Fe^+3 + SCN^- ==> FeSCN^+2
Use an ICE chart.
I assume those are molarities? You don't have any units. If 2.5 x 10^-3 is not molar then you must adjust the following.
initial:
Fe^+3 = 2.5 x 10^-3 M x 1 mL = 2.5 x 10^-3 millimoles.
SCN^- = same or 2.5 x 10^-3 millimoles.
FeSCN^+2 = 0
equilibrium:
FeSCN^+2 = 4.44 x 10^-5 M x 10 mL = 4.44 x 10^-4 millimoles
Final concn:
Fe^+3 = 2.5 x 10^-3 mmoles to start - 4.44 x 10^-4 mmoles used = 2.06 x 10^-3 mmoles Fe^+3 remaining unreacted. The concn then is mmoles/mL = 2.06 x 10^-3/10 mL = ??
SCN is the same.
m1v1 = m2v2 works ok for dilution problems but this is a dilution as well as a reaction.
Check my thinking. Check my arithmetic.
In a run involving 1.00 mL 2.50 * 10^-3 FE and 1.00 mL of 2.50 * 10^-3 SCN solutions made up to a total volume of 10.0 mL, the concentration of complex formed at equilibrium was found to be 4.44 * 10^-5. What are the equilibrium concentrations of Fe^3+ and SCN^-?
I tried m1v1=m2v2 but its wrong.
1 answer
1 answer