24 June, 22:44

# 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 cuvet. The results are shown in the table. Sample Absorbance (625 nm) A 0.419B 0.666C 0.045What is the molar absorptivity (?) of the indophenol product?=M-1cm-1What is the concentration of ammonia in the lake water?[NH3]lake water=M

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1. 25 June, 00:13
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Answer:Sample Absorbance (625 nm)

A 0.536

B 0.783

C 0.045

Therefore, I will use these data to solve your question. If you have other absorbances values, just follow my steps and plug in different numbers.

First, 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)

Because, n net absorbance due to complex blank absorbance must be removed.

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

(you can plug in different numbers in this step)

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

So, mA = 2.733 x 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

Lake water vol = 10 ml out of 25,

Concentration of ammonia in lake water = 2.733 x10^-6 x 1000/10 = 2.733 x 10^-4 M

Then, A = 0.491 = e x 1 x 1.093 x10^-4

e = 4492 M-1cm-1