Asked by Ellie
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.
I tried m1v1=m2v2 but its wrong.
Answers
Answered by
DrBob222
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.
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.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.