A standard solution of FeSCN2+ is prepared by combining 9.00 mL of 0.200 M Fe(NO3)3 w/1.00 mL of 0.0020 M KSCN. The equilibrium concentration of FeSCN2+ ([FeSCN2+]std) for this standard solution is assumed to be ? M.

Wouldn't that be 0.002 x 1/11 = ??

To find the equilibrium concentration of FeSCN2+ ([FeSCN2+]std) for the standard solution, we would need to consider the balanced chemical equation and use the information given.

The balanced chemical equation for the reaction between Fe3+ and SCN- to form FeSCN2+ is:
Fe3+ + SCN- → FeSCN2+

From the given information, we know:
Volume of Fe(NO3)3 solution (V1) = 9.00 mL
Molarity of Fe(NO3)3 solution (M1) = 0.200 M
Volume of KSCN solution (V2) = 1.00 mL
Molarity of KSCN solution (M2) = 0.0020 M

To find the equilibrium concentrations, we need to use the concept of stoichiometry and the dilution formula. Here are the steps to calculate it:

1. Convert the volumes to liters:
V1 = 9.00 mL = 9.00 x 10^-3 L
V2 = 1.00 mL = 1.00 x 10^-3 L

2. Use the dilution formula to calculate the final concentration of Fe(NO3)3 (C1):
C1 = (M1 x V1) / (V1 + V2)
= (0.200 M x 9.00 x 10^-3 L) / (9.00 x 10^-3 L + 1.00 x 10^-3 L)
= (0.0018 mol/L) / (0.010 L)
= 0.18 M

3. Now, we can use the balanced equation to determine the stoichiometry between Fe(NO3)3 and FeSCN2+:
According to the balanced equation, the stoichiometric ratio between Fe(NO3)3 and FeSCN2+ is 1:1. This means that for every 1 mol of Fe(NO3)3 consumed, 1 mol of FeSCN2+ is formed.

4. Since the stoichiometry between Fe(NO3)3 and FeSCN2+ is 1:1, the equilibrium concentration of FeSCN2+ ([FeSCN2+]std) will be equal to the concentration of Fe(NO3)3.

Therefore, the equilibrium concentration of FeSCN2+ ([FeSCN2+]std) for the standard solution is 0.18 M.