M = 1 x 10^(-7) M + [CO3--]
Rearranging the equation, we find:
[CO3--] = 0.8 x 1e-02 M - 1 x 10^(-7) M
Now that we know the concentration of CO3--, we can substitute it into the Ksp expression for FeCO3:
Ksp = [Fe++][CO3--]
10^(-10.7) = [Fe++] x (0.8 x 1e-02 M - 1 x 10^(-7) M)
Next, we solve for [Fe++]:
[Fe++] = 10^(-10.7) / (0.8 x 1e-02 M - 1 x 10^(-7) M)
Evaluating this expression will give us the maximum amount of Fe++ that can be present in water without causing the precipitation of FeCO3.