10.0 mL of Na2CO3 solution is titrated against 0.2 M HCl solution. The following titre values were o

Q: Why do we ignore the 4.8 mL and 4.9 mL readings?

In any experiment, initial readings can have slight human or instrumental errors. The readings that repeat consistently (concordant readings) are considered the most accurate and reliable for calculations.

Q: Why is the n-factor of $\text{Na}_2\text{CO}_3$ taken as 2?

A complete neutralization of sodium carbonate by hydrochloric acid yields $\text{NaCl}$, $\text{H}_2\text{O}$, and $\text{CO}_2$. One molecule of $\text{Na}_2\text{CO}_3$ consumes two molecules of HCl (or accepts two $H^+$ ions), making its n-factor 2.

Q: Can I just use $M_1V_1 = M_2V_2$?

No! That formula only works when both reactants have an n-factor of 1 (like NaOH and HCl). For reactants with different n-factors, you must use the equivalence formula: $M_1V_1n_1 = M_2V_2n_2$.

Q: What does 'mM' mean in the question?

'mM' stands for millimolar. Just like 1 meter is 1000 millimeters, 1 Molar (M) is 1000 millimolar (mM). You multiply your final molarity by 1000 to get the answer in mM.

Q: What if the question mentioned phenolphthalein indicator?

If phenolphthalein was used, the titration would stop at the formation of $\text{NaHCO}_3$ (half-neutralization). In that specific case, the n-factor of $\text{Na}_2\text{CO}_3$ would be 1, and the volume of HCl consumed would be exactly half.

Question

10.0 mL of $	ext{Na}_2	ext{CO}_3$ solution is titrated against 0.2 M HCl solution. The following titre values were obtained in 5 readings. 4.8 mL, 4.9 mL, 5.0 mL, 5.0 mL and 5.0 mL based on these readings and convention of titrimetric estimation of concentration of $	ext{Na}_2	ext{CO}_3$ solution is ……… mM (Round off to the nearest integer).

Sigma Percentile · Accepted Answer

### The Art of Titration

Imagine you are standing in a brightly lit chemistry laboratory. In front of you is a classic titration setup.

You have a conical flask containing exactly $10.0 	ext{ mL}$ of a sodium carbonate ($	ext{Na}_2	ext{CO}_3$) solution, but its concentration is a mystery.

Suspended above it is a burette filled with a standard $0.2 	ext{ M}$ hydrochloric acid ($	ext{HCl}$) solution. Our mission is to uncover the hidden concentration of the base using the precision of volumetric analysis.

### Decoding the Titre Values

Before we dive into the math, we must look at the experimental data. The problem provides five different titre readings: $4.8 	ext{ mL}$, $4.9 	ext{ mL}$, $5.0 	ext{ mL}$, $5.0 	ext{ mL}$, and $5.0 	ext{ mL}$.

Why so many? In any physical experiment, the first few readings often contain slight human or instrumental errors.

To ensure absolute accuracy, chemists rely on the **concordant reading**—the value that repeats consistently. Here, $5.0 	ext{ mL}$ is our golden number. We will use $V_{	ext{HCl}} = 5.0 	ext{ mL}$ for all our calculations.

### The Master Equation: Law of Equivalence

The entire magic of titration rests on one beautiful principle: the Law of Equivalence.

At the exact moment the indicator changes color (the endpoint), the acid has perfectly neutralized the base. Mathematically, this means the milliequivalents of the acid equal the milliequivalents of the base.

$$ m_{eq}(	ext{Na}_2	ext{CO}_3) = m_{eq}(	ext{HCl}) $$

To expand this, we use the formula involving molarity ($M$), volume ($V$), and the crucial $n$-factor ($n$):

$$ M_1 	imes V_1 	imes n_1 = M_2 	imes V_2 	imes n_2 $$

### Unlocking the n-factors

This is where many students make a fatal silly mistake. You cannot simply equate moles; you must account for the $n$-factor!

Hydrochloric acid ($	ext{HCl}$) is a monoprotic acid. It releases exactly one $	ext{H}^+$ ion per molecule, so its $n$-factor is $1$.

Sodium carbonate ($	ext{Na}_2	ext{CO}_3$), however, is a salt of a weak acid that can accept two $	ext{H}^+$ ions to fully convert into carbonic acid (which then breaks down into water and carbon dioxide). Therefore, its $n$-factor is **$2$**.

### The Final Calculation

Now, we simply substitute our known values into the equivalence equation.

For the base ($	ext{Na}_2	ext{CO}_3$), we have an unknown molarity $M$, a volume of $10.0 	ext{ mL}$, and an $n$-factor of $2$. For the acid ($	ext{HCl}$), we have a molarity of $0.2 	ext{ M}$, a volume of $5.0 	ext{ mL}$, and an $n$-factor of $1$.

$$ M_{	ext{Na}_2	ext{CO}_3} 	imes 10.0 	imes 2 = 0.2 	imes 5.0 	imes 1 $$

Let's simplify the right side: $0.2 	imes 5.0$ is exactly $1.0$. On the left side, $10.0 	imes 2$ gives us $20$.

$$ 20 	imes M_{	ext{Na}_2	ext{CO}_3} = 1.0 $$

Dividing both sides by $20$, we find the molarity:

$$ M_{	ext{Na}_2	ext{CO}_3} = \frac{1.0}{20} = 0.05 	ext{ M} $$

### The Final Twist

We have the molarity, but the question sets a final trap. It asks for the concentration in **millimolar (mM)**, not molar (M).

Just as there are $1000$ millimeters in a meter, there are $1000$ millimolar units in one molar unit. We must multiply our result by $1000$.

$$ 	ext{Concentration} = 0.05 	imes 1000 = 50 	ext{ mM} $$

And there we have it! The unknown concentration of the sodium carbonate solution is exactly **$50 	ext{ mM}$**.

Enter Numerical Value:

Visualized Solution

\text{Titration Setup}

\text{Concordant Volume}

\text{Law of Equivalence}

\text{Determining n-factors}

\text{Substitution}

\text{Calculation}

\text{Final Answer in mM}

\text{The Way Forward}

The Sigma Insight: Stoichiometric and Volumetric Calculations

The Art of Titration

Decoding the Titre Values

The Master Equation

Unlocking the n-factors

The Final Calculation

The Final Twist

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