5.00 mL of 0.10 M oxalic acid solution taken in a conical flask is titrated against NaOH from a bure

Q: Why do we ignore the first two readings of $12.5$ and $10.5$ mL?

The first few readings in a titration are usually rough estimates to find the approximate endpoint. We only use the 'concordant' readings—the ones that are identical—because they represent the precise, repeatable volume.

Q: Why is the n-factor of oxalic acid 2?

Oxalic acid ($\text{H}_2\text{C}_2\text{O}_4$) is a dibasic acid, meaning each molecule can donate two protons ($\text{H}^+$) during the neutralization reaction.

Q: Do I need to convert the volumes from mL to Liters?

No, you don't have to. Since volume appears on both sides of the equation $N_1 V_1 = N_2 V_2$, the conversion factor ($1000$) would simply cancel out. Using mL on both sides is perfectly fine.

Q: Why was phenolphthalein used as the indicator?

This is a titration between a weak acid (oxalic acid) and a strong base (NaOH). The equivalence point will be slightly basic (pH > 7), which falls perfectly within phenolphthalein's color change range.

Q: What if we used molarity directly instead of normality?

If you use molarity directly ($M_1 V_1 = M_2 V_2$), you will get the wrong answer because it ignores the stoichiometry of the reaction. One mole of oxalic acid reacts with two moles of NaOH. Using normality automatically accounts for this.

Question

5.00 mL of 0.10 M oxalic acid solution taken in a conical flask is titrated against NaOH from a burette using phenolphthalein indicator. The volume of NaOH required for the appearance of permanent faint pink color is tabulated below for five experiments. What is the concentration, in molarity, of the NaOH solution ?

Sigma Percentile · Accepted Answer

### The Art of Titration

Imagine you are in a chemistry lab, carefully letting drops of sodium hydroxide fall into a flask of oxalic acid. The goal? To find the exact concentration of the base. This process is called **titration**, and it relies on a beautiful principle: the law of equivalence.

When the indicator (phenolphthalein) turns a faint pink, it signals the **equivalence point**. At this exact moment, the acid has completely neutralized the base.

### Decoding the Data

Before we jump into calculations, we must look at the experimental data. The table shows five different readings for the volume of NaOH used.

Notice how the first reading is $12.5 	ext{ mL}$, the second is $10.5 	ext{ mL}$, but the last three are exactly $9.0 	ext{ mL}$. Why is this? In titrations, the first few attempts are often rough estimates. We only trust the **concordant readings**—the values that repeat consistently. Therefore, the true volume of NaOH used is $9.0 	ext{ mL}$.

### The Law of Equivalence

The fundamental equation governing this reaction is the law of equivalence:

$$ 	ext{Equivalents of Acid} = 	ext{Equivalents of Base} $$

We can express this using normality ($N$) and volume ($V$):

$$ N_1 V_1 = N_2 V_2 $$

But we are given molarity ($M$), not normality. The bridge between them is the **n-factor** ($n_f$), which represents the number of exchangeable protons or hydroxide ions.

$$ N = M 	imes n_f $$

### Unpacking the n-factors

Oxalic acid ($	ext{H}_2	ext{C}_2	ext{O}_4$) is a **dibasic acid**. It can donate two protons ($	ext{H}^+$) per molecule. Thus, its n-factor is $2$.

Sodium hydroxide ($	ext{NaOH}$) is a **monoacidic base**. It provides one hydroxide ion ($	ext{OH}^-$) per molecule. Thus, its n-factor is $1$.

### The Final Calculation

Now, we substitute our known values into the equivalence equation:

$$ (M_{	ext{oxalic}} 	imes n_{f,	ext{oxalic}}) 	imes V_{	ext{oxalic}} = (M_{	ext{NaOH}} 	imes n_{f,	ext{NaOH}}) 	imes V_{	ext{NaOH}} $$

Plugging in the numbers:

$$ (0.10 	imes 2) 	imes 5.00 = (M_{	ext{NaOH}} 	imes 1) 	imes 9.0 $$

Simplifying the left side:

$$ 0.20 	imes 5.00 = M_{	ext{NaOH}} 	imes 9.0 $$

$$ 1.0 = M_{	ext{NaOH}} 	imes 9.0 $$

Solving for the molarity of NaOH:

$$ M_{	ext{NaOH}} = \frac{1.0}{9.0} \approx 0.111... 	ext{ M} $$

Rounding to two significant figures, the concentration of the NaOH solution is **$0.11 	ext{ M}$**.

Enter Numerical Value:

Visualized Solution

$Concordant Volume$

$Law of Equivalence$

$Normality and n-factor$

$Substitution$

$Calculation$

$Final Answer$

$Reflection$

The Sigma Insight: Stoichiometric and Volumetric Calculations

The Art of Titration

Decoding the Data

The Law of Equivalence

Unpacking the n-factors

The Final Calculation

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