Phosphorus in urine can be determined by treating with molybdenum (VI) and then
ID: 880793 • Letter: P
Question
Phosphorus in urine can be determined by treating with molybdenum (VI) and then reducing the phosphomolybdo complex with aminonaphtholsulfonic acid to give the characteristic molybdenum blue color. This absorbs at 660 nm. A patient excreted 1570 mL of urine in 24 hours and the pH of the urine was 6.5. A 1.00 mL aliquot of the urine was treated with molybdate reagent and aminophtholsulfonic acid and was diluted to a volume of 50.0 mL. A series of phosphate standards were similarly treated. The absorbance of the solutions at 660 nm, measured against the blank were as follows:
SOLUTION
ABSORBANCE
1.00 ppm P
0.205
2.00 ppm P
0.410
3.00 ppm P
0.615
4.00 ppm P
0.820
urine sample
0.652
Calculate the number of grams of phosphorus excreted per day.
SOLUTION
ABSORBANCE
1.00 ppm P
0.205
2.00 ppm P
0.410
3.00 ppm P
0.615
4.00 ppm P
0.820
urine sample
0.652
Explanation / Answer
A1/A2 = C1/C2
where,
A1 = 0.82
A2 = 0.652
C1 = 4 ppm
C2 = unknown
feed values,
C2 = 4 x 0.652/0.82 = 3.1805 ppm
1 ppm = 0.001 g/L
So,
3.1805 = 3.1805 x 10^-3 g/L of phosphorous is in 0.05 L of solution
Total volume of urine 1570 mL = 1.57 L
So, we have 3.1805 x 10^-3 x 1.57 = 0.10 g of phosphorous excreated per day