Chemistry concentration

Question

Analysis of iron in water from farm runoff was determined according to the following. A 25.0 mL sample of the runoff was acidified and then treated with excess KSCN to form a red complex. This solution was then diluted to 100.0 mL and put in a variable pathlength cuvet. As a comparison, a 10.0 mL reference sample containing 6.80x104 M Fe3* with acid and an excess of KSCN and diluted to 50.0 mL. This reference sample was placed in cuvet with a 1.00 cm pathlength. The runoff sample had the same absorbance as the reference when the pathlength of the runoff cell was 2.48 cm. Calculate the concentration of iron in the original (undiluted) runoff sample.

Answer 1

Answer: the original, undiluted runoff is 2.19e-4 M in iron, which is about 12.2 mg/L (12.2 ppm) Fe.

Two notes on the posted text before any chemistry. The reference concentration prints as "6.80x104 M" -- the exponent lost its minus sign when the question was posted, and it has to be 6.80e-4 M, because 6.80e4 M is not a concentration that can exist. Likewise "Fe3*" is Fe3+. Everything below uses those readings, so check them against your own worksheet.

The idea behind the whole problem. Beer's law is A = e*b*c, where e is the molar absorptivity, b the pathlength and c the concentration. Both cuvets hold the same red species, Fe(SCN)2+, measured at the same wavelength, so e is identical in the two cells and cancels. Set the two absorbances equal:

A(runoff) = A(reference)
e * b(runoff) * c(runoff) = e * b(ref) * c(ref)
c(runoff) = c(ref) * b(ref) / b(runoff)

Notice what you never need: the value of e, and the absorbance reading itself. Neither appears anywhere in the arithmetic.

Step 1 -- what is actually sitting in the reference cuvet. The 6.80e-4 M stock was diluted 10.0 mL up to 50.0 mL, so the stock concentration is not what the light passes through:

c(ref) = 6.80e-4 M * (10.0 mL / 50.0 mL)
       = 6.80e-4 M * 0.200
       = 1.36e-4 M

Step 2 -- match the absorbances. The two cells read the same when the runoff cell was 2.48 cm and the reference cell 1.00 cm:

c(runoff, in the 100.0 mL flask) = 1.36e-4 M * (1.00 cm / 2.48 cm)
                                 = 1.36e-4 M * 0.4032
                                 = 5.48e-5 M

Step 3 -- undo the runoff dilution. That 5.48e-5 M describes the diluted solution in the flask, not the sample that came off the field. The original 25.0 mL was made up to 100.0 mL, a factor of 4.00, so the undiluted runoff is four times more concentrated:

c(original) = 5.48e-5 M * (100.0 mL / 25.0 mL)
            = 5.48e-5 M * 4.00
            = 2.19e-4 M Fe3+

In mass units that is 2.19e-4 mol/L * 55.85 g/mol = 0.0122 g/L, or 12.2 mg/L. Every input carries three significant figures, so the answer does too.

The free sanity check. The runoff cell had to be made longer (2.48 cm against 1.00 cm) to reach the same absorbance, so the runoff solution must be the more dilute of the two -- and 5.48e-5 M is indeed below 1.36e-4 M. If your diluted runoff came out more concentrated than the reference, you inverted the pathlength ratio. One line, and it catches the single commonest error in this problem.

The trap the problem is built around: the two dilutions run in opposite directions. The reference step (10.0 up to 50.0 mL) makes a solution weaker, so you multiply by 10.0/50.0, a number less than one. The runoff step runs backwards -- you are going from the diluted flask back to the original sample -- so you multiply by 100.0/25.0, a number greater than one. Write "diluting makes it smaller, undoing a dilution makes it bigger" beside each line and neither can flip. Inverting that last factor alone gives 1.37e-5 M, which is off by a factor of 16.

Why anyone would use a variable-pathlength cuvet at all. This is a null method. You are not reading a number off an absorbance scale, you are adjusting b until two cells look identical, and identical is something an instrument judges far better than it measures an absolute value. That makes the result independent of the photometric accuracy of the spectrophotometer and independent of e, which is exactly where an ordinary calibration-curve method is weakest. It also lets you match a sample that would otherwise fall outside the reliable 0.1 to 1.0 absorbance window without diluting it yet again.

Two experimental details in the prompt are doing real work, and they are fair game for a follow-up question. The excess KSCN drives Fe3+ + SCN- -> Fe(SCN)2+ essentially to completion in both solutions, so the same fraction of the iron is coloured in each and e genuinely is common to the two cells; with limiting thiocyanate the comparison would mean nothing. The acidification keeps the iron as free Fe3+ rather than hydroxo complexes or solid Fe(OH)3, which are not red and would quietly remove analyte from the measurement before it was ever made.

For context on the answer: 12.2 mg/L is roughly forty times the 0.3 mg/L secondary drinking-water standard for iron. That is the sort of number that makes farm runoff worth analysing in the first place.