I know the answer should be 1.9ppm but I need to know how to calculate this. 45.

Question

I know the answer should be 1.9ppm but I need to know how to calculate this.

45. A UV-Vis method was developed for routine determination of iron in ground water. 5.00 ml of ground water was treating with excess of KSCN to form an iron complex (1:1 ratio) and the iron was determined based on the following results. Determine the concentration of iron in ground water in ppm. Volume 0 sample taken, ml Absorbance A at 580 nm (1.00-cm cells) Reagent Volume Used, ml 2.5 ppm ligand H2O 5.00 0.00 20.0 25.0 0.723 5.00 1.00 20.0 24.0 0.917

Explanation / Answer

Ans. Given-

I. 1.0 mL of 2.5 ppm Fe3+ is mixed with 5.0 mL sample and diluted to a final volume of 50.0 mL. The absorbance of this spiked solution is 0.917.

II. 5.0 mL sample is diluted to a final volume of 50.0 mL. The absorbance of this spiked solution is 0.723.

# Step 1: Increase in [Fe3+] in sample due to spiking can be calculated using-

C1V1 (Standard Fe3+) = C2V2 (spiked sample)

            Or, 2.5 ppm x 1.0 mL = C2 x 50.0 mL

Or, C2 = (2.5 ppm x 1.0 mL) / 50.0 mL

Hence, C2 = 0.05 ppm

Therefore, increase in [Fe3+] in the spiked solution due to addition of standard Fe3+ (spiking) = 0.05 ppm.

# Step 2: Given,

Absorbance of un-spiked sample = 0.723

Absorbance of spiked sample = 0.917

# Increase in absorbance due to spiking = 0.917 – 0.723 = 0.194

Since, the increase in Abs by 0.194 unit is solely due to addition of standard Fe3+, an Abs of 0.194 units is equivalent to 0.05 ppm Fe3+ - the increase in [Fe3+] due to spiking.

So,

            0.194 unit Abs is equivalent to 0.05 ppm Fe3+

     Or, 0.723 -           -                      - (0.05 / 0.194) x 0.723

                                                            = 0.18634 ppm

# Therefore, [Fe3+] in 50.0 mL of un-spiked sample = 0.18634 ppm

# Step 3: As mentioned, 5.0 mL sample is diluted to a final volume of 50.0 mL. The absorbance of this spiked solution is 0.723.

Now, using C1V1 (original 5.0 mL sample) = C2V2 (50.0 mL diluted un-spiked sample)

            Or, C1 x 5.0 mL = 0.18634 ppm x 50.0 mL

            Or, C1 = (0.18634 ppm x 50.0 mL) / 5.0 mL

            Hence, C1 = 1.8634 ppm

Therefore, [Fe3+] in 5.0 mL original sample = 1.8643 ppm = 1.9 ppm

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