1. a) Calculate the volume of
1. a) Calculate the volume of 2.00 X 10^-3 M SCN^- required toprepare 50 mL of 2.00 X 10^-4 M SCN^-.
b) Calculate [FeSCN^2+] for each standard solution:
| Sample ID | 2.00 X 10^-4 M SCN^- (mL) (limiting reagent) | H2O (mL) | 0.200 M Fe(NO3)3 (mL) (excess reagent) |
| 1 (blank) | 0 | 9.00 | 1.00 |
| 2 | 1.00 | 8.00 | 1.00 |
| 3 | 3.00 | 6.00 | 1.00 |
| 4 | 5.00 | 4.00 | 1.00 |
| 5 | 7.00 | 2.00 | 1.00 |
| 6 | 9.00 | 0 | 1.00 |
Assume that all SCN^- ions react (mol SCN^- = mol FeSCN^2+ ).Thus the calculation of [FeSCN^2+] is: mol FeSCN^2+ / L of totalsolution.
Answer:
a)
Concentration of available solution of SCN– =
Let the volume of the available solution needed to make therequired solution be 
Target concentration of the required solution = 
Target volume of the required solution = 
Now, we know that the number of moles of SCN– will besame before and after dilution.
Hence, the product of concentration and volume will also be samebefore and after dilution. Hence, we can write
Hence, 5.0 mL of the given solution is required to make50 mL of the required solution.
b)
Note that SCN– is the limiting reactant andFe(NO3)3 is the excess reagent.
The reaction for the complex formation can be written as
Hence, the number of moles of SCN– present in thesolution will be equal to the number of moles of complexFe(SCN)2+ formed.
Given that the concentration of SCN- solution used = 
Number of moles of SCN- can be calculated by multiplying theconcentration of SCN- solution used and the volume used in litres.Hence, we can create the following table to calculate thefinal concentrations of complex formed.
| Sample ID | Volume of SCN–(L) | Moles of SCN– =Moles of Fe(SCN)2+ | Total Volume(L) |  ,M |
| 1 | 0 |  |
 |
 |
| 2 | 0.001 |  |
 |
 |
| 3 | 0.003 |  |
 |
 |
| 4 | 0.005 |  |
 |
 |
| 5 | 0.007 |  |
 |
 |
| 6 | 0.009 |  |
 |
 |
Note: the answers are in red.
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