Map deb Sapling Learning Treatment of ammonia with phenol in the presence of hyp
ID: 1007645 • Letter: M
Question
Map deb Sapling Learning Treatment of ammonia with phenol in the presence of hypochlorite vields indophenol, a blue product absorbing light at 625 nm, which can be used for the spectrophotometric determination of ammonia. OCI NH3 OC indophenol anion 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 x 10 4 Mammonia solution and dilute to 25.0 mL (sample B). As a reagent blank, you mixt 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: Sample Absorbance (625 nm) 0.466 B 0.713 0.045 is the molar absorptivity (E)of the indophenol product, and what is the concentration of ammonia in the Number Number E- 367.3 NH MMA cm 2.8x 10 lake waterExplanation / Answer
In this case, we are spiking the concentration of NH3 in the sample solution (lake water) by adding NH3 solution as standard from outside. Let the lake water contain x M NH3.
Total volume of the solution in both cases = 25.0 mL
The concentration of NH3 in the lake water (sample A) is (10 mL)*(x M)/(25 mL) = 0.4x M …..(1)
Now in sample B, 2.50 mL of 5.50*10-4 M NH3 solution was added from outside. Therefore, moles of NH3 in sample B = [(2.50 mL) (5.50*10-4 M) + (10 mL)*(x M)]/(25 mL) = (1.375*10-3 + 10x)/25 …….(2)
Now, Beer’s law is followed; thus we must have
A = .C.l where A = absorbance of indophenol solution; is the molar absorptivity of indophenol and C is the concentration of NH3; l is the path length of the solution = 1.00 cm. is a constant for a particular compound.
Therefore, absorbance is directly proportional to concentration. Thus, we have,
A1 = 0.466 = .(0.4x).(1.00) …..(3)
A2 = 0.713 = .[(1.375*10-3 + 10x)/25].(1.00) …..(4)
Taking ratio of (4) and (3),
(0.713/0.466) = [(1.375*10-3 + 10x)/25]/0.4x
===> 1.5 = (1.375*10-3 + 10x)/10x
===> 15x = 1.375*10-3 + 10x
===> 5x = 1.375*10-3
===> x = 2.75*10-4 2.8*10-4
The concentration of NH3 in lake water = 2.8*10-4 M (there seems to be some error with the order here; following all the given data, I could arrive at an order that is 10 orders of magnitude less than the one given in the answer) (ans)