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

Question

Treatment of ammonia with phenol in the presence of hypochlorite vields indophenol, a blue product absorbing light at 625 nm, which can be used for the spectrophotometric determination of ammonia OCI H NH3 indophenol anion 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 x10-4 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) 0.352 0.599 0.045 What is the molar absorptivity (E) of the indophenol product, and what is the concentration of ammonia in the lake water? Number Number M-1cm-11 [NH = 3 Jlake water

Explanation / Answer

Let the concentration of ammonia, NH3 in lake water be x M.

Millimoles of NH3 (lake water) in samples A and B = (10 mL)*(x M) = 10x mmole.

Total volume of solutions A and B = 25.0 mL; molar concentration of NH3 in lake water = (10x mmole)/(25.0 mL) = 0.4x M.

We add 2.50 mL of 5.50*10-4 M NH3 solution to solution B; therefore, millimoles of NH3 added = (2.50 mL)*(5.50*10-4 M) = 0.001375 mmole.

Millimoles of NH3 in sample B = (10x + 0.001375) mmole; the concentration of NH3 in sample B is (10x + 0.001375) mmole/25.0 mL = (10x + 0.001375)/25.0 M.

Find out the corrected absorbance for solutions A and B by subtracting the absorbance of the blank from the given absorbance. The corrected absorbances are

Solution A: 0.352 – 0.045 = 0.307

Solution B: 0.599 – 0.045 = 0.554

Use Beer’s law to write down the expression for the absorbance. Beer’s law is

A = *C*l where A = absorbance; = molar absorptivity; C = molar concentration and l = path length of the solution.

Given l = 1.00 cm, we must have

0.307 = *(0.4x M)*(1.00 cm)

======> = 0.307/(0.4x M.cm) ……(1)

0.554 = *[(10x + 0.001375)/25.0 M]*(1.00 cm)

======> = 0.554*(25.0)/[(10x + 0.001375) M.cm] …..(2)

is a property of a particular compound and hence, is constant; therefore, we must have,

0.307/(0.4x M.cm) = 0.554*(25.0)/(10x + 0.001375)

====> 0.7675/x = 13.85/(10x + 0.001375)

====> 0.7675*(10x + 0.001375) = 13.85x

====> 7.675x + 0.0010553125 = 13.85x

====> 6.175x = 0.0010553125

====> x = 1.7090*10-4 1.71*10-4

The concentration of NH3 in lake water is 1.71*10-4 M (ans).

Put the value of x in expression 1 and obtain

= 0.307/(0.4*1.71*10-4 M.cm) = 4488.30 M-1cm-1 (ans).

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