Question

How do you find A.4

Answer 1

Part1: Preparation of Standard Solutions and Determination of Absorbances of the Standard Solutions: Preparation of Standard Solutions – known concentration of FeSCN2+

Fe³⁺    +    SCN-   à     [Fe(SCN)]²⁺

[FeSCN2+]/ [Fe3+][ SCN-] = Kc

In this set of beakers the concentration of Fe3+ ions is overwhelming greater than the concentration of SCN- ions. This shifts the equilibrium to the right so far that ALL the SCN- ions will react to become FeSCN2+ ions.

So in this set of beakers the concentration of the FeSCN²⁺ ions will be equal to the initial concentration of the SCN^- ions. This initial concentration can be calculated using MV=MV to find the new concentration of ions in the new volume of 25.0mL.

With known concentration of FeSCN²⁺ we can calculate the absorbance of the solution using the measured percent absorbance. Graphing the Absorbance as a function of concentration will yield a line that will allow us to find the concentration of other solutions that we do not already know the concentration.

M1V1= M2V2  

0.001 (1ml) = x (25);   x = 0.00004 M

0.001 (2ml) = x (25);   x = 0.00008 M

0.001 (3ml) = x (25);   x = 0.00012 M

0.001 (4ml) = x (25);   x = 0.00016 M

0.001 (5ml) = x (25);   x = 0.00020 M

Initial concentration of SCN^- ions and final concentration of FeSCN²⁺

Solution	Volume of 0.001M NaSCN (ml)	Initial concentration of SCN-	Final concentration of FeSCN2+ in 25 ml (Answer: A4)
1	0	0	0
2	1	0.00004 M	0.00004 M
3	2	0.00008 M	0.00008 M
4	3	0.00012 M	0.00012 M
5	4	0.00016 M	0.00016 M
6	5	0.00020 M	0.00020 M

From this known concentration of FeSCN²⁺ and absorbance value at lambda max we can plot a standard curve. Then by using equation for straight line we can calculate concentration of FeSCN²⁺ for any unknown sample. (Answer for A4 is last column in above table)