Question

Based on previous failed attempts, I think the answer is around 5.80x10^-4.

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

Let m moles of nitrite were present in the original sample.

Absorbance is given by beer lambert's law:

$A=\varepsilon \: l\: c$

0.3 = $\varepsilon \: l$ x ( m / 23 )

$\varepsilon \: l$ = 6.9 / m

Now,

Moles of nitrite from standard solution added = 10^-3 x 10^-3 = 10^-6

Total moles of Nitrite after dilution = m + 10^-6

$A=\varepsilon \: l\: c$

0.48 = ( 6.9 / m ) x ( m + 10^-6 / 52)

3.6174 m =  m + 10^-6

m = 3.82 x 10^-7 moles

Concentration in original solution = 3.82 x 10^-7 moles / 0.051 L

Concentration in original solution = 7.49 x 10^-6 M