Use the table of complex ion formation constants to calculate the molar concentration of free Fe^2+ in 0.510M aqueous solution of potassium hexacyanoferrate(II) K4(Fe(CN)6) at 25 degrees celsius.
I know the answer is 1.96*10^-6 m, but I'm at a loss on how to get there.
I've written down the information from the formation constant chart that i need, but again I don't know how exactly to use it.
To calculate the molar concentration of free Fe^2+ in the aqueous solution of potassium hexacyanoferrate(II) (K4(Fe(CN)6)), you can use the formation constant chart.
1. First, find the complex ion formed between Fe^2+ and CN^- in the chart. The complex ion is [Fe(CN)6]^4-.
2. Identify the formation constant (Kf) for the formation of the complex ion [Fe(CN)6]^4-. Let's assume the value is given as Kf = 1.1 x 10^15.
3. Write down the formation reaction:
Fe^2+ + 6CN^- -> [Fe(CN)6]^4-
4. Define the equilibrium concentrations of the reactants and products. In this case, let's assume the concentration of [Fe(CN)6]^4- is x M, and the initial concentration of Fe^2+ is 0.510 M.
[Fe^2+] = 0.510 M (initial concentration)
[CN^-] = 6x M (molar concentration of CN^- contributed by [Fe(CN)6]^4-)
5. Since Fe^2+ reacts with all 6 CN^- ions, the change in concentration of Fe^2+ is -x M, and the change in concentration of CN^- is subtracted to reflect the reaction stoichiometry: -6x M.
[Fe^2+] = (0.510 - x) M (final concentration of Fe^2+)
[CN^-] = 6x M (final concentration of CN^-)
6. As per the formation constant expression, the equilibrium expression can be written as:
Kf = ([Fe(CN)6]^4-)/([Fe^2+][CN^-]^6)
Substitute the values:
1.1 x 10^15 = (x)/((0.510 - x)*(6x)^6)
7. Solve the equation for x. This can be done using algebraic methods or a numerical solver. The solution will give you the concentration of [Fe^2+] in M.
x = 1.96 x 10^-6 M
Therefore, the molar concentration of free Fe^2+ in the 0.510 M aqueous solution of potassium hexacyanoferrate(II) at 25 degrees Celsius is 1.96 x 10^-6 M.