Help with Absorbance question

Question

Treatment of ammonia with phenol in the presence of hypochlorite yields indophenol, a blue product that absrobs 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, three samples are prepared. In sample A, 10.0 mL of lake water is mixed with 5 mL of phenol solution and 2 mL of sodium hypochlorite solution, and diluted to 25.0 mL in a volumetric flask. In sample B, 10.0 mL of lake water is mixed with 5 mL of phenol solution, 2 mL of sodium hypochlorite solution, and 2.50 mL of a 5.50×10−4 M ammonia solution, and diluted to 25.0 mL. Sample C is a reagent blank. It contains 10.0 mL of distilled water, 5 mL of phenol solution, and 2 mL of sodium hypochlorite solution, diluted to 25.0 mL. The absorbance of the three samples is then measured at 625 nm in a 1.00 cm cuvette. The results are shown in the table.

SampleAbsorbance (625 nm)
A	0.3620.362
B	0.6090.609
C	0.0450.045

What is the molar absorptivity (𝜖) of the indophenol product?

What is the concentration of ammonia in the lake water?

Answer 1

To calculate molar absorptivity (𝜖) for the indophenol product, we will be using Lambert-Beer law equation, as shown below:

A = C × L × 𝜖

where A is absorbance, C is concentration of the sample (M), and L is optical path length (cm), and 𝜖 is molar absorptivity (M^-1 cm^-1).

First, we need to subtract the Absorbances of sample B and A with the Absorbance of the blank sample C.

A_B - A_C = 0.609-0.045 = 0.564

A_A - A_C = 0.362-0.045 = 0.317

Moles of ammonia in the sample A is unknown (x)

To calculate the concentration of ammonia in the lake water sample, we need to do the following:

Total amount of moles of ammonia in the sample B (from the lake water sample as well as added ammonia) is x + 5.5 × 10^-4 M × 0.0025 L = x + 1.38 x 10^-6 moles

A_{A /} A_{B =} x / x + 1.38 x 10^-6 moles

(0.317/0.564) = x / x + 1.38 x 10^-6 moles

5.62 x 10^-1 = x / x + 1.38 x 10^-6 moles

5.62 x 10^-1 x + 7.76 x 10^-7 moles = x

x = 1.88 x 10^-6 moles of ammonia in the sample A

Concentration of ammonia in the sample A: [NH₃] = 1.88 x 10^-6 moles / 0.001 L = 1.88 x 10^-4 M

molar absorptivity is: 𝜖 = A/ C × L

𝜖 = 0.317/7.51 x 10^-5 M cm

𝜖 = 4.491 × 10³ M^-1 cm^-1 or simply 4491

 

Answer 2

Net absorbance values:

A (10.0 mls lake water) = 0.3620 - 0.0450 = 0.317

B (10.0 mls lake water + 5.50x10^-4 M NH₃) = 0. 6090 - 0.0450 = 0.564

note: molar ratio of NH₃ to indolphenol is 1:1, thus....

Concentration of NH₃ in B = 2.5 ml x 1 L / 1000 ml x 5.50x10^-4 mol / L / 0.025 L = 5.50x10^-5 M

Absorbance of 5.50x10^-5 M = 0.564 - 0.317 = 0.247

Molar absorptivity (𝜖) of indophenol = 0.247 / 5.50x10^-5 M = 4491 M^-1 cm^-1 (assuming a 1 cm cuvette)

[NH₃] in lake water:

A = 𝜖cl = 4491 M^-1 cm^-1

0.317 = (4491 M^-1 cm^-1) (c)(1 cm)

c = 7.06x10^-5 M in the cuvette

sample in cuvette is 10 ml lake water diluted to 25 mls, or a 2.5x dilution

[NH₃] in original lake water sample = 7.06x10^-5 M x 2.5 = 1.77x10^-4 M