Question

Ethanoic acid was added to four test tubes
containing the following chemicals
(i) Sodium carbonate
(ii) Blue litmus solution
(iii) Lime water
(iv) Distilled water.
Which among these is/are the correct option(s) for
carrying out a characteristic test for identification
of a carboxylic acid (ethanoic acid) in the
laboratory?
(a) i only
(b) ili only
(c) i and ii only
(d) iii and iv​

Answer 1

$\huge{\textbf{\underline{\pink{Answer:-}}}}$

(c). i and ii only

$\huge{\textbf{\underline{\pink{Step-by-Step Explanation:-}}}}$

Ethanoic acid was added to four test tubes containing the following chemicals are:

(i) Sodium carbonate→$Na_2CO_3(s)$

(ii) Blue litmus solution

(iii) Lime water→$Ca(OH)_2(aq)$

(iv) Distilled water→$H_2O(l)$

When $CH_3COOH(aq)$ was added to the sodium carbonate solution it produces or releases $CO_2(g)$ gas which proves or indicates that it is a carboxylic acid.

When $CH_3COOH(aq)$ was added to the blue litmus solution, it turns blue colour into the red colour because it is acidic in nature and it proves or indicates that it is a carboxylic acid.

When $CH_3COOH(aq)$ was added to the lime water solution, ethanoic acid neutralizes the lime water which is known as calcium hydroxide solution to form a salt and water. It doesn't indicates that it is a carboxylic acid.

When $CH_3COOH(aq)$ was added to the distilled water, the ethanoic acid becomes dilute ethanoic acid which doesn't indicates that it is a carboxylic acid.

So therefore, when the ethanoic acid is added to the sodium carbonate and blue litmus solution it gives the result that ethanoic acid is a carboxylic acid.