Question

You wish to determine the amount of lead in a drinking water sample (500mL). It is possible that sulfate in the sample (as well as other ions) will produce a matrix effect that will compromise the analysis, so you use a standard addition calibration method. You transfer 15 mL of your sample into each of five 50 mL volumetric flasks. You add the following volumes of a 5 ppm lead standard to the 5 flasks : 0, 5, 10, 15, 25 mL. The flasks are brought to volume with DI water. Upon analysis, the data indicated below are obtained. Determine the concentration of lead in the original sample.

Volume (mL)                      Detector Response

0                                             1223

5                                           2257

10                                       4564

15                                      6867

25                                        10841

Answer 1

Ans. Step 1: Determining [Pb] in 50.0 mL aliquot using standard addition method.

The trendline equation for the emission intensity vs concertation graph is “y = 4769.2x + 381.2” in the form of y = m x + c. where, y-axis = detector’s signal, x-axis = [Pb] in ppm; Y-Intercept = c ; Slope = m

Now, Concertation of unknown = (Intercept / Slope) concertation units

= (381.2 / 4769.2) ppm

= 0.07993 ppm

Hence, [Pb] in 50 mL aliquot = 0.07793 ppm

Step 2: Calculating [Pb] in original sample.

[Note: C1V1 and C2V2 used in excel are different from those used in step 2.]

Given, 15.0 mL (V1) of original water sample is diluted to 50.0 mL (V2) final volume.

Now, using C1V1 = C2V2   - equation 1

C1= Concentration, and V1= volume of initial solution 1         ; Original sample

C2= Concentration, and V2 = Volume of final solution 2         ; 50 mL aliquots

Putting the values in equation 1-

            C1 x 15.0 mL = 0.07993 ppm x 50.0 mL

            Or, C1 = (0.07993 ppm x 50.0 mL) / 15.0 mL

            Or, C1 = 0.266 ppm

Therefore, in original water sample, [Pb] = 0.266 ppm