Question

Absorbance measurements were taken on a series of Fe standards complexed with 1,10-phenanthroline. A plot of absorbance vs. concentration yielded a linear curve with the equation y = 1008x + 0.0395 (where y is the absorbance and x represents the Fe concentration in moles/L). Using this information calculate the Fe concentration (in ppm) in a 30.0-mL aliquot of ground water that produced an absorbance of 0.643.

Answer 1

y = 1008x + 0.0395

We will plug in the absorbance value to find x, concentration of Fe in moles/L

0.643 = 1008(X) + 0.0395

X = 0.0005987 moles/L

Volume of solution = 30.0-mL = 0.03 L

No. of moles of Fe = 0.0005987 moles/L 0.03 L = 0.00001796 mol (Molarity times volume)

Molecular Weight of Fe = 55.845 g/mol

Now let’s convert the moles of Fe to grams of Fe:

0.00001796 mol Fe x ( 55.845 g / 1 mol ) = 0.001003 g Fe

Now we know that

ppm = (grams of solute / grams of solution ) x 10^6

grams of solute = 0.001003g (Fe)

grams of solvent = density of water x volume of water = 1 g/mL x 30 mL =30g

Mass of solution = mass of solvent(Water) + mass of solute(Fe) = 30g + 0.001003g = 30.001003g

Putting our numbers in the ppm formula gives: (0.001003g / 30.001003g) x 10^6 = 33.4 ppm

This is our final answer.