Q.Formic acid is stronger than acetic acid. Explain.

Only a carboxylic acid is acidic enough to release CO2 from NaHCO3, so effervescence identifies acetic acid.

Carboxylic acids are strong enough acids to decompose sodium bicarbonate, releasing carbon dioxide gas (brisk effervescence):

CH3COOH + NaHCO3 --> CH3COONa + H2O + CO2 (up-arrow)

Alkanes, alkenes and ethyl alcohol are not acidic enough to do this (alcohols and phenols do not react with NaHCO3). The effervescence test therefore confirms the -COOH group, i.e. acetic acid.

The ortho -OH's strong -I effect plus H-bond-stabilised conjugate base make salicylic acid the strongest acid of the four.

Comparing substituted benzoic acids: an ortho- or para- electron-withdrawing group increases acidity (stabilises the carboxylate conjugate base), while an electron-donating group decreases acidity.

• Salicylic acid (o-OH): the $-OH$ group is strongly electron-withdrawing by induction at the ortho position (its resonance donation is largely used up in its own ring conjugation from that close a position), and the conjugate base is further stabilised by intramolecular hydrogen bonding — making salicylic acid distinctly MORE acidic than benzoic acid ($pK_a \approx 2.97$ vs $4.2$).
• Anthranilic acid (o-NH$_2$): $-NH_2$ is a strong electron donor by resonance, decreasing acidity — weaker acid than benzoic acid.
• Benzoic acid: reference acid, no substituent effect.
• p-Anisic acid (p-OCH$_3$): $-OCH_3$ donates electron density into the ring by resonance from the para position, decreasing acidity — weaker acid than benzoic acid.

Salicylic acid is therefore the most acidic of the four.

Use NaHCO3: formic acid effervesces (CO2), formaldehyde does not.

Both formic acid (HCOOH) and formaldehyde (HCHO) contain a -CHO type group, so both reduce Tollen's reagent, Fehling's solution and Benedict's solution — these cannot distinguish them.

However, formic acid is a carboxylic acid, so it reacts with sodium bicarbonate to liberate CO2 with brisk effervescence:

HCOOH + NaHCO3 -> HCOONa + H2O + CO2

Formaldehyde, not being acidic, gives no such effervescence. Hence sodium bicarbonate distinguishes them.

Both reduce Tollens'/Fehling's/Benedict's, so those cannot distinguish them; only NaHCO3 does (acid vs non-acid).

Formic acid (HCOOH) has both an acidic -COOH and a reducing -CHO-like character, while formaldehyde (HCHO) is only a reducing aldehyde. Therefore:

• Tollens' reagent: BOTH give silver mirror (both are reducing) - no distinction.
• Fehling's / Benedict's solution: BOTH reduce them (formic acid also reduces) - no distinction.
• Sodium bicarbonate: formic acid, being a carboxylic acid, reacts to give brisk CO2 effervescence: HCOOH + NaHCO3 -> HCOONa + H2O + CO2. Formaldehyde is not acidic and gives NO effervescence.

So NaHCO3 cleanly distinguishes them.

Carboxylic acid strength is governed by how well the conjugate base's negative charge is stabilised: electron-donating alkyl groups (+I effect) destabilise the carboxylate and WEAKEN the acid, while electron-withdrawing groups (like $-Cl$, -I effect) stabilise the carboxylate and STRENGTHEN the acid.

Propanoic acid ($CH_3CH_2COOH$, $pK_a\approx4.87$) has the longest alkyl chain (strongest +I donation) and is the weakest of the four. Acetic acid ($CH_3COOH$, $pK_a\approx4.76$) has a shorter alkyl group, so is slightly stronger. Formic acid ($HCOOH$, $pK_a\approx3.75$) has NO alkyl group at all (just H), removing the destabilising +I effect entirely, making it noticeably stronger than acetic acid. Chloroacetic acid ($ClCH_2COOH$, $pK_a\approx2.86$) has a strongly electron-withdrawing $-Cl$ directly stabilising the carboxylate by induction, making it the strongest of the four by a wide margin.

So, in increasing order of acid strength: $CH_3CH_2COOH < CH_3COOH < HCOOH < ClCH_2COOH$ — exactly option (a).