Question

Ethanoic acid and ethanol can be distinguished by litmus as well as sodium
bicarbonate test.
(a) True
(b) False

Answer 1

Hint: We know that litmus test is used to test whether the given compound has any acidic or basic content present in it, whereas sodium bicarbonate test can be a test in which we look for carboxyl groups.

Complete Step by step answer:
(i) Litmus test:
As we said earlier, litmus test is used to identify whether it is acid or basic. Also, we know that litmus paper itself is a type of pH paper made by treating the paper with natural dyes from lichens. So, ethanol which is a very weak acid with $p{K_a}$ typically around 17. From this we understood that it is an extremely weak acid. Hence the litmus itself is stronger acid than ethanol therefore dissociation of alcohol won’t produce enough protons to shift the equilibrium between two coloured forms of the indicator, so we will see no colour change.
But in case of ethanoic acid as we know it has a tendency to donate protons and act as acids. Hence they will show the litmus test. Also we can identify the carboxyl group with this test.
They turn blue litmus to red since it is an acid.
Hence we can say that litmus test can distinguish between ethanol and ethanoic acid.

(ii) Sodium bicarbonate test:
Sodium carbonate tests are specifically to identify the carboxyl groups present in the compound. Sodium bicarbonate when reacted with carboxylic acid to give sodium salt and liberates carbon dioxide

$C{H_3}COOH + NaHC{O_3} \to C{H_3}COONa + \;C{O_2} + {H_2}O$

We can see a brisk effervescence produced due to the carbon dioxide.
In ethanol we cannot see this brisk effervescence since they do not produce carbon dioxide when reacted with sodium bicarbonate. Also, ethanol does not contain carboxyl groups. Hence, we can say that sodium bicarbonate test can distinguish between ethanol and ethanoic acid.

Hence the correct option is A i.e true.

Note: We can also distinguish between ethanol and ethanoic acid using a strong alkali such as (NaOH). Ethanol does not react with NaOH whereas the ethanoic acid has strong alkali to give salt and water.
$C{H_3}COOH + NaOH \to C{H_3}COONa + {H_2}O$