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icang [17]
3 years ago
10

Taking advantage of their large differences in pKa values, describe how a mixture of phenol and benzoic acid in diethyl ether so

lution could be separated. How could a mixture of phenol and cyclohexanol in diethyl ether solution be separated?
Chemistry
1 answer:
Marat540 [252]3 years ago
6 0

Answer:

By adding bicarbonate.

Explanation:

The pka of the phenol (C₆H₅OH) is 10 and the pka of the benzoic acid (C₆H₅COOH) is 4, which means that the benzoic acid is a stronger acid than phenol, so if we want to separate phenol from benzoic acid in diethyl ether we need to first use a weak base that will react with benzoic acid and not with the phenol:  

C₆H₅-COOH + HCO₃⁻  ⇄  C₆H₅-COO⁻  +  H₂CO₃

C₆H₅-OH + HCO₃⁻  ⇄  no reaction

The reaction of the benzoic acid with bicarbonate will produce the benzoate ion that will be soluble in the aqueous layer, while the phenol will remain dissolved in the organic layer, so we can separate the two of them by the separation of the two immiscible layers.      

Having the two layers separated, the benzoic acid can be recovered from the aqueous layer by adding HCl:

C₆H₅-COO⁻ + HCl  ⇄  C₆H₅-COOH + Cl⁻

<u>This acid will precipitate from the aqueous solution, and the solid can be isolated by filtration</u>.  

The phenol in the organic layer can be dissolved into an aqueous layer by the adding of a strong base like NaOH:

C₆H₅-OH + OH⁻  ⇄  C₆H₅-O⁻ + H₂O

The phenoxide ion soluble in the aqueous layer can be recovered later by the adding of HCl, which will form the original phenol:

C₆H₅-O⁻ + HCl  ⇄  C₆H₅-OH + Cl⁻  

<u>The precipitated phenol can be isolated by filtration. </u>

This way we can separate a mixture of phenol and benzoic acid in diethyl ether solution.  

I hope it helps you!

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Answer:

Option (d) is correct

N³⁻ > F⁻ > Mg²⁺ > Si⁴⁺

Explanation:

Total electrons for all the species = 10

So these are <u>iso electronic</u> with each other.

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Ionic radii ∝ \frac{Magnitude of Negative Charge}{Magnitude of Positive Charge}

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How many moles of NH3 can be produced from 12.0 mol of H2 and excess N2? Express your answer numerically in moles. View Availabl
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Answer:

A) 8.00 mol NH₃

B) 137 g NH₃

C) 2.30 g H₂

D) 1.53 x 10²⁰ molecules NH₃

Explanation:

Let us consider the balanced equation:

N₂(g) + 3 H₂(g) ⇄ 2 NH₃(g)

Part A

3 moles of H₂ form 2 moles of NH₃. So, for 12.0 moles of H₂:

12.0molH_{2}.\frac{2molNH_{3}}{3molH_{2}} =8.00molNH_{3}

Part B:

1 mole of N₂ forms 2 moles of NH₃. And each mole of NH₃ has a mass of 17.0 g (molar mass). So, for 4.04 moles of N₂:

4.04molN_{2}.\frac{2molNH_{3}}{1molN_{2}} .\frac{17.0gNH_{3}}{1molNH_{3}} =137gNH_{3}

Part C:

According to the <em>balanced equation</em> 6.00 g of H₂ form 34.0 g of NH₃. So, for 13.02g of NH₃:

13.02gNH_{3}.\frac{6.00gH_{2}}{34.0gNH_{3}} =2.30gH_{2}

Part D:

6.00 g of H₂ form 2 moles of NH₃. An each mole of NH₃ has 6.02 x 10²³ molecules of NH₃ (Avogadro number). So, for 7.62×10⁻⁴ g of H₂:

7.62 \times 10^{-4} gH_{2}.\frac{2molNH_{3}}{6.00gH_{2}} .\frac{6.02\times 10^{23}moleculesNH_{3}  }{1molNH_{3}}=1.53\times10^{20}moleculesNH_{3}

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