Treatment of ammonia with phenol in the presence of hypochlorite yields indophen
ID: 971809 • Letter: T
Question
Treatment of ammonia with phenol in the presence of hypochlorite yields indophenol, a blue product absorbing light at 625 nm, which can be used for the spectrophotometric determination of ammonia. To determine the ammonia concentration in a sample of lake water, you mix 10.0 mL of lake water with 5 mL of phenol solution and 2 mL of sodium hypochlorite solution and dilute to 25.0 mL in a volumetric flask (sample A). To a second 10.0 mL solution of lake water you add 5 mL of phenol. 2 mL of sodium hypochlorite, and 2.50 mL of a 5.50 times 10^-4 M ammonia solution and dilute to 25.0 mL (sample B). As a reagent blank, you mix 10.0 mL of distilled water with 5 mL of phenol, 2 mL of sodium hypochlorite and dilute to 25.0 mL (sample C). You measure the following absorbances using a 1.00 cm cuvet: What is the molar absorptivity (epsilon) of the indoPhenol product, and what is the concentration of ammonia in the lake water?Explanation / Answer
From Beer-Lambert law, A = e * c * l
for solution B, concentration of NH3 is given as 5.50 x 10^-4 M
initi's initial volume taken is 2.5 ml and after dilution it is 25 ml
so, we will use M1V1 = M2V2 to get the diluted concentration.
M2 = M1V1/V2 = (5.50 x 10^-4 M x 2.5 ml)/25 ml = 5.5 x 10^-5 M
absorbance,A is the difference between recorded value and blank value.
So, A = 0.678 - 0.045 = 0.633
e = A/(c*l)
e = 0.633/(5.5 x 10^-5 Mx 1.0 cm)
e = 1.15 x 10^4 M^-1 cm^-1
Now, for solution (A), absorbance,A would be the difference of sample A and sample C values.
A = 0.431 - 0.045 = 0.386
now we have the values of A, e and l. so, concentration,c could easily be calculated.
c = A/(e*l)
c = 0.386/(1.15 x 10^4 M^-1 cm^-1 x 1.0 cm)
c = 3.36 x 10^-5 M
This calculate concentraion is the concentration of diluted solution. so by using M1V1 = M2V2, we could calculate the concentration of NH3 in the lake.
M1 x 10 ml = 3.36 x 10^-5 M x 25 ml
M1 = 3.36 x 10^-5 M x 25ml/10ml = 8.4 x 10^-5 M
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