A student mixes 5.0 mL of 0.00200 M Fe(NO3)3 with 5.0 mL 0.00200 KSCN. She finds that the

concentration of FeSCN2+ in the equilibrium mixture is 0.000125 M. Follow these steps to
determine the corresponding experimental value of Kc for the reaction of Fe3+ and SCN–
to produce
this complex ion. Show your calculations for each step below and then place the appropriate
value(s) in the equilibrium (or 'ICE') table near the bottom of the page.
Step 1. Calculate the molarity of Fe3+, SCN–
, and FeSCN2+ initially present after mixing the two
solutions, but prior to any reaction taking place. (M1V1 = M2V2)

[Fe(NO3)3] = 0.00200M x (5.0 mL/10.0 mL) = ?

(KSCN) = 0.00200 x (5.0 mL/10.0 mL) = ?
(FeSCN)^2+ = 0

Did you try inserting these numbers into the M1V1 - M2V2 equation the prof gave you in the problem. That will give you the same numbers I have above. Try it.

Step 1. Calculate the molarity of Fe3+, SCN-, and FeSCN2+ initially present after mixing the two solutions, but prior to any reaction taking place. (M1V1 = M2V2)

Given:
Volume of Fe(NO3)3 solution (V1) = 5.0 mL = 0.005 L
Volume of KSCN solution (V2) = 5.0 mL = 0.005 L
Molarity of Fe(NO3)3 solution (M1) = 0.00200 M
Molarity of KSCN solution (M2) = 0.00200 M

Using the equation M1V1 = M2V2:
(0.00200 M) * (0.005 L) = M2 * (0.005 L)
0.000010 mol Fe3+ = M2 * (0.005 L)

Since Fe(NO3)3 is a 1:1 ratio with Fe3+, we have:
0.000010 mol Fe3+ = M2 * (0.005 L)
M2 = 0.000010 mol / 0.005 L
M2 = 0.00200 M

Similarly, since KSCN is a 1:1 ratio with SCN-, we have:
0.000010 mol SCN- = M2 * (0.005 L)
M2 = 0.000010 mol / 0.005 L
M2 = 0.00200 M

So, initially, the molarity of Fe3+ (M1) and SCN- (M2) is 0.00200 M, and the molarity of FeSCN2+ is 0 M (since no reaction has occurred yet).

Step 1:

Given:
Initial volume of Fe(NO3)3 solution (V1) = 5.0 mL
Initial concentration of Fe(NO3)3 solution (M1) = 0.00200 M
Initial volume of KSCN solution (V2) = 5.0 mL
Initial concentration of KSCN solution (M2) = 0.00200 M

To calculate the molarity of Fe3+, SCN-, and FeSCN2+ initially present, we'll use the equation M1V1 = M2V2.

Molarity of Fe3+ in initial mixture:
M1V1 = M2V2
(0.00200 M) (5.0 mL) = M2 (5.0 mL)
M2 = (0.00200 M) (5.0 mL) / (5.0 mL)
M2 = 0.00200 M

Molarity of SCN- in initial mixture:
M1V1 = M2V2
(0.00200 M) (5.0 mL) = M2 (5.0 mL)
M2 = (0.00200 M) (5.0 mL) / (5.0 mL)
M2 = 0.00200 M

Molarity of FeSCN2+ in initial mixture:
M1V1 = M2V2
(0 M) (5.0 mL) = M2 (5.0 mL)
M2 = (0.00200 M) (5.0 mL) / (5.0 mL)
M2 = 0 M

Therefore, the molarities of Fe3+, SCN-, and FeSCN2+ initially present in the mixture are:
[Fe3+] = 0.00200 M
[SCN-] = 0.00200 M
[FeSCN2+] = 0 M

To calculate the molarity of Fe3+, SCN–, and FeSCN2+ initially present, we can use the formula M1V1 = M2V2. Here's how to do it:

Step 1: Calculate the moles of Fe(NO3)3.
Given: Volume of Fe(NO3)3 solution (V1) = 5.0 mL
Concentration of Fe(NO3)3 solution (M1) = 0.00200 M

Convert the volume of Fe(NO3)3 to liters:
Volume (in liters) = 5.0 mL * (1 L / 1000 mL) = 0.005 L

Calculate the moles of Fe(NO3)3:
Moles = Concentration * Volume
Moles of Fe(NO3)3 = 0.00200 M * 0.005 L = 0.00001 moles

Step 2: Calculate the moles of KSCN.
Given: Volume of KSCN solution (V1) = 5.0 mL
Concentration of KSCN solution (M1) = 0.00200 M

Convert the volume of KSCN to liters:
Volume (in liters) = 5.0 mL * (1 L / 1000 mL) = 0.005 L

Calculate the moles of KSCN:
Moles = Concentration * Volume
Moles of KSCN = 0.00200 M * 0.005 L = 0.00001 moles

Step 3: Calculate the moles of FeSCN2+ initially present.
The reaction equation for the formation of FeSCN2+ is:
Fe3+ + SCN– → FeSCN2+

Since the stoichiometry of the reaction is 1:1, the moles of FeSCN2+ will be the same as the moles of Fe(NO3)3 and KSCN.

Moles of FeSCN2+ = Moles of Fe(NO3)3 = Moles of KSCN = 0.00001 moles

Now that we have calculated the moles of Fe3+, SCN–, and FeSCN2+ initially present, we can fill in the 'ICE' table for the equilibrium calculation.