To neutralize completely $$20$$ mL of $$0.1$$ M aqueous solution of phosphorous acid $$(H_3PO_3)$$, the volume of $$0.1$$ M aqueous KOH solution required is

The trick to this problem lies in understanding the structural formula and the true basicity (n-factor) of phosphorous acid ($$\text{H}_3\text{PO}_3$$).

Although it contains three hydrogen atoms, its structure reveals that only the hydrogens bonded directly to an oxygen atom can dissociate in water as $$\text{H}^+$$ ions. The single hydrogen bonded directly to the central phosphorus atom is non-acidic:

• Therefore, $$\text{H}_3\text{PO}_3$$ is a dibasic acid ($$n\text{-factor} = 2$$).
• $$\text{KOH}$$ is a strong monacidic base, meaning it releases one $$\text{OH}^-$$ ion ($$n\text{-factor} = 1$$).

$$\text{H}_3\text{PO}_3 + 2\text{KOH} \rightarrow \text{K}_2\text{HPO}_3 + 2\text{H}_2\text{O}$$

The stoichiometry reveals that $$1\text{ mole}$$ of $$\text{H}_3\text{PO}_3$$ requires exactly $$2\text{ moles}$$ of $$\text{KOH}$$ for complete neutralization.

• Method: Law of Equivalents

  At complete neutralization, the equivalents of acid must equal the equivalents of base:

  $$\text{Normality of Acid } (N_1) \times V_1 = \text{Normality of Base } (N_2) \times V_2$$

  Since $$\text{Normality} = \text{Molarity} \times n\text{-factor}$$:

  $$(\text{Molarity}_1 \times n_1) \times V_1 = (\text{Molarity}_2 \times n_2) \times V_2$$




• Substitute the given variables:

  • Acid ($$\text{H}_3\text{PO}_3$$): $$M_1 = 0.1 \text{ M}$$, $$n_1 = 2$$, $$V_1 = 20 \text{ mL}$$
  • Base ($$\text{KOH}$$): $$M_2 = 0.1 \text{ M}$$, $$n_2 = 1$$, $$V_2 = ?$$

  $$(0.1 \times 2) \times 20 = (0.1 \times 1) \times V_2$$

  $$0.2 \times 20 = 0.1 \times V_2$$

  $$4 = 0.1 \times V_2$$

  $$V_2 = \frac{4}{0.1} = 40 \text{ mL}$$

The required volume of the $$0.1 \text{ M}$$ $$\text{KOH}$$ solution is exactly $$40 \text{ mL}$$.

Answer: Option C — 40 mL