M = mols/L of solution.
mols Fe(NO3)3 = M x L, then
M = mols/total L (11 mL or 0.011 in this case). OR another way, is dilution,
0.002 x 5/11 = ??
0.003 x 3/11 = ??
Mix 5mL of 0,002M Fe(NO3)3; 3mL of H2O; and 3mL of 0,002M KSCN....
how can you find the initial concentration of Fe and SCN??
4 answers
thank you very much!
The equilibrium reaction is:
Fe+3(aq) + SCN-(aq) <=> FeSCN+2(aq)
If the reaction to the right goes to completion, the final concentration of Fe+3(aq) can be found as follows:
A. Assuming a reverse reaction occurs and equilibrium is established:
1. We can find [Fe+3] by determining [FeSCN+2] from experimental data colorimetricaly. Consult the instructions for the experiment.
2. If the equilibrium constant is known, it can be determined mathematically. Long, tedious procedure.
B. Assuming the reaction goes to completion, with SCN- the limiting reagent:
(0.0050L)(0.0020mol/L) = 1.0x10^-5 moles Fe+3
(1.0x10^-5 moles Fe+3 ) / (0.0100 L) = 1.0x10^-3 M Fe+3 (initial)
(.0030 L)(0.0020 mol/L) = 6.0x10^-6 mol SCN-
(6.0x10^-6 mol SCN-) / (0.0100 L) = 6.0x10^-4 M SCN- (initial)
(1.0x10^-5 moles Fe+3) - (6.0x10^-6 mol FeSCN-)* = 4.0x10^-6 mol
* That is moles Fe+3 reacting with equal moles of SCN-
4.0x10^-6 mol / 0.0100 = 4.0x10^-4 M final Fe+3
The above answer assumed completion of the reaction to the right. Since that is not true, the actual value of remaining Fe+3 will be less than 4.0x10^-4 M. (Go back to Part A)
Fe+3(aq) + SCN-(aq) <=> FeSCN+2(aq)
If the reaction to the right goes to completion, the final concentration of Fe+3(aq) can be found as follows:
A. Assuming a reverse reaction occurs and equilibrium is established:
1. We can find [Fe+3] by determining [FeSCN+2] from experimental data colorimetricaly. Consult the instructions for the experiment.
2. If the equilibrium constant is known, it can be determined mathematically. Long, tedious procedure.
B. Assuming the reaction goes to completion, with SCN- the limiting reagent:
(0.0050L)(0.0020mol/L) = 1.0x10^-5 moles Fe+3
(1.0x10^-5 moles Fe+3 ) / (0.0100 L) = 1.0x10^-3 M Fe+3 (initial)
(.0030 L)(0.0020 mol/L) = 6.0x10^-6 mol SCN-
(6.0x10^-6 mol SCN-) / (0.0100 L) = 6.0x10^-4 M SCN- (initial)
(1.0x10^-5 moles Fe+3) - (6.0x10^-6 mol FeSCN-)* = 4.0x10^-6 mol
* That is moles Fe+3 reacting with equal moles of SCN-
4.0x10^-6 mol / 0.0100 = 4.0x10^-4 M final Fe+3
The above answer assumed completion of the reaction to the right. Since that is not true, the actual value of remaining Fe+3 will be less than 4.0x10^-4 M. (Go back to Part A)
The total volume of the mixture is:
5 + 3 + 3 = 11 mls = 0.011 L
I used 0.010 L by mistake. It should be replaced by 0.011 L. This will case a small change in the concentrations of Fe+3 and SCN-
5 + 3 + 3 = 11 mls = 0.011 L
I used 0.010 L by mistake. It should be replaced by 0.011 L. This will case a small change in the concentrations of Fe+3 and SCN-
3 answers