You wish to measure the iron content of the well water on the new property you a

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 Fe" solution with a concentration of 7.76 x 10* M. You treat 14.0 mL of this reference with HNO3 and excess KSCN to form a red complex, and dilute the reference to 50.0 mL. The diluted reference is placed in a cell with a 1.00-cm light path. You then take 10.0 mL of the well water, treat with HNOs and excess KSCN, and dilute to 100.0 mL. This diluted sample is placed in a variable pathlength cell. The absorbance of the neference and the sample solutions match when the pathlength is 2.62 cm. What is the concentration of iron in the well water? For each solution, the zero is set with a blank Number [rewell water-1 ] M

Explanation / Answer

Ans. Step 1: Standard Fe3+ solution: 14.0 mL of 7.76 x 10-4 M is diluted to a final volume of 50.0 mL.

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

            Or, 7.76 x 10-4 M x 14.0 mL = C2 x 50.0 mL

            Or, C2 = (7.76 x 10-4 M x 14.0 mL) / 50.0 mL = 2.1728 x 10-4 M

Thus, [Fe3+] in final diluted aliquot (whose Abs is taken) = 2.1728 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 2.1728 x 10-4 M x 1.0 cm                        - equation 2

# Step 3: Water sample from Well

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

10.0 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 10.0 mL) / 100.0 mL = 0.10X

Hence, [Fe3+] in final diluted water sample (whose Abs is taken) = 0.10X

Putting the values in equation 1 for diluted water sample-

            A = e x 0.10X x 2.62 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 2.1728 x 10-4 M x 1.0 cm = e x 0.10X x 2.62 cm        

            Or, 0.10X = (2.1728 x 10-4 M x 1.0 cm) / 2.62 cm

            Or, X = 8.2931 x 10-5 M / 0.10

            Hence, X = 8.2931 x 10-4 M

Therefore,, [Fe] in original well water sample = X = 8.2391 x 10-4 M

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.