Consider titration of NaOH solution versus 1.25 M oxalic acid solution. At the end point following b

Q: Why do we take the average of the burette readings instead of just using one?

In any experimental setup, random errors can occur while reading the meniscus. Taking an average of multiple concordant readings minimizes these random errors and gives a more accurate volume.

Q: Why did we multiply the molarity of oxalic acid by 2?

The law of equivalence works with normality, not molarity. Normality is Molarity $\times$ n-factor. Since oxalic acid ($H_2C_2O_4$) can donate two $H^+$ ions, its n-factor is 2.

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

No! The equation $M_1V_1 = M_2V_2$ is only valid for dilution or when both the acid and base have an n-factor of 1. Here, the acid has an n-factor of 2, so you must use $N_1V_1 = N_2V_2$ or $n_1M_1V_1 = n_2M_2V_2$.

Q: What is the n-factor of NaOH?

NaOH is a strong base that dissociates to give one $OH^-$ ion per formula unit. Therefore, its n-factor (acidity) is 1.

Q: Why did we round off the final answer to 6?

The question explicitly asks for the molarity rounded off to the nearest integer. Since 5.63 is closer to 6 than to 5, we round it up to 6.

Question

Consider titration of NaOH solution versus $1.25	ext{ M}$ oxalic acid solution. At the end point following burette readings were obtained.
(i) $4.5	ext{ mL}$
(ii) $4.5	ext{ mL}$
(iii) $4.4	ext{ mL}$
(iv) $4.4	ext{ mL}$
(v) $4.4	ext{ mL}$
If the volume of oxalic acid taken was $10.0	ext{ mL}$, then the molarity of the NaOH solution is ......... M. (Rounded off to the nearest integer)

Sigma Percentile · Accepted Answer

The world of volumetric analysis is built on precision, patience, and the beautiful symmetry of chemical equivalence. In this problem, we step into the shoes of a chemist performing a classic acid-base titration. Our goal? To uncover the hidden concentration of a sodium hydroxide ($	ext{NaOH}$) solution using a standard solution of oxalic acid ($	ext{H}_2	ext{C}_2	ext{O}_4$).

### The Lab Setup and The Power of Averages

Imagine the setup: a burette filled with the unknown $	ext{NaOH}$ solution stands tall above a conical flask. Inside the flask rests exactly $10.0 	ext{ mL}$ of $1.25 	ext{ M}$ oxalic acid. As we carefully open the stopcock, drops of base fall into the acid until the indicator signals the end point.

But a good chemist never relies on a single reading. To eliminate random human errors—perhaps a slight misjudgment of the meniscus—we perform the titration five times. The readings obtained are $4.5 	ext{ mL}$, $4.5 	ext{ mL}$, $4.4 	ext{ mL}$, $4.4 	ext{ mL}$, and $4.4 	ext{ mL}$.

To find the most accurate volume of $	ext{NaOH}$ consumed, we calculate the average:
$$V_1 = \frac{4.5 + 4.5 + 4.4 + 4.4 + 4.4}{5} = 4.44 	ext{ mL}$$

This $4.44 	ext{ mL}$ is the true volume of the base that reacted completely with our acid.

### The Concept of Equivalents

Before we can equate the acid and the base, we must speak their common language: **Normality**. Molarity tells us the number of molecules, but Normality tells us the number of *reactive units* (like $	ext{H}^+$ or $	ext{OH}^-$ ions).

Oxalic acid ($	ext{H}_2	ext{C}_2	ext{O}_4$) is a dibasic acid. Each molecule is generous enough to donate two protons ($	ext{H}^+$). Therefore, its n-factor ($n_2$) is $2$.

To convert its molarity to normality, we multiply by the n-factor:
$$N_2 = M_2 	imes n_2 = 1.25 	ext{ M} 	imes 2 = 2.5 	ext{ N}$$

This means our $1.25 	ext{ M}$ oxalic acid behaves as a $2.5 	ext{ N}$ solution in terms of its neutralizing power.

### The Master Equation

At the equivalence point of a titration, the fundamental law of equivalence dictates that the number of equivalents of the acid must perfectly match the number of equivalents of the base. This gives us our master equation:
$$N_1 V_1 = N_2 V_2$$

Here, $N_1$ and $V_1$ represent the normality and volume of the $	ext{NaOH}$ solution, while $N_2$ and $V_2$ represent the oxalic acid.

### The Final Calculation

Let's substitute our known values into the master equation. We know the volume of $	ext{NaOH}$ ($V_1 = 4.44 	ext{ mL}$), the normality of oxalic acid ($N_2 = 2.5 	ext{ N}$), and the volume of oxalic acid ($V_2 = 10.0 	ext{ mL}$).

$$N_1 	imes 4.44 = 2.5 	imes 10.0$$

$$N_1 	imes 4.44 = 25$$

Solving for $N_1$:
$$N_1 = \frac{25}{4.44} \approx 5.63 	ext{ N}$$

We have found the normality of the sodium hydroxide solution. But the question asks for its **molarity**.

Sodium hydroxide ($	ext{NaOH}$) is a monoacidic base; it releases exactly one hydroxide ion ($	ext{OH}^-$) per formula unit. Thus, its n-factor ($n_1$) is $1$. For any substance with an n-factor of $1$, its molarity is numerically equal to its normality.

$$M_1 = \frac{N_1}{n_1} = \frac{5.63}{1} = 5.63 	ext{ M}$$

Finally, the problem asks us to round off the answer to the nearest integer. Since $5.63$ is closer to $6$ than to $5$, our final, beautifully derived answer is **$6$**.

Enter Numerical Value:

Visualized Solution

Titration Setup

Average Volume of NaOH

n-factor of Oxalic Acid

Normality of Oxalic Acid

Law of Equivalence

Substituting Values

Calculating Normality of NaOH

Molarity of NaOH

The Sigma Insight: Stoichiometric and Volumetric Calculations

The Lab Setup and The Power of Averages

The Concept of Equivalents

The Master Equation

The Final Calculation

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