Animated Solution for Chemistry - Ionic Equilibrium: Four species are listed below I. ${\text{HCO}}_3^{-}$ II. ${\text{H}}_3{\text{O}}^{+}$ III. ${\text{HSO}}_4^{-}$ IV. ${\text{HSO}}_3\text{F}$ Which one of the following is the correct sequence of their acid strength?

When you are presented with a list of seemingly unrelated chemical species—like the bicarbonate ion (${\text{HCO}}_3^{-}$), the hydronium ion (${\text{H}}_3{\text{O}}^{+}$), the bisulfate ion (${\text{HSO}}_4^{-}$), and fluorosulfuric acid (${\text{HSO}}_3\text{F}$)—and asked to rank their acidic strength, it can feel overwhelming. How do you compare a neutral-looking complex acid with a positively charged ion and two negatively charged anions?

The secret lies not in memorizing endless tables of ${\text{pK}}_a$ values, but in understanding the fundamental nature of acids and bases.

Conjugate Acid-Base Theory

Our biggest weapon in this chemical battle is the Bronsted-Lowry concept. According to this theory, an acid is a proton (${\text{H}}^{+}$) donor, and a base is a proton acceptor. When an acid donates its proton, what remains is called its conjugate base.

The golden rule of this theory is beautifully simple: The stronger the acid, the weaker and more stable its conjugate base.

Why? Because if a conjugate base is highly stable, it has very little desire to grab a proton back. This means the parent acid is more than happy to let that proton go, making it a strong acid. Therefore, to rank the acidic strength of our four species, we just need to rank the basic strength of their conjugate bases and reverse the order!

Let's systematically remove one proton (${\text{H}}^{+}$) from each of our given species to reveal their conjugate bases:

1. ${\text{HCO}}_3^{-}\stackrel{-{\text{H}}^{+}}{\to }{\text{CO}}_3^{2-}$ (Carbonate ion) 2. ${\text{H}}_3{\text{O}}^{+}\stackrel{-{\text{H}}^{+}}{\to }{\text{H}}_2\text{O}$ (Water) 3. ${\text{HSO}}_4^{-}\stackrel{-{\text{H}}^{+}}{\to }{\text{SO}}_4^{2-}$ (Sulfate ion) 4. ${\text{HSO}}_3\text{F}\stackrel{-{\text{H}}^{+}}{\to }{\text{SO}}_3{\text{F}}^{-}$ (Fluorosulfate ion)

Now, let's evaluate the basic strength of these resulting species.

The fluorosulfate ion (${\text{SO}}_3{\text{F}}^{-}$) is incredibly stable. The highly electronegative fluorine atom exerts a massive inductive pull ($-I$ effect), dispersing the negative charge across the entire molecule. Because it is so stable, it is an extremely weak base.

Next is water (${\text{H}}_2\text{O}$), which is a neutral molecule and a very weak base compared to most anions.

The sulfate ion (${\text{SO}}_4^{2-}$) carries a $-2$ charge. While resonance stabilizes it, it is still more basic than neutral water.

Finally, the carbonate ion (${\text{CO}}_3^{2-}$) also carries a $-2$ charge but is less stable than sulfate (since carbon is less electronegative than sulfur and has fewer resonance structures). Thus, it is the strongest base among the four.

So, the increasing order of basic strength is: ${\text{SO}}_3{\text{F}}^{-}<{\text{H}}_2\text{O}<{\text{SO}}_4^{2-}<{\text{CO}}_3^{2-}$

Reversing the Order

Since acidic strength is the exact inverse of basic strength, we simply flip our sequence to find the final answer:

Matching this with our original Roman numerals: $\text{I}<\text{III}<\text{II}<\text{IV}$

This perfectly aligns with option (c).

It is worth taking a moment to appreciate fluorosulfuric acid (${\text{HSO}}_3\text{F}$). It belongs to a terrifyingly powerful class of chemicals known as superacids. By definition, a superacid is an acid with an acidity greater than that of $100\%$ pure sulfuric acid.

The substitution of an $-\text{OH}$ group in sulfuric acid with a highly electronegative fluorine atom drastically increases the acid's ability to donate a proton. Understanding these extreme ends of the pH spectrum not only helps you solve JEE questions but also gives you a deeper appreciation for the incredible forces at play in chemical bonding!