The volume (in mL) of 0.1 N NaOH required to neutralise 10 mL of 0.1 N phosphonic acid is ..........

Q: Why didn't we use the n-factor of Phosphonic acid ($H_3PO_3$)?

Because the concentration was given in Normality ($0.1\text{ N}$). Normality is defined as Molarity $\times$ n-factor. The n-factor is already accounted for in the value $0.1\text{ N}$. If it were given as $0.1\text{ M}$, we would have needed to multiply by its n-factor (which is $2$).

Q: What is the basicity of Phosphonic acid ($H_3PO_3$)?

It is dibasic (n-factor = $2$). Even though it has three hydrogen atoms, only the two hydrogens attached to oxygen atoms are ionizable. The third hydrogen is directly attached to the phosphorus atom and does not ionize.

Q: Can I use $M_1V_1 = M_2V_2$ for this titration?

No, $M_1V_1 = M_2V_2$ is only valid when both the acid and the base have an n-factor of $1$ (like $HCl$ and $NaOH$). For general titrations, always use $N_1V_1 = N_2V_2$ or $n_1M_1V_1 = n_2M_2V_2$.

Q: What if the volume was given in liters instead of mL?

As long as you use the same unit of volume on both sides of the equation $N_1V_1 = N_2V_2$, the units will cancel out. If you use mL on one side, the calculated volume on the other side will also be in mL.

Q: Is Phosphonic acid the same as Phosphorous acid?

Yes! Phosphonic acid is the IUPAC name for Phosphorous acid ($H_3PO_3$).

Question

The volume (in $	ext{mL}$) of $0.1	ext{ N NaOH}$ required to neutralise $10	ext{ mL}$ of $0.1	ext{ N}$ phosphonic acid is ............ .

Sigma Percentile · Accepted Answer

### The Elegance of the Law of Equivalence

Imagine you are standing in a chemistry lab, looking at a classic titration setup. In the conical flask below, you have $10	ext{ mL}$ of a $0.1	ext{ N}$ solution of phosphonic acid ($H_3PO_3$). Suspended above it is a burette filled with a $0.1	ext{ N}$ solution of sodium hydroxide ($NaOH$). Your mission is simple: determine the exact volume of the base required to completely neutralize the acid.

At first glance, you might start thinking about writing out the balanced chemical equation. You might recall that phosphonic acid is a dibasic acid, meaning it has two ionizable hydrogen atoms, and its n-factor is $2$. You might start worrying about stoichiometry and mole ratios. But wait! There is a much more elegant and direct way to solve this.

### The Master Equation

This is where the **Law of Equivalence** comes to our rescue. The law states that for a complete neutralization reaction, the total equivalents of the acid must perfectly equal the total equivalents of the base.

Mathematically, the number of equivalents is simply the product of Normality ($N$) and Volume ($V$). Therefore, we can write our master equation as:

$$N_1 V_1 = N_2 V_2$$

Here is the beautiful catch in this problem: the concentrations are already given in **Normality**. Normality is defined as Molarity multiplied by the n-factor ($N = M 	imes n_f$). Because the question provides the concentration as $0.1	ext{ N}$, the n-factor of phosphonic acid has *already* been accounted for! We don't need to multiply or divide anything by $2$. The value $0.1	ext{ N}$ is ready to be plugged directly into our equation.

### Final Calculation

Let's substitute our known values into the equivalence equation. For our base ($NaOH$), we have $N_1 = 0.1	ext{ N}$, and $V_1$ is our unknown. For our acid ($H_3PO_3$), we have $N_2 = 0.1	ext{ N}$ and $V_2 = 10	ext{ mL}$.

$$0.1 	imes V_1 = 0.1 	imes 10$$

The $0.1$ on both sides of the equation beautifully cancel each other out, leaving us with a direct and satisfying result:

$$V_1 = 10	ext{ mL}$$

It is that simple! Exactly **$10	ext{ mL}$** of the $0.1	ext{ N NaOH}$ solution is required to completely neutralize the phosphonic acid.

Always remember to check the units of concentration in titration problems. If the question had given $0.1	ext{ M}$ phosphonic acid instead, you would have had to multiply it by its n-factor of $2$ to find the normality ($0.2	ext{ N}$) before using the equivalence formula. Paying attention to these small details is what separates a good student from a great one!

Animated Solution for Chemistry - Basic Concepts in Chemistry: The volume (in $mL$ ) of $0.1 N NaOH$ required to neutralise $10 mL$ of $0.1 N$ phosphonic acid is ............ .

Enter Numerical Value:

Visualized Solution

\text{Titration Setup}

\text{Law of Equivalence}

\text{Substituting the Values}

\text{Calculating Volume}

\text{Conclusion}

\text{What if Molarity was given?}

The Sigma Insight: Stoichiometric and Volumetric Calculations

The Elegance of the Law of Equivalence

The Master Equation

Final Calculation

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Animated Solution for Chemistry - Basic Concepts in Chemistry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The volume (in mL) of 0.1 N NaOH required to neutralise 10 mL of 0.1 N phosphonic acid is ............ .