Suppose that the reaction of Fe3+ and SCN- produces Fe(SCN)2 (+). 5.00mL of 2.0mM Fe3+ is mixed with 5.00 mL of 2.0mM SCN-. The student finds the equilibrium concentrations of Fe(SCN)2(+) to be 0.3mM.

1.) What is the initial number of moles of each species present?

2.) What is the equilibrium number of moles of each species present?

3.) What is the equilibrium concentration of each species present

4.) What is the value of the equilibrium constant?

1. millimols = mL x M = 5 mL x 0.002M = 0.01 mmols or 1E-5 mols.

mols SCN = same

2.
...........Fe^3+ + 2SCN^- ==> Fe(SCN)2^+
I..........1E-5....1E-5........0
C...........-x......-2x........x
E.........1E-5-x...1E-5-2x......x
x from the problem is 3E-4M and that x 0.01L (10 mL) = 3E-6 mols. So
Fe = 1E-5-3E-6 = ? mols
SCN = 1E-5 - 2*3E-6) = ?mols
complex = 3E-6 mols

3. Fe = mols/L
SCN = mols/L
complex = 3E-4 M from the problem.

4. Substitute the E values (of concentration--not mols) into Keq expression and solve for Keq.

Note: Your teacher, trying to make things easier for you by going through steps 1,2 has actually made it more complicated and easier to make a mistake. The easy way to do this is to calculate (Fe^2+), (SCN^-) from the beginning and use that with the concn of the complex already given and solve for Keq.

In order to determine the answers to these questions, we need to use the given information and apply the principles of stoichiometry. Let's break down each question and explain how to calculate the answers step by step.

1.) What is the initial number of moles of each species present?

To calculate the initial number of moles, we need to use the given volume and concentration of each species. The equation for this reaction shows that for every 1 mole of Fe3+, we get 1 mole of Fe(SCN)2(+). Similarly, for every 1 mole of SCN-, we also get 1 mole of Fe(SCN)2(+).

Given:
Volume of Fe3+ = 5.00 mL = 5.00 x 10^(-3) L
Volume of SCN- = 5.00 mL = 5.00 x 10^(-3) L
Concentration of Fe3+ = 2.0 mM = 2.0 x 10^(-3) mol/L
Concentration of SCN- = 2.0 mM = 2.0 x 10^(-3) mol/L

Using the formula: moles = concentration x volume (in liters)

The initial number of moles for Fe3+ and SCN- would be:
moles of Fe3+ = (2.0 x 10^(-3) mol/L) x (5.00 x 10^(-3) L) = 1.0 x 10^(-5) moles
moles of SCN- = (2.0 x 10^(-3) mol/L) x (5.00 x 10^(-3) L) = 1.0 x 10^(-5) moles

Therefore, the initial number of moles for Fe3+ and SCN- is 1.0 x 10^(-5) moles each.

2.) What is the equilibrium number of moles of each species present?

According to the reaction equation, for every 1 mole of Fe(SCN)2(+), we form 1 mole of Fe(SCN)2(+). Therefore, the equilibrium number of moles for Fe(SCN)2(+) will also be 1.0 x 10^(-5) moles.

3.) What is the equilibrium concentration of each species present?

To calculate the equilibrium concentration, we divide the number of moles of each species by the total volume of the solution.

Equilibrium concentration of Fe3+ = (1.0 x 10^(-5) moles) / (10.00 x 10^(-3) L) = 1.0 x 10^(-3) M
Equilibrium concentration of SCN- = (1.0 x 10^(-5) moles) / (10.00 x 10^(-3) L) = 1.0 x 10^(-3) M
Equilibrium concentration of Fe(SCN)2(+) = (1.0 x 10^(-5) moles) / (10.00 x 10^(-3) L) = 1.0 x 10^(-3) M

Therefore, the equilibrium concentration of Fe3+, SCN-, and Fe(SCN)2(+) is 1.0 x 10^(-3) M.

4.) What is the value of the equilibrium constant?

The equilibrium constant (Kc) can be determined using the equilibrium concentrations of the reactants and products. Since the equation for the reaction is:

Fe3+ + SCN- -> Fe(SCN)2(+)

The equilibrium constant expression would be:

Kc = [Fe(SCN)2(+)] / ([Fe3+] x [SCN-])

Given the equilibrium concentration of Fe(SCN)2(+) is 0.3 mM = 0.3 x 10^(-3) M,
and the equilibrium concentration of Fe3+ and SCN- is 1.0 x 10^(-3) M each,

The value of the equilibrium constant (Kc) would be:
Kc = (0.3 x 10^(-3) M) / ((1.0 x 10^(-3) M) x (1.0 x 10^(-3) M))
Kc = 0.3 x 10^3

Therefore, the value of the equilibrium constant (Kc) is 0.3 x 10^3.