Question

How do I find the concentration of KSCN & FeSCN2+ from the following data?

Solution	Beaker 1	Beaker 2	Beaker 3	Beaker 4	Beaker 5
Volume of 0.200 M Fe(NO3)3. mL	2.5 mL	2.5 mL	2.5 mL	2.5 mL	2.5 mL
Volume of 0.0020 M KSCN, mL	.50 mL	.75 mL	1.0 mL	1.25 mL	1.5 mL
Volume of 0.50 M HNO3, mL	7 mL	6.75 mL	6.5 mL	6.25 mL	6 mL
Total volume, mL	10 mL	10 mL	10 mL	10 mL	10 mL
Initial [Fe(NO3)3] in the mixture, M	.05 M	.05 M	.005 M	.05 M	.05 M
Initial [KSCN] in the mixture, M
[FeSCN 2+], M

Answer 1

Initial concentrations of KSCN in mixture KSCN] - [KSCN](volumeof KSCN) total volume Beaker 1 (0.0020M (0.50 mL) [KSCN].max = 10.0 mL = 0.0001M Beaker 2 (0.0020M)(0.75 mL) [KSCN ) with 10.0 mL = 0.00015M Beaker 3 [KSCN).- (0.0020M)(1.0mL) 10.0mL = 0.00020M Beaker 4 [KSCN)- (0.0020M)(1.25 mL) Jinitial 10.0 mL = 0.00025M Beaker 5 (0.0020M)(1.50 mL) [KSCN Jital = 10.0 mL = 0.00030 M

Fe+ (aq) +SCN- (aq)=[FeSCN]** (aq) [FeSCN]* (aq) this complex has very very high formation constant. That means all the limiting reactants will be converted into complex. From the concentrations of Fe3and KSCN the limiting reactant should be KSCN as it has very low concentration when compare with Fe3+ ion. Henc the concentrations of [FeSCN]* complex is equal to the concentrations of KSCN in each beaker. Beaker 1 [[FeSca]* )= 0.0001M Beaker 2 [[FeSCN)* ] = 0.00015 M Beaker 3 [[FeSCA]* ] = 0.00020M Beaker 4 [[FeSCN}* ] = 0.00025 M Beaker 5 [[FeSCN}* ] = 0.00030 M