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??
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 = ??
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)
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-
To determine the initial concentrations of Fe and SCN, you will need to use the information given in the question and apply the principles of stoichiometry and solution dilution.
Here's the step-by-step process:
1. Calculate the moles of Fe(NO3)3 using its given concentration and volume:
- Moles of Fe(NO3)3 = concentration x volume
= 0.002 M x 0.005 L
= 0.00001 moles
2. Calculate the moles of KSCN using its given concentration and volume:
- Moles of KSCN = concentration x volume
= 0.002 M x 0.003 L
= 0.000006 moles
3. Since Fe(NO3)3 and KSCN react in a 1:1 ratio to form a product, the number of moles of SCN generated will be equal to the number of moles of Fe(NO3)3 or KSCN used. Therefore, the moles of SCN are also 0.000006 moles.
4. Since H2O does not react, its volume does not affect the concentration calculations.
5. To find the initial concentration of Fe(NO3)3:
- Initial concentration of Fe(NO3)3 = Moles of Fe(NO3)3 / Total volume of the mixture
= 0.00001 moles / (0.005 L + 0.003 L)
= 0.00001 moles / 0.008 L
= 0.00125 M
6. To find the initial concentration of KSCN (which is also the concentration of SCN):
- Initial concentration of KSCN = Moles of KSCN / Total volume of the mixture
= 0.000006 moles / (0.005 L + 0.003 L)
= 0.000006 moles / 0.008 L
= 0.00075 M
Therefore, the initial concentration of Fe(NO3)3 is 0.00125 M, and the initial concentration of SCN is 0.00075 M.