Question

My answer is incorrect. The correct answer is 7.14x10^-4, but I am not sure how to calculate it.

The method of standard additions was used to determine nitrite in a soil sample. A 1.00-ml portion of the sample was mixed with 22.00 mL of a colorimetric reagent, and the nitrite was converted to a colored product that produced a blank-corrected absorbance of 0.300. To 51.00 mL of the original sample, 1.00 mL of a standard solution of 1.00×10 M nitrite was added. The same color-forming procedure was followed, and the new absorbance was 0.480. What was the concentration of nitrite in the original undiluted sample? Concentration = 3.11x10^-5 M An error has been detected in your answer. Check for typos, miscalculations etc. before submitting your answer.

Answer 1

Given Aroome portion of the sample was mixed with 22.00 ml ót a cobrimednie reagent A ber blank- (orreted abgor bance of 0.300.70 The original Sample = 51.00mL standard solution - 100 mb Nitrite = 1.00 x16 *M The new absorbance was 0.480. let the concentration of nitrite in the original Sample be a. moles of nitrite in 100 ml Sample - volume x loncentration | : ' xa = 0.001 a mol 1001 Total volume after mixing with colorimetric agent =1+22.00 :23 m2 -0.0231 troetel volume after mixing with colorimetric agent $ 0.0232

concentration of nitrite obter mining with colorimetric agent = moles . Total volume = 0.001a 0.023 = 0.043a concentration of nitrile after mixing with colorime-tic agent = 0.043a From Beer's law, ... Absorbance A = 20 Where & is molar absorptivity cix the concentration' and lis path length =Y 0.300 = EX0.04 30xd = 0.300 = 0.043a el = a: 0.300 0.04381 a = 6.9/41

Alter adding Standard Solution. moles of nitrite added to 5100 mL of sample = volume xconcentration = 100. 1000 -X0.001 = 1x10-6 mol moles of nitrite originally in 50 ml Sample = volumex concentration = Godoxa = 0.051a . Moles of nitrite Originally in some Sample-0-051a total volume after mising = 517) = 52 m2 20.0522 it total volume after mining = 0.0522! mining = 0.052 L New concentration of solution = Total moles Total volume

- 0.05 latlx106 0.0512 0.051001 0.0512 = 0.99610 0.99610+ 1.961x165 moles of nitrite in 100 m2 of new Sample e volume x concentration 59000 * 10.996 kr 1961X155) = 0.00099612 +1.961x10mg Total volume after mixing with colorimetric agent - H+22 = 23 m2 20.0232 I'mcentration of nitsite oftex mia ing with colorimetric agent = te volume = 10.00099612 +1.961416 B 0.023 = 0.04330 a + 8.526x10-7

from Beer's law, Absorbance (A JE &c where ea molar absorptivity -> => => c= concentration la path length 0.480 = € 10.04330 +B.5268107) xd .0.480 = 0.04330 a El +8.526x1075l 10.043300 +8:526x1038420-480 = Q = 0:480 0.0433091 -8.52631007 => a = 11.0854 ed - 2x10-5 This only = 1.0854 2105 = 6.9 – 11.0854 + 2x105 El-o => -4.1854 +2x105clco. =7 281055 81 3.4.1854

4.1854 2x10-5 El = 209,270 a as good 209270 I a= 3.29x105m z