Treatment of ammonia with phenol in the presence of hypochlorite yields indophen

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. determine the ammonia concentration in a sample of lake water, you mix 10.0 mL of lake water with 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? Epsilon = Number M^-1 cm^-1 [NH_3]_lake water = Number M

Explanation / Answer

Absorbance - blank = actual absorbance

For sample A, absorbance = 0.360 - 0.045 = 0.315 [Ax]

For sample B, absorbance = 0.607 - 0.045 = 0.562 [A]

Sample B, [NH3] added = 5.50 x 10^-4 M x 2.50 ml/25 ml = 5.50 x 10^-5 M [Cs]

So total ammonia in solution B = Cs + Cx

Cx = NH3 concentration in lake water

Sample A : without additional NH3 added [Cx]

So using method of standard addition,

Cx/(Cx + Cs) = Ax/A

Cx/(Cx + 5.50 x 10^-5) = 0.315/0.562

0.562Cx - 0.315Cx = 1.7325 x 10^-5

NH3 concentration in 25 ml diluted lake sample = Cx = 7.0142 x 10^-5 M

NH3 concentration in original 10 ml lake sample = 7.0142 x 10^-5 M x 25/10 = 1.75355 x 10^-4 M

molar absorptivity = absorbance/concentration x path length

                             = 0.315/7.0142 x 10^-5 x 1

                             = 4491 M-1.cm-1

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