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.
To determine the equilibrium concentration of FeSCN2+, we need to understand the reaction that occurs when Fe(NO3)3 and KSCN are combined.
The reaction between Fe(NO3)3 and KSCN can be represented by the following balanced equation:
Fe(NO3)3 + 3KSCN → Fe(SCN)3 + 3KNO3
We can see from the equation that the stoichiometry is 1:1 between Fe(SCN)3 and Fe(SCN)2+. This means that for every 1 mole of Fe(SCN)3 formed, 1 mole of Fe(SCN)2+ is also formed. Therefore, we can assume that the concentration of Fe(SCN)2+ is equal to the concentration of Fe(SCN)3 in the solution.
Given:
Volume of Fe(NO3)3 solution (V1) = 9.00 mL = 0.00900 L
Concentration of Fe(NO3)3 solution (C1) = 0.200 M
Volume of KSCN solution (V2) = 1.00 mL = 0.00100 L
Concentration of KSCN solution (C2) = 0.0020 M
Using the principles of stoichiometry and dilution, we can determine the molarity of Fe(SCN)3 and Fe(SCN)2+.
First, we calculate the moles of Fe(NO3)3:
moles of Fe(NO3)3 = volume (L) x concentration (M)
moles of Fe(NO3)3 = 0.00900 L x 0.200 M
moles of Fe(NO3)3 = 0.0018 mol
Next, we calculate the moles of KSCN:
moles of KSCN = volume (L) x concentration (M)
moles of KSCN = 0.00100 L x 0.0020 M
moles of KSCN = 0.000002 mol
Since the stoichiometry is 1:1 between Fe(NO3)3 and KSCN, the moles of Fe(SCN)3 formed is equal to the moles of Fe(NO3)3 or KSCN used. Therefore, the moles of Fe(SCN)3 is 0.0018 mol.
Now, we can calculate the concentration of Fe(SCN)3 (which is also the concentration of Fe(SCN)2+):
concentration of Fe(SCN)3 = moles/volume
concentration of Fe(SCN)3 = 0.0018 mol / 0.010 L (since V1 + V2 = 0.009 L + 0.001 L = 0.010 L)
concentration of Fe(SCN)3 = 0.18 M
Therefore, the equilibrium concentration of FeSCN2+ ([FeSCN2+]std) for this standard solution is assumed to be 0.18 M.