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 × 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: Absorbance (625nm) Sample A: .361 Sample B: .608 Sample C: .045 What is the molar absorptivity (ε) of the indophenol product, and what is the concentration of ammonia in the lake water?

Answer 1

Assumption:ammonia is completely converted to indophenol.

Hence amount of indophenol formed upon adding phenol and sodium hypochlorite (each added in same quantities for sample A,B and blank) will be proportional to the amount of ammonia present in the sample.

Indophenol hence is a colourimetric probe, and according to BEER-LAMBER law,its absorbance will be directly proportional to it's concentration,

A=cl ,=molar absorptivity of indophenol

              c=concentration of indophenol & l=path length of the cuvette=1cm; constant in this case.

So absorbance of each sample measured will be proportional to the ammonia concentration in it.

Sample	Absorbance	Corrected Absorbance
C (Blank)	.045	0
A	.361	.316
B	.608	.563

The blank, doesn't have ammonia in it, but still gave a non-zero absorbance, which may be due to instrumentation error or phenol or any impurities(lees likely though).

So we need to substract this value from absorbances corresponding to samples A & B to get the absorbance soley due to indophenol.

n=no. of moles of ammonia in 10 mL of lake water.

no of moles of ammonia in each sample:

• Sample A=n

Sample B=n+n'(n'= no. of moles of ammonia in 2.5 mL of 5.50 × 10–4 M ammonia solution.)

n'=5.50 × 10–4 moles/L × 2.5 × 10-3 L=1.375 ×10-6 moles=1.375 m

SInce Samples A & B have same volume, concentration of .indophenol and hence ammonia, will be proportional of no. of moles of indophenol (ammonia in disguise!) present.

Using Beer Lambert's law:

A(SampleA)/A(SampleB)=.316/.563=.561=n/(n+n'), n'/n=0.782,

n=n'/.782=1.375 m/.782=1.756 m

Concentration (Molarity) of ammonia in lake water=1.756 x 10-6 moles/(10 x 10-3L)=0.176mM.

for sample A, assuming 1 mole of ammonia gives same quantity of indophenol(which is indeed true, this reaction is called Berthelot's reaction, ans in this, 1 equivalent/mol of ammonia gives same of indophenol, if it is the limiting reagent),

concentration of indophenol=.176x10-3M.

using A=cl,   =.316/(.176x10-3Mx1cm)=1795.4 M cm-1.