Question

The urine sample containing phosphate was treated with molybdenum blue reagent to produce a species absorbing at 820 nm with an absorbance of 0.428. Then 1.00 mL of a 5.00 g/L phosphate standard was added to 104.0 mL of the urine sample. The spiked urine sample gave an absorbance of 0.517. Find the original concentration of phosphate in g/L in the urine sample. Correct sigfigs!

Answer 1

concentration of standard in the solution = 5 g/L x 1 ml/105 ml

= 0.04762 g/L

let Cx be the concentration of phosphate in the urine sample

then,

Cx/(Cx + 0.04762) = 0.428/0.517

0.517Cx = 0.428Cx + 0.204

Cx = 2.292 g/L

Is the needed concentration of phosphate