10. The method of standard addition was used to determine nitrite in a soi sampi
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Question
10. The method of standard addition was used to determine nitrite in a soi sampic. A 1.00-mL portion of the sanmple was mixod with 24.00m the nitrite couverted to a colorcd product whase absorbauce was determined to h ti.300t 50 00 ml of the original sample 1.00 InL of the standard solution of 1 was added ellowing the same procedur 0.530. what is the conecntrat on of the nitrite in the original undiluted sumple? i of a colorimetric reagent and -3 nine c as hefore. The absorbanec was imeasured to beExplanation / Answer
Let the concentration of nitrite in the original sample be a.
Moles of nitrite in 1.00 mL sample = volume x concentration
= 1.00/1000 x a = 0.001a mol
Total volume after mixing with colorimetric agent = 1 + 24 = 25 mL = 0.025 L
Concentration of nitrite after mixing with colorimetric agent = moles/total volume
= 0.001a/0.025 = 0.04a
From Beer's law, absorbance A = cl where is molar absorptivity, c is concentration and l is path length
0.300 = x 0.04a x l => a = 7.5/l
After adding standard solution:
Moles of nitrite added to 50 mL of sample = volume x concentration
= 1.00/1000 x 0.001 = 1 x 10-6 mol
Moles of nitrite originally in 50 mL sample = volume x concentration
= 50/1000 x a = 0.05a mol
Total volume after mixing = 50 + 1 = 51 mL = 0.051 L
New concentration of solution = total moles/total volume
= (0.05a + 1 x 10-6)/0.051 = 0.9804a + 1.961 x 10-5
Moles of nitrite in 1.00 mL of new sample = volume x concentration
= 1.00/1000 x (0.9804a + 1.961 x 10-5) = 0.0009804a + 1.961 x 10-8
Total volume after mixing with colorimetric agent = 1 + 24 = 25 mL = 0.025 L
Concentration of nitrite after mixing with colorimetric agent = moles/total volume
= (0.0009804a + 1.961 x 10-8)/0.025 = 0.039216a + 7.844 x 10-7
From Beer's law, absorbance A = cl where is molar absorptivity, c is concentration and l is path length
0.530 = x (0.039216a + 7.844 x 10-7) x l => a = 13.515/l - 2 x 10-5
Thus 7.5/l = 13.515/l - 2 x 10-5
l = 300750
a = 7.5/300750 = 2.49 x 10-5 M
Concentration of nitrite in the original undiluted sample = 2.49 x 10-5 M
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