How do I set up my equation to determine what the pH will be when the concentration of Fe^3+ is reduced to 10^-7? The previous part of the question had me find what the pH would be for the Fe(OH)3 to precipitate at 0.20 M. which i found was 1.37
Ksp = (Fe^+3)(OH^-)^3
You know Ksp, substitute 10^-7 for Fe^+3 and solve for OH^-, then convert to pH.
I take it 10^-7 means 10^-7 M.
yes Im sorry, it was for M
To determine the pH when the concentration of Fe^3+ is reduced to 10^-7, we can use the concept of equilibrium and the dissociation of water. The equation you used to find the pH when Fe(OH)3 precipitates at 0.20 M was likely the expression for the solubility product constant (Ksp), which is given by:
Ksp = [Fe^3+][OH^-]^3
Since Fe(OH)3 is a strong base, it will completely dissociate in water into Fe^3+ and OH^- ions. Therefore, when Fe(OH)3 precipitates, the concentration of Fe^3+ is equal to the initial concentration of Fe(OH)3, which is 0.20 M.
Now, to determine the pH when the concentration of Fe^3+ is reduced to 10^-7, we can set up an equation using the dissociation of water. The dissociation constant of water (Kw) is given by:
Kw = [H+][OH^-]
At equilibrium, the concentration of H+ ions is equal to the concentration of OH^- ions (neutral solution). Thus, when the concentration of H+ ions is x, the concentration of OH^- ions is also x.
Since the concentrations of H+ and OH^- ions are equal and relate to the concentration of Fe^3+ through the stoichiometry of the reaction, we can write the following equation:
Ksp = [Fe^3+][OH^-]^3 = (10^-7)(x)^3
Now we can substitute the previously calculated value of Ksp (1.37) and solve for x:
1.37 = (10^-7)(x)^3
To find x, take the cube root of both sides:
x = (1.37/10^-7)^(1/3)
Using a calculator, evaluate the expression inside the parentheses and then raise it to the power of 1/3. This will give the concentration of H+ ions at equilibrium. To find the pH, take the negative logarithm (base 10) of the concentration:
pH = -log[H+]
Let me know if you need any further assistance with the calculations or if you have any other questions!