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Margarita [4]
3 years ago
10

The sulfuric acid–catalyzed reaction of isopentyl alcohol with acetic acid to form isopentyl acetate is performed. This reaction

does not proceed to completion. After the reaction is heated under reflux for an hour, the product mixture is washed with aqueous sodium bicarbonate solution, and the organic layer is thoroughly dried. A distillation of the organic layer is then performed. Given the boiling points listed below, select the compound that will distill first.
Compound Boiling point, ℃
acetic acid 118
isopentyl acetate 142
isopentyl alcohol 130
sulfuric acid water 290
Chemistry
1 answer:
murzikaleks [220]3 years ago
6 0

Answer:

The correct option is acetic acid

Explanation:

Distillation is the process of separating a mixture of substances based on differences in boiling points. During distillation, the compound with the lowest/least boiling point is distilled and collected first and then the one with the next least boiling point and it goes on like that.

From the explanation above, <u>acetic acid has the least boiling point (in the organic layer) with 118°C and thus will distill first</u>. This is then followed by isopentyl alcohol (130°C) and then isopentyl acetate (142°C) and finally sulfuric acid water (290°C).

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There is some info missing. I think this is the original question.

<em>Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is  4.50  ×  10 ⁻⁴.</em>

<em />

Step 1: Given data

Initial concentration of the acid (Ca): 0.249 M

Acid dissociation constant (Ka): 4.50  ×  10 ⁻⁴

Step 2: Write the ionization reaction for nitrous acid

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Step 3: Calculate the concentration of nitrite in the equilibrium ([A⁻])

We will use the following expression.

[A^{-} ] = \sqrt{Ca \times Ka } = \sqrt{0.249 \times 4.50 \times 10^{-4}  } = 0.0106 M

Step 4: Calculate the percent ionization of nitrous acid

We will use the following expression.

\alpha = \frac{[A^{-} ]}{[HA]} \times 100\% = \frac{0.0106M}{0.249} \times 100\% = 4.26\%

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