Sapling Learning Treatment of ammonia with phenol in the presence of hypochlorit
ID: 542179 • Letter: S
Question
Sapling Learning 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. Ma oCI To determine the ammonia concentration in a sample of lake water, you mix 100 mL of lake water with ration in a sample of lake water, you mix lake water with lorite solution and dilute to 25.0 mL in a volumetric 2 mL of sodium indophenol anion 5 mL of phenol solution and 2 mL of sodium flask (sample A). To a hypochlorite, and 2.50 ml of a 5,50 x10-4M 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 civet Sample Absorbance (625 nm) 0436 0.683 0.045 What is the molar absorptivity (e) of the indophenol product, and what is the concentration of ammonia in the lake water? Number IMI"cm il [NH,]lake water = y HintExplanation / Answer
The reagent blank contains no ammonia; hence, we must subtract the absorbance due to the reagent blank from the absorbance for solutions A and B. The corrected absorbances are
A: 0.436 – 0.045 = 0.391
B: 0.683 – 0.045 = 0.638
Let the lake water contain c M = c mol/L NH3. We take 10.0 mL of the lake water; hence, the millimoles of NH3 in the lake water is (10 mL)*(c mol/L) = 10c mmole.
The total volume of the solution for the both solutions A and B is 25.0 mL; hence, the concentration of NH3 in the diluted solution A is (10c mmole)/(25.0 mL) = 0.4c M.
We add 2.50 mL of 5.50*10-4 M NH3 solution to prepare solution B. The total number of millimoles of NH3 is (10c + 2.50*5.50*10-4) mmole = (10c + 0.001375) mmole.
The concentration of NH3 in solution B is (10c + 0.001375)mmole/(25.0 mL) = (10c + 0.001375)/25 M.
Now, write the Beer’s law expression for both the solutions. Beer’s law is expressed as
A = *[..]*l where = molar absorptivity of the compound, [..] = molar concentration of the compound and l = path length of the solution.
Given than l = 1.00 cm, we have,
0.391 = *(0.4c M)*(1.00 cm) …..(1)
0.638 = *(10c + 0.001375)/(25)*(1.00 cm) …..(2)
Divide (2) by (1) and get
0.638/0.391 = [(10c + 0.001375)/(25)]/(0.4c)
===> 1.63171 = (10c + 0.001375)/(10c)
===> 16.3171c = 10c + 0.001375
===> 6.3171c = 0.001375
===> c = 2.15798*10-4 2.16*10-4
The concentration of ammonia in lake water, i.e, [NH3]lake water = 2.16*10-4 M (ans).
Put the value of [NH3]lake water in (1) and get
0.391 = *(2.16*10-4 M)*(1.00 cm)
====> = (0.391)/(2.16*10-4 M).(1.00 cm) = 1810.185185 M-1 cm-1 1810.2 M-1 cm-1 (ans).
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