Animated Solution for Chemistry - Basic Concepts in Chemistry: The exact volumes of $1\text{ M NaOH}$ solution required to neutralise $50\text{ mL}$ of $1{\text{ M H}}_3{\text{PO}}_3$ solution and $100\text{ mL}$ of $2{\text{ M H}}_3{\text{PO}}_2$ solution, respectively, are

Neutralization problems often seem incredibly straightforward. You might be tempted to just plug the values into the classic dilution formula $M_1{V}_1=M_2{V}_2$ and call it a day. But wait! That is the biggest trap in stoichiometry. When dealing with acids and bases, we must always equate their equivalents, not just their moles. This means we desperately need to know the $n$-factor (basicity) of the acids involved.

The $n$-factor of an acid is simply the number of ionizable $H^{+}$ ions it can release in an aqueous solution. For the oxyacids of phosphorus, there is a golden rule: only the hydrogen atoms attached to highly electronegative oxygen atoms can be released as $H^{+}$. The hydrogen atoms attached directly to the central phosphorus atom are non-ionizable because the $P-H$ bond is almost non-polar.

Let's draw the structure of $H_{3}P{O}_3$. It consists of one $P=O$ bond, two $P-OH$ bonds, and one $P-H$ bond. Since there are exactly two $P-OH$ bonds, it can release two protons. Thus, it is a dibasic acid with an $n$-factor of $2$.

Using the equivalence equation $M_1{V}_1{n}_1=M_2{V}_2{n}_2$, we substitute the given values for the acid and the base ($NaOH$, which has an $n$-factor of $1$):

Solving this gives $V=100\text{ mL}$.

Now, let's look at $H_{3}P{O}_2$. Its structure features one $P=O$ bond, one $P-OH$ bond, and two $P-H$ bonds. Because only one hydrogen is attached to an oxygen atom, it is a monobasic acid with an $n$-factor of $1$. Don't let the $H_3$ in the formula fool you!

Substituting into the equivalence equation again:

Solving this gives $V=200\text{ mL}$.

The required volumes of $NaOH$ are $100\text{ mL}$ and $200\text{ mL}$, respectively. This perfectly matches option (c). The key takeaway here is to always draw the structures of phosphorus oxyacids before jumping into calculations!