Question

This is for the verification of beer-lambert law experiment.

Volume of copper sulphate solution /mL	Volume of Concentrate d ammonia /mL	Final volume     of solution /mL	[CuSO4]/m ol dm-3	[CuSO4]/pp m	[CuSO4]/mg dm-3	Transmit tance /%	Absorbance
0	25	25				101.5	0.006
1	24	25				95.9	0.018
2	23	25				89.9	0.046
3	22	24				86.7	0.062
4	21	25				81.4	0.090

Can someone show one sample example of how to determine the three ways the following concentrations can be calculated.

Original Concentration of CuSO4.5H2) = 0.1015 mol/dm3 . 25mL of this was then taken and dissolved in water up to 250mL flask.

Concentration of the ammonia= 17.64mol/L. Ammonia was not diluted.( at 600nm)

OD values for calibration? I am not too sure what that means.

Answer 1

Given that the original concentration of CuSO₄.5 H₂O is 0.1015 mol / dm³

     25 mL of this solution is diluted to 250 mL or it is 10 times [ or 1/10th ] dilution.

Hence the final concentraion in the250 mL flask = 0.1015 * 1 / 10 = 0.01015 mol / L

In all the samples the final volume = 25 mL

In first sample the volume of CuSO₄ . 5 H₂O sample added = 0

In the second sample it is 1 mL in 25 mL of the solution ie 1/25 dilution

Therefore the concentration is 0.01015x 1/25 = 0.000406 mol / dm³

we know that the mol wt of CuSO4 = 63.55 + 32 + 64 = 159.55

therefore mass of CuSO₄ in sample 1 = 0.000406 x 159.55 = 0.0648 g/dm³

                                                                                     = 64.8 mg / dm³ [since 1 g = 1000 mg]

                                                                                    = 64.8 mg / L       [ since 1 dm³ = 1 L]

                                                                                      = 64.8 ppm [ since mg/L = ppm]

Thus sample 2 contains

        CuSO4 = 0.000406 mol / dm³

                  = 64.8 mg / dm³

                 = 64.8 ppm

Similarly for sample 3 , the volume taken is 2 mL.where the concentration gets doubled, therefore

sample 3 contains

CuSO4 = 0.000812 mol / dm³

                  = 130 mg / dm³

                 = 130 ppm

similarly sample 4 contains

CuSO4 = 0.001218 mol/ dm³

                  = 194.4 mg / dm³

                 = 194.4 ppm

sample 5 contains

CuSO4 = 0.001624 mol/ dm³

                  = 259.2 mg / dm³

                 = 259.2 ppm

The value of OD is the ability of the component to slow the transmission of light. Higher the OD, lower is the transmission.

OD = -log T it is the negative log of transmission.

Absorbance also = - log T = - log %T/100

                           = - [ log %T - log 100]

                       = 2 - log %T

In common practice people use the term OD in place of Absorbance.