Procedure for determination: 1) To 50.0 mL of Unknown solution containing nitrit

ID: 817325 • Letter: P

Question

Procedure for determination:

1) To 50.0 mL of Unknown solution containing nitrite is added 1.00 mL of sulfanilic acid solution

2) After 10 min, 2.00 mL of 1-aminonaphthalene solution and 1.00 mL of buffer are added

3) After 15 min, the absorbance is 520 nm in a 5.00-cm cell

The following solutions were analyzed:

A) 50.0 mL of food extract known to contain no nitrite (that is, a negligible amount); final absorbance = 0.153

B) 50.0 mL of food extract suspected of containing nitrite; final absorbance = 0.622

C) Same as B, but with 10.0 microliters of 7.50 x 10^-3 M NaNO2, added to the 50.0-mL sample; final absorbance = 0.967

Calculate the molar absorptivity, epsilon, of colored product. The 5.00 cm cell was used

How many micrograms of No2 were present in 50.0 mL of food extract?

Beer's Law

A=ebc

A) A = 0.153

B) A = 0.622

let concentration of nitrite be C

absorbence due to nitrite - 0.153 = 0.469

by beer's law

0.469 = e * 5 * C

C) A = 0.967

as 10.0 microliters of 7.50 x 10^-3 M NaNO2, added to the 50.0-mL sample

concentration of nitrile in sample now = (0.05*C +7.5*10^-8)/0.05 = H

absorbence due to nitrite - 0.153 = 0.814

A= e* 5 * H

0.814 = (0.469/5*C) * 5 * (0.05*C +7.5*10^-8)/(0.05)

0.814 *C*0.05 = 0.469 * (0.05*C +7.5*10^-8)

0.0407*C = 0.02345 * C + 3.5175 *10^(-8)

(0.01725) * C = 3.5175 *10^(-8)

C = 2.04 * 10^(-6) mol L-1

grams of No2 were present in 50.0 mL of food extracT = 2.04 * 10^(-6) * 0.05 * 68.9953 = 7.03 * 10(-6) grams

= 7.03 micrograms

0.469 = e * 5 * C

0.469 = e * 5 * 2.04 * 10^(-6)

e = 45980.4 L mol-1 cm-1

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