Question

A 2.00-mL urine sample was treated with molybdenum blue reagent to produce a species absorbing at 820nm, after which the sample was diluted to 100mL. A 25.00mL aliquot of this solution gave an absorbance of 0.428. Addition of 1.00 mL of a standard solution containing 0.0500 mg/mL of phosphate to a second 25.00-mL aliquot gave an absorbance of 0.517. Use these data to calculate the concentration of phosphate in milligrams per milliliter of the original urine sample.

Answer 1

This is standard addition method

absorbance due to 25 ml aliquot solution without standard (Ax) = 0.428

after adding standard

concentration of standard in solution (Cs) = 1 ml x 0.05 mg/ml/26 ml = 0.0019 mg/ml

absorbance of solution with standard (As) = 0.517

let Cx be the concentration of phosphate in unknown solution

Using,

Cx/(Cx + Cs) = Ax/As

feeding values,

Cx/(Cx + 0.0019) = 0.428/0.517

0.517Cx - 0.428Cx = 0.0019

Cx = 0.0213 mg/ml phosphate in 25 ml aliquot

So,

concentration of phosphate in original 2 ml solution = 0.0213 x 100/25 = 0.0852 mg/ml