Question

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

Answer 1

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

Explanation:

Answer 2

Molarity of ammonia in lake water is 0.00508 $\times\;10^5$ M.

A. Lake warm water + Phenol + Sodium hypochlorite = 0.419

B. Lake warm water + Phenol + Sodium hypochlorite + Ammonia = 0.666

C. Distilled Water + Phenol + Sodium hypochlorite = 0.045 (Blank)

The given cuvette has the path length = 1 cm.

The absorbance of sample A and sample B has been calculated by subtracting the blank.

Absorbance of A = 0.419 - 0.045 = 0.374

Absorbance of B = 0.666 - 0.045 = 0.621

The absorbance of sample B has been affected by the ammonia concentration.

The concentration of Ammonia in sample B =  $\rm 5.50\;\times\;10^-^4$ M (2.50 ml)

The moles of ammonia in 25 ml solution = Molarity $\times$ volume (L)

The moles of ammonia in 25 ml solution =  $\rm 5.50\;\times\;10^-^4$ M $\times$ 0.025 L

The moles of ammonia in 25 ml solution = 1.3 $\rm \times\;10^-^5$ moles.

The moles of ammonia in sample B = moles of Ammonia in sample A + moles of ammonia in sample B

The moles of ammonia in sample B = moles of Ammonia in sample A + 1.3 $\rm \times\;10^-^5$ moles.

According to Beer's law,

Absorbance = molar coefficient $\times$ concentration $\times$ path length

The path length has been 1 cm, and the coefficient has been same, the ratio of the absorbance of two samples has been equal to their concentration.

$\rm \dfrac{Absorbance\;of\;A}{Absorbance\;of\;B}\;=\;\dfrac{Concentration\;of\;A}{Concentration\;of\;B}$

$\rm \dfrac{0.374}{0.621}\;=\;\dfrac{moles\;of\;A}{1.3\;\times\;10^-^5\;+\;moles\;of\;A}$

Moles of A =  0.508 $\rm \times\;10^5$

The moles of A has been the concentration of ammonia in lake water.

The volume of lake water has been = 10 ml

Molarity of ammonia = moles $\times$ volume (L)

Molarity of ammonia in lake water = 0.508 $\rm \times\;10^5$ $\times$ 0.01

Molarity of ammonia in lake water = 0.00508 $\times\;10^5$ M

The coefficient of molar absorptivity can be calculated as:

Absorbance = molar coefficient $\times$ concentration $\times$ path length

0.374 = molar coefficient $\times$ 0.00508 $\times\;10^5$ M $\times$ 1

Molar coefficient of Ammonia = 73.62 $\rm \times\;10^-^5\;M^-^1\;cm^-^1$.

Molarity of ammonia in lake water is  0.00508 $\times\;10^5$ M.

For more information about the concentration, refer to the link:

https://brainly.com/question/1600559