Question

1. a) Calculate the volume of 2.00 X 10^-3 M SCN^- required to prepare 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 total solution.

Answer 1

a)

Concentration of available solution of SCN^- =

Let the volume of the available solution needed to make the required 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 be same before and after dilution.

Hence, the product of concentration and volume will also be same before and after dilution. Hence, we can write

Hence, 5.0 mL of the given solution is required to make 50 mL of the required solution.

b)

Note that SCN^- is the limiting reactant and Fe(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 the solution will be equal to the number of moles of complex Fe(SCN)²⁺ formed.

Given that the concentration of SCN- solution used =

Number of moles of SCN- can be calculated by multiplying the concentration of SCN- solution used and the volume used in litres. Hence, we can create the following table to calculate the final 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.