[tarik3001]

MTS determined the Fe concentration of a sample of acid mine drainage collected from a runoff stream north of New Eagle, PA.  A sample volume of 0.5000 mL was pipetted into a beaker, treated with analytical reagents to produce an indigo color, transferred to a volumetric flask and diluted to 25.00 mL (analysis solution).  The absorbance of the analysis solution was subsequently measured, and the concentration of Fe in the analysis solution was found to be 2.856 x 10-4 mole Fe L-1.  Calculate the molar concentration of Fe, and the ug Fe mL-1, in the original acid mine drainage sample.  Then, calculate pFe for the original acid mine drainage sample.

2.856 x 10^-4 mold Fe L^-1 Fe * (25ml/5ml) = .0142M Fe
.01428M Fe (58.85 g/mol) = .7975g Fe
.7975g Fe * (1000ug Fe Ml^-1) = 797.5 ugFeml^-1

pFe= -log[Fe]
pFe=-log[2.856 x 10^ -4]
pFe=3.544

Thanks!

[Borek]

2.856 x 10^-4 mold Fe L^-1 Fe * (25ml/5ml) = .0142M Fe

Nothing on the left is correct, but miraculously number on the right is right.

.01428M Fe (58.85 g/mol) = .7975g Fe

Everything on the left is OK, but - another miracle - number on the right is not right.

pFe=-log[2.856 x 10^ -4]

This is not original concentration in the sample.

[DrCMS]

2.856 x 10^-4 mold Fe L^-1 Fe * (25ml/5ml) = .0142M Fe

Nothing on the left is correct, but miraculously number on the right is right.

Well the concentration in the analysed sample on the left is correct and the number on the right is not quite right but on the next line down is correct.

.01428M Fe (58.85 g/mol) = .7975g Fe

Everything on the left is OK, but - another miracle - number on the right is not right.

I think you need to look at the Atomic weight used which on the left which is wrong but when corrected makes the number on the right correct.

I think either we have a very sloppy chemist here in tarik3001 or someone who's coppied the answers from someone else and is trying to work out how they got the correct answer.