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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:

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Explanation:

The pH of the solution = 2.46

pH=-\log[H^+]

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HA\rightleftharpoons H^++A^-

Initially

0.0144         0      0

At equilibrium

(0.0144-x)       x       x

The expression if an dissociation constant is given by :

K_a=\frac{[A^-][H^+]}{[HA]}

K_a=\frac{x\times x}{(0.0144-x)}

x=[H^+]=0.003467 M

K_a=\frac{0.003467 \times 0.003467 }{(0.0144-0.003467 )}

K_a=1.099\times 10^{-3}

The value of dissociation constant of the monoprotic acid is 1.099\times 10^{-3}.

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3 years ago
What is the pH of a<br> 2.5 x 10-5 M solution<br> of HCI?
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Answer:

The pH of the solution is 4.60.

Explanation:

The pH gives us an idea of the acidity or basicity of a solution. More precisely, it indicates the concentration of H30 + ions present in said solution. The pH scale ranges from 0 to 14: from 0 to 7 corresponds to acid solutions, 7 neutral solutions and between 7 and 14 basic solutions. It is calculated as:

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A flexible container at an initial volume of 5.120 L contains 8.500 mol of gas.
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Answer:

30.05 mol

Explanation:

solving the proportion

V1 / n1 = V2 / n2

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Troyanec [42]

Answer:

The concentration of the copper sulfate solution is 83 mM.

Explanation:

The absorbance of a copper sulfate solution can be calculated using Beer-Lambert Law:

A = ε . c . <em>l</em>

where

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