Animated Solution for Chemistry - Basic Concepts in Chemistry: To neutralise completely $20\text{ mL}$ of $0.1\text{ M}$ aqueous solution of phosphorous acid (${\text{H}}_3{\text{PO}}_3$), the volume of $0.1\text{ M}$ aqueous $\text{KOH}$ solution required is

Imagine you are standing in a chemistry laboratory. In front of you is a conical flask containing exactly $20\text{ mL}$ of a $0.1\text{ M}$ aqueous solution of phosphorous acid (${\text{H}}_3{\text{PO}}_3$). Above it, a burette is filled with a $0.1\text{ M}$ solution of potassium hydroxide ($\text{KOH}$). Your mission is to find out exactly how much of this $\text{KOH}$ solution is needed to completely neutralize the acid in the flask.

At first glance, this might seem like a straightforward stoichiometry problem. But beware! This question contains one of the most classic and frequently tested traps in the entire JEE chemistry syllabus. Let's break it down step by step and uncover the hidden secret of phosphorous acid.

When you look at the molecular formula of phosphorous acid, ${\text{H}}_3{\text{PO}}_3$, what is the first thing that comes to your mind? You see three hydrogen atoms, and it is incredibly tempting to assume that all three of them are acidic. If you make this assumption, you would say the basicity (or $n$-factor) of the acid is $3$.

To understand the true nature of this acid, we must look beyond its formula and examine its molecular structure. In the structure of ${\text{H}}_3{\text{PO}}_3$, the central phosphorus atom is bonded to one oxygen atom via a double bond ($\text{P=O}$), two hydroxyl groups ($\text{-OH}$), and one hydrogen atom directly ($\text{P-H}$).

Why does this structural arrangement matter so much? It all comes down to electronegativity.

Oxygen is a highly electronegative element. When hydrogen is bonded to oxygen, the oxygen atom pulls the shared electron pair towards itself, creating a highly polar $\text{O-H}$ bond. This polarity allows the hydrogen atom to easily break away as an ${\text{H}}^{+}$ ion (a proton) in an aqueous solution.

On the other hand, the electronegativity of phosphorus is very similar to that of hydrogen. Because there is almost no electronegativity difference, the $\text{P-H}$ bond is essentially non-polar. The hydrogen atom is held tightly and cannot be released as a proton.

Therefore, out of the three hydrogen atoms in ${\text{H}}_3{\text{PO}}_3$, only the two attached to oxygen atoms are ionizable. This makes phosphorous acid a dibasic acid, meaning its $n$-factor is exactly $2$.

Now that we have successfully navigated the trap and determined the correct $n$-factor for our acid, the rest of the problem is a beautiful application of the Law of Chemical Equivalence.

For a complete neutralization reaction, the fundamental rule is that the number of equivalents of the acid must be perfectly equal to the number of equivalents of the base.

We can express the number of equivalents as the product of Normality ($N$) and Volume ($V$). This gives us our master equation:

But wait, our concentrations are given in Molarity ($M$), not Normality. How do we bridge this gap? We use the simple relationship:

Let's substitute our known values into the master equation.

For our acid (${\text{H}}_3{\text{PO}}_3$): - Molarity ($M_1$) = $0.1\text{ M}$ - $n$-factor = $2$ - Volume ($V_1$) = $20\text{ mL}$

So, the Normality of the acid ($N_1$) is $0.1\times 2=0.2\text{ N}$.

For our base ($\text{KOH}$): - Molarity ($M_2$) = $0.1\text{ M}$ - $n$-factor = $1$ (since it releases one ${\text{OH}}^{-}$ ion) - Volume ($V_2$) = ?

So, the Normality of the base ($N_2$) is $0.1\times 1=0.1\text{ N}$.

Plugging these into our equivalence equation:

The $0.1$ on both sides cancels out elegantly, leaving us with:

And there we have it! We need exactly $40\text{ mL}$ of the $0.1\text{ M KOH}$ solution to completely neutralize the phosphorous acid.

By understanding the molecular structure and not just blindly trusting the chemical formula, you have successfully solved a classic JEE problem. Always remember the trend for phosphorus oxyacids: ${\text{H}}_3{\text{PO}}_4$ is tribasic, ${\text{H}}_3{\text{PO}}_3$ is dibasic, and ${\text{H}}_3{\text{PO}}_2$ is monobasic. Keep this in your arsenal, and you will never fall for this trap again!