Solution-
Let us assume the original sample have the concentration of nitrite be a.
So the moles of nitrite in 1.00 mL sample = volume x concentration
= 1.00/1000*a = 0.001a mol
Let the total volume after mixing with colorimetric agent = 1 + 22 = 23 mL = 0.023 L
Concentration of nitrite after mixing with colorimetric agent = moles/total volume
= 0.001a/0.023 = 0.04a
Now using the Beer's law, absorbance A = εcl [here ε = molar absorptivity; c = concentration ; l = path length
0.300 = ε x 0.04a x l => a = 7.5/εl
After adding standard solution:
The moles of nitrite added to 51 mL of sample = volume x concentration
= 1.00/1000 x 0.001
= 1 x 10^-6 mole
Moles of nitrite originally in 51 mL sample = volume x concentration
= 51/1000 x a = 0.051a mol
Now the total volume after mixing = 51 + 1 = 52 mL = 0.052 L
New concentration of solution = total moles/total volume
= (0.051a + 1*10^-6)/0.052 = 0.980a + 1.92e-5
Now the moles of nitrite in 1.00 mL of new sample = volume x concentration
= 1.00/1000 x (0.980a + 1.92e-5) = 0.00098 a + 1.92e-8
So the total volume after mixing with colorimetric agent = 1 + 22 = 23 mL = 0.023 L
Concentration of nitrite after mixing with colorimetric agent = moles/total volume
= (0.00098 a + 1.92e-8)/0.025 = 0.0392a + 7.68e-7
Using the Beer's law, absorbance A = εcl [here ε = molar absorptivity, c = concentration , l = path length]
=>0.480 = ε x (0.0392a + 7.68e-7) x l
=> 0.0392aεl+ 7.68e-7εl= 0.480
=> εl(0.0392a+7.68e-7)= 0.480
=> (0.0392a+7.68e-7)= (0.480/ εl)
=> (0.480/ εl)-7.68e-7 = 0.0392a
=> a = 13.515/εl - 1.96e-5
=>7.5/εl-13.515/εl= - 1.96e-5
=> 6.015 /εl=1.96e-5
=>εl = 306887.75
a = 7.5/306887.75
= 2.44* 10^-5 M Answer
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