Question

You wish to measure the iron content of the well water on the new property you are about to buy. You prepare a reference standard Fe3 solution with a concentration of 6.52 × 10-4 M. You treat 13.0 mL of this reference with HNO3 and excess KSCN to form a red complex, and dilute the reference to 60.0 mL. The diluted reference is placed in a cell with a 1.00-cm light path. You then take 30.0 mL of the well water, treat with HNO3 and excess KSCN, and dilute to 100.0 mL. This diluted sample is placed in a variable pathlength cell. The absorbance of the reference and the sample solutions match when the pathlength is 3.99 cm. What is the concentration of iron in the well water? For each solution, the zero is set with a blank. Please give in correct sig figs!!

Answer 1

Ans. Step 1: Standard Fe³⁺ solution: 13.0 mL of 6.52 x 10^-4 M is diluted to a final volume of 60.0 mL.

Now, using    C1V1 (standard Fe³⁺ solution) = C2V2 (diluted Fe³⁺ solution)

            Or, 6.52 x 10^-4 M x 13.0 mL = C2 x 60.0 mL

            Or, C2 = (6.52 x 10^-4 M x 13.0 mL) / 60.0 mL = 1.4127 x 10^-4 M

Thus, [Fe³⁺] in final diluted aliquot (whose Abs is taken) = 1.4127 x 10^-4 M

Step 2: Beer-Lambert’s Law, A = e C L             - equation 1,              

where,

                                                A = Absorbance

                                                e = molar extinction coefficient (M^-1cm^-1)

                                                L = path length (in cm)

                                                C = Molar concentration

Putting the values for standard Fe³⁺ solution in equation 1-

            A = e x 1.4127 x 10^-4 M x 1.0 cm                        - equation 2

# Step 3: Water sample from Well

Let [Fe³⁺] in original water sample be X.

30.0 mL of original well water sample is diluted to a final volume of 100.0 mL

Again, using C1V1 (original well water) = C2V2 (diluted well water)

            Or, C2 = (X x 30.0 mL) / 100.0 mL = 0.30X

Hence, [Fe³⁺] in final diluted water sample (whose Abs is taken) = 0.30X

Putting the values in equation 1 for diluted water sample-

            A = e x 0.30X x 3.99 cm                            - equation 3

Step 4: Given,

Both the diluted water sample and diluted standard Fe³⁺ solutions give same absorbance at different path length.

Note that molar extinction coefficient e remains constant.

Now,

            Comparing equation 2 and 3-

            e x 1.4127 x 10^-4 M x 1.0 cm = e x 0.30X x 3.99 cm

            Or, 0.30X = (e x 1.4127 x 10^-4 M x 1.0 cm) / (e x 3.99 cm)

            Or, 0.30X = 3.5406 x 10^-5 M

            Or, X = (3.5406 x 10^-5 M) / 0.30

            Hence, X = 1.180 x 10^-4 M

Therefore,, [Fe] in original well water sample = X = 1.18 x 10^-4 M