Question

Test tubes 1-8 contain the following amounts of 0.05M [KI] and 0.002M [I2]. The absorbance of the solutions were then measured at a wavelength of 430nm. Calculate the final [I2].

Test Tube	0.05 M KI (mL)	0.002M I2 (mL)	Final [I2]	Absorbance
1	10	0		0.0
2	8	2		0.369
3	6	4		0.631
4	4	6		0.781
5	3	7		0.847
6	2	8		0.914
7	1	9		0.966
8	0	10		1.023

Answer 1

To calculate final I2 concentration, the following formula will be used:

M1V1 = M2V2

M1 = molarity of I2 initially = 0.002 M

V1 = volume of 0.002 M I2 used (given in table)

M2 = final concentration of I2 (to be found out)

V2 = final volume. In all the cases, it is 10 ml, which is made by mixing KI and I2.

For example in test tube 1,

M1V1 = M2V2

0.002*0 = M2*10

M2 = 0 M

Similarly other concentrations can be calculated.

Test Tube	0.05 M KI (mL)	0.002M I2 (mL)	Final [I2]	Absorbance
1	10	0	0	0.0
2	8	2	0.0004	0.369
3	6	4	0.0008	0.631
4	4	6	0.0012	0.781
5	3	7	0.0014	0.847
6	2	8	0.0016	0.914
7	1	9	0.0018	0.966
8	0	10	0.0020	1.023