Question

There are six glass cups. all of them were filled by 100 mL sample solutions that contain Pb²⁺ ion with unknown concentration. Then into it added successively 0 ml; 0,1 ml; 0,2 ml; 0,3 ml; 0,4 ml; and 0,5 ml standard solutions 10 mg/L Pb²⁺. Table below shows the absorbances measurement of the solution in all six cups.

Volume	0	0.1	0.2	0.3	0.4	0.5
Absorbansi	0.27	0.37	0.53	0.65	0.75	0.68

Calculate the concentration of Pb²⁺in the above sample if we assume the solution's volume is constant 100 ml.

Answer 1

We need to find out the concentration of the added Pb²⁺ standard in the samples using the dilution equation.

M₁*V₁ = M₂*V₂

where M₁ = concentration of the added standard = 10 mg/L; V₁ = volume of added standard and V₂ = volume of the sample used for analysis = 100 mL.

Take the third solution as an example. We take 0.2 mL of 10 mg/L Pb²⁺ standard and add to 100.0 mL solution.

Plug in values and obtain

(0.2 mL)*(10 mg/L) = (100.0 mL)*M₂

====> M₂ = 0.02 mg/L

Prepare the table as below.

Solution #	Volume of 10 mg/L Pb2+ added (mL)	Concentration of added Pb2+ in the sample	Absorbance
1	0.0	0.00	0.27
2	0.1	0.01	0.37
3	0.2	0.02	0.53
4	0.3	0.03	0.65
5	0.4	0.04	0.75
6	0.5	0.05	0.68

Plot absorbance vs concentration of the added Pb²⁺ standard.

Use the regression equation to find out the concentration of Pb²⁺ in the sample. The sample is believed to give no absorbance on its own; hence, put y = 0 and write

0 = 9.4571x + 0.3052

====> -9.4571x = 0.3052

====> x = 0.3052/(-9.4571) = -0.03227 ≈ -0.0323

The concentration cannot be negative; hence, ignore the negative sign and obtain the concentration of Pb²⁺ in the original solution as 0.0323 mg/L (ans).