Animated Solution for Chemistry - Ionic Equilibrium: The decreasing order of electrical conductivity of the following aqueous solution is 0.1 M formic acid (A), 0.1 M acetic acid (B), 0.1 M benzoic acid (C).

Imagine you have three beakers, each containing a $0.1\text{ M}$ aqueous solution of a different weak acid: formic acid, acetic acid, and benzoic acid. If you were to dip a pair of electrodes into each beaker and measure their electrical conductivity, you would notice distinct differences. But why? They all have the exact same concentration!

The secret lies in the microscopic world of ions. Electrical conductivity in an aqueous solution is not determined by the total number of acid molecules present, but by the number of molecules that have actually broken apart—or dissociated—into ions. These ions are the charge carriers. The more charge carriers you have, the higher the electrical conductivity.

For weak acids, the extent of this dissociation is measured by the degree of ionization ($\alpha$). According to Ostwald's dilution law, for a weak acid at a given concentration $C$, the degree of ionization is directly proportional to the square root of its acid dissociation constant ($K_a$):

Since the concentration $C$ is constant ($0.1\text{ M}$) for all three solutions, the conductivity becomes a direct competition of their $K_a$ values. In simpler terms: The stronger the acid, the higher the conductivity.

Now, the problem transforms from a physics question about conductivity into a classic organic chemistry puzzle: How do we determine which of these three acids is the strongest?

The golden rule for comparing acidic strength is to remove the acidic proton ($H^{+}$) and analyze the stability of the resulting conjugate base.

When an acid donates a proton, it leaves behind a negative charge on the oxygen atom, forming a carboxylate ion ($R-CO{O}^{-}$). Nature abhors concentrated charge. If the $R$ group attached to the carboxylate ion can help spread out or "disperse" this negative charge, the conjugate base becomes highly stable. A stable conjugate base means the original acid is very willing to give up its proton, making it a strong acid.

Conversely, if the $R$ group pushes even more electron density onto the already negative oxygen, it intensifies the charge, creating massive electrostatic repulsion. This destabilizes the conjugate base, making the original acid very weak.

Let's put our three contenders to the test.

The Destabilizing $+I$ Effect

Let's start with acetic acid ($C{H}_{3}COOH$). When it loses a proton, it forms the acetate ion ($C{H}_{3}CO{O}^{-}$).

The group attached to the carboxylate carbon is a methyl group ($-C{H}_3$). Alkyl groups are known for their electron-donating inductive effect ($+I$ effect). The methyl group acts like an electronic pump, pushing electron density through the sigma bonds toward the carboxylate group.

But the oxygen atoms are already bearing a negative charge! Pushing more electrons toward them is like trying to force more air into an overinflated balloon. It creates intense electronic repulsion, severely destabilizing the acetate ion. Because its conjugate base is so unstable, acetic acid is very reluctant to lose its proton. Thus, it is the weakest acid among the three.

The $s{p}^2$ Carbon's $-I$ Effect

Next, we look at benzoic acid ($Ph-COOH$). Its conjugate base is the benzoate ion ($Ph-CO{O}^{-}$).

Here, the carboxylate group is attached directly to a phenyl ring. The carbon atom of the benzene ring that connects to the carboxylate group is $s{p}^2$ hybridized.

Why does hybridization matter? An $s{p}^2$ hybridized carbon has more s-character ($33.3\%$) compared to the $s{p}^3$ hybridized carbon ($25\%$) of the methyl group in acetic acid. More s-character means the electrons are held closer to the nucleus, making the $s{p}^2$ carbon significantly more electronegative.

Because of this higher electronegativity, the phenyl ring exerts a mild electron-withdrawing inductive effect ($-I$ effect) relative to an alkyl group. It gently pulls some of the electron density away from the negatively charged carboxylate group, helping to disperse the charge.

While the phenyl ring can also participate in resonance ($+R$ effect), the $-I$ effect of the $s{p}^2$ carbon dominates in this specific position, making the benzoate ion more stable than the acetate ion. Therefore, benzoic acid is a stronger acid than acetic acid.

The Perfect Reference

Finally, we arrive at formic acid ($HCOOH$). Its conjugate base is the formate ion ($HCO{O}^{-}$).

Look at what is attached to the carboxylate group: just a single hydrogen atom. In the study of inductive effects, hydrogen is the universal reference point. It has zero inductive effect ($I=0$).

The formate ion doesn't have an electron-withdrawing group to help stabilize it, but more importantly, it doesn't have an electron-donating group to destabilize it! Without the detrimental $+I$ effect of a methyl group, the negative charge on the formate ion is perfectly happy resonating between the two oxygen atoms without any added electronic pressure.

Because it lacks any destabilizing groups, the formate ion is the most stable conjugate base of the three. Consequently, formic acid is the strongest acid in this lineup.

We have successfully ranked the stability of the conjugate bases: Formate ion $>$ Benzoate ion $>$ Acetate ion

This directly translates to their acidic strengths ($K_a$): Formic acid (A) $>$ Benzoic acid (C) $>$ Acetic acid (B)

And bringing it all back to our original macroscopic observation, since conductivity is directly proportional to acidic strength for weak acids at the same concentration, the decreasing order of electrical conductivity is: $A>C>B$

The beauty of chemistry lies in how the invisible, subatomic tug-of-war of electrons perfectly dictates the measurable, physical properties of the world around us.