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:

Sample	Absorbance(625 nm)
A	.385
B	.632
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

Let the concentration of ammonia in the lake water be c mol/L. In the first experiment, we use 10.0 mL of the lake water and dilute to a final volume of 25.0 mL.

Use the dilution equation:

M₁*V₁ = M₂*V₂

===> (10 mL)*(c mol/L) = (25.0 mL)*M₂

===> M₂ = (10*c)/(25.0) mol/L = 0.4c mol/L.

Determine the number of millimoles of ammonia in 10.0 mL of the lake water = (10.0 mL)*(c mol/L) = 10c mmol.

Determine the number of millimoles of ammonia solution added to sample B = (2.50 mL)*(5.50*10^-4 M) = 0.001375 mmol.

Determine the concentration of ammonia in sample B = (millimoles of ammonia)/(final volume of solution) = (10c + 0.001375) mmol/(25.0 mL) = (10c + 0.001375)/25 M (1 mol/L = 1 M).

Find out the corrected absorbance for both the solutions by subtracting the absorbance due to the blank.

Sample A = 0.385 – 0.045 = 0.340

Sample B = 0.632 – 0.045 = 0.587

Write down the Beer’s law expression for both.

0.340 = ε*(0.4c)*(1.00 cm) …..(1)

0.587 = ε*[(10c + 0.001375)/25]*(1.00 cm) …..(2)

Divide (2) by (1).

0.587/0.340 = [(10c + 0.001375)/25]/(0.4c) = (10c + 0.001375)/10c

===> 1.7265 = (10c + 0.001375)/10c

===> 17.265c = 10c + 0.001375

===> 7.265c = 0.001375

===> c = 0.00018926 = 1.8926*10^-4 ≈ 1.90*10^-4

The concentration of ammonia in the lake water is 1.90*10^-4 M (ans).

Put the value of c in (1) and obtain

0.340 = ε*(0.4*1.90*10^-4 M)*(1.00 cm)

===> 0.340 = ε*7.6*10^-5 =M.cm

===> ε = (0.340)/(7.6*10^-5 M.cm) = 4473.6842 M^-1cm^-1 ≈ 4473.68 M^-1cm^-1

The molar absorptivity of ammonia is 4473.68 M^-1 cm^-1 (ans).