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. 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 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 (e) of the indophenol product, and what is the concentration of ammonia in the lake water?

Explanation / Answer

we see 1 mole of Nh3 gives 1 mole product

in B moles of NH3 = moles of NH3 in A + (5.5 x10^-4 x2.5/1000) = 1.375 x10^6 + mA

( mA = moles of Nh3 in A) vol of B = 25 = vol of A

now A = el C = eC ( since l = 1cm)

in net absorbance due to complex blanck absorbance must be removed hence

here A(A) = 0.536-0.045 = 0.491 , A(B) = 0.783-0.045 = 0.738

A2/A1 = C2/C1 , A(B)/A(A) = ( 1.375x10^-6 +mA)/(mA) = 0.738/0.491

mA = 2.733 x 10^-6

mA = 2.733x 10^-6 = moles of Nh3 in A ( lake water )

hence [NH3] water ( 2.733 x10^-6 ) x 1000/25 = 1.093 x 10^-4 M

now lake water vol = 10 ml out of 25 ,

hence conc in lake water = 2.733 x10^-6 ) x 1000/10 = 2.733 x 10^-4 M

now A = 0.491 = e x 1 x 1.093 x10^-4

e = 4492 M-1cm-1

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