Map a Treatment of ammonia with phenol in the presence of hypochlorite yields in
ID: 561287 • Letter: M
Question
Map a 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 oci indophenol anion 95 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 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 100 Sample Absorbance (625 nm) 0.417 0.664 0045 What is the molar absorptivity (e) of the indophenol product, and what is the concentration of ammonia in the lake water? Number Number lake waterExplanation / Answer
The blank solution is (sample C) .....absorbance = 0.045
The standard solution is (sample B) .....absorbance = 0.664
Therefore, the absorbance of 5.50 x 10-4 M ammonia solution = Absorbance of sample B - Blank solution absorbance
= 0.664 - 0.045
= 0.619
we have [NH3] = 5.50 x 10-4 M
absorbance A = 0.619
path length l = 1 cm
Therefore, from Beer's law;
A = .c.l
0.619 = . 5.50 x 10-4 M . 1cm
= 1.125 x 103 M-1 cm-1
From the lake water sample (sample C), we have absorbance = 0.417
We have to remove the background absorbance = 0.417 - 0.045
= 0.372
A = .c.l
0.372 = 1.125 x 103 M-1 cm-1. c . 1cm
c = 3.3 x 10-4 M
[NH3]lake water = 3.3 x 10-4 M