Question

For a water in equilibrium with the atmosphere, a pH of 2.0, and a total soluble iron concentration of 1 mg/L, calculate the concentrations of Fe2+(aq) and Fe3+(aq).

Answer 1

Fe^2+(aq) = 4.51 × 10^-11 M

Fe3+(aq) = 1.79 × 10^5 M.

Step-by-step explanation:

So, the first thing to do is to to write out the chemical reaction showing the ionic reaction of the chemical species present in the chemical reaction;

4Fe^2+ (aq) + O2(g) + 4H^+ <---------------------------> 4Fe^3+(aq) + 2H2O(l).

The next thing to do is to determine or calculate for G° and the equilibrium constant,kc. Thus, the value of G° = 4(-4.6) + 2(-237.18) - {4(-78.87)} = -177.28 kJ/mol.

Therefore, the equilibrium constant = e^-∆G°/RT = 1.19 × 10^31.

Therefore the value for Fe^2+ and Fe^3+ can be determine as;

NB=> Recall that [Fe^3+]/[Fe^2+] = kc [(Po2) (H^+)^4]^1/4.

Also, 1/ 55850 = 1.79 × 10^-5 M.

[Fe^2+] + [Fe^3+] = 1.79 × 10^-5. -------(1).

Therefore, [Fe^3+]/[Fe^2+] = kc [(Po2) (H^+)^4]^1/4. = [Fe^3+]/[Fe^2+] = [ 1.19 × 10^31 × 0.2095 × (10^2)^2] ^1/4 = 3.97 × 10^5.

Therefore, [Fe^3+]/[Fe^2+] = 3.97 × 10^5.

[Fe^3+] = 3.97 × 10^5[Fe^2+].

Hence, using the equation (1) above we can determine the value of the species.

[Fe^2+] + [Fe^3+] = 1.79 × 10^-5

[Fe^2+] + [Fe^2+] = 1.79 × 10^-5.

[Fe^2+] + 3.97 × 10^5[Fe^2+] = 1.79 × 10^-5.

[Fe^2+] = 4.51 × 10^-11 M.

Therefore, we will have [Fe^3+]/[Fe^2+] = 3.97 × 10^5.

Thus, [Fe^3+] = [Fe^2+] × 3.97 × 10^5. = 1.79 × 10^5 M.