10.0 mL of 0.05 M KMnO4 solution was consumed in a titration with 10.0 mL of given oxalic acid dihyd

Q: Why is the n-factor of $KMnO_4$ taken as 5?

In an acidic medium, the Manganese (Mn) in $KMnO_4$ is in a $+7$ oxidation state. During the redox reaction, it gets reduced to $Mn^{2+}$. The difference in oxidation state is $7 - 2 = 5$. Therefore, it accepts 5 electrons per molecule, making its n-factor 5.

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

In oxalic acid ($H_2C_2O_4$), the oxidation state of Carbon is $+3$. It oxidizes to form Carbon Dioxide ($CO_2$), where Carbon is in a $+4$ state. Since there are two Carbon atoms in one molecule of oxalic acid, the total change in oxidation state is $2 \times (4 - 3) = 2$. Hence, it loses 2 electrons per molecule, making its n-factor 2.

Q: Do I need to convert the volume from mL to Liters before using the formula?

No, it is not strictly necessary as long as you use the same unit on both sides of the equation $N_1V_1 = N_2V_2$. If both volumes are in mL, you are essentially equating milliequivalents, and the $10^{-3}$ conversion factor cancels out on both sides.

Q: What is the difference between Molarity and Strength?

Molarity is the concentration expressed in 'moles of solute per liter of solution' (mol/L). Strength is the concentration expressed in 'grams of solute per liter of solution' (g/L). You can easily convert between them using the formula: $\text{Strength} = \text{Molarity} \times \text{Molar Mass}$.

Q: Why did we multiply by 126 instead of 90 for the molar mass?

The question specifically mentions 'oxalic acid dihydrate', which has the chemical formula $H_2C_2O_4 \cdot 2H_2O$. When you weigh out this solid to make the solution, the water molecules are part of the mass. Anhydrous oxalic acid ($H_2C_2O_4$) has a molar mass of 90 g/mol, but the two water molecules add $2 \times 18 = 36$ g/mol, bringing the total molar mass to 126 g/mol.

Question

10.0 mL of 0.05 M $	ext{KMnO}_4$ solution was consumed in a titration with 10.0 mL of given oxalic acid dihydrate solution. The strength of given oxalic acid solution is ...... $	imes 10^{-2} 	ext{ g/L}$.
(Round off to the nearest integer)

Sigma Percentile · Accepted Answer

### The Purple Indicator: Mastering Redox Titrations

Imagine you are standing in a chemistry laboratory, holding a burette filled with the vibrant, deep purple solution of Potassium Permanganate ($KMnO_4$). Below it sits a conical flask containing a colorless solution of Oxalic Acid Dihydrate ($H_2C_2O_4 \cdot 2H_2O$). This is a classic redox titration, a beautiful dance of electrons where the purple color acts as its own self-indicator.

Our mission is to find the exact strength of the oxalic acid solution in grams per liter. Let's break down the chemistry step-by-step.

### The Master Equation: Law of Equivalence

In any titration, the fundamental principle that governs the reaction at the equivalence point is the **Law of Equivalence**. It states that the number of equivalents of the oxidizing agent must exactly equal the number of equivalents of the reducing agent.

Mathematically, this is expressed as:
$$N_1 V_1 = N_2 V_2$$

Since Normality ($N$) is the product of Molarity ($M$) and the n-factor ($n$), we can rewrite this as:
$$(n_1 	imes M_1) V_1 = (n_2 	imes M_2) V_2$$

### Decoding the n-factors

This is where many students stumble. The n-factor represents the number of electrons transferred per molecule during the redox reaction.

For **Potassium Permanganate** in an acidic medium, the Manganese ion reduces from a $+7$ oxidation state to a $+2$ oxidation state. 
$$Mn^{+7} + 5e^- ightarrow Mn^{+2}$$
This means it gains 5 electrons, so its n-factor ($n_1$) is **$5$**.

For **Oxalic Acid**, the Carbon atoms oxidize from a $+3$ state to a $+4$ state in Carbon Dioxide ($CO_2$). Since there are two carbon atoms per molecule of oxalic acid, the total electron loss is 2.
$$C_2O_4^{2-} ightarrow 2CO_2 + 2e^-$$
Thus, its n-factor ($n_2$) is **$2$**.

### Calculating the Molarity

Now, we substitute our known values into the equivalence equation. We know the volume of both solutions is $10.0 	ext{ mL}$, and the molarity of $KMnO_4$ is $0.05 	ext{ M}$.

$$5 	imes 0.05 	imes 10 = 2 	imes M_2 	imes 10$$

The volumes ($10 	ext{ mL}$) beautifully cancel out on both sides:
$$0.25 = 2 	imes M_2$$
$$M_2 = 0.125 	ext{ M}$$

We have successfully found the molarity of the oxalic acid solution!

### The Dihydrate Trap and Final Strength

The question asks for the **strength** of the solution in $	ext{g/L}$. To convert molarity (moles per liter) to strength (grams per liter), we must multiply by the molar mass.

Here lies a crucial detail: the chemical is **Oxalic Acid Dihydrate** ($H_2C_2O_4 \cdot 2H_2O$). We must include the mass of the two water molecules of crystallization.

$$	ext{Molar Mass} = 2(1) + 2(12) + 4(16) + 2(18) = 126 	ext{ g/mol}$$

Now, we calculate the strength:
$$	ext{Strength} = 	ext{Molarity} 	imes 	ext{Molar Mass}$$
$$	ext{Strength} = 0.125 	ext{ mol/L} 	imes 126 	ext{ g/mol} = 15.75 	ext{ g/L}$$

Finally, the question requests the answer in the format of $... 	imes 10^{-2} 	ext{ g/L}$. By shifting the decimal point two places to the right, we get:
$$15.75 = 1575 	imes 10^{-2}$$

Our final integer answer is **$1575$**. Always read the chemical names carefully, as missing the 'dihydrate' would have led to an incorrect molar mass of $90 	ext{ g/mol}$ and a completely wrong answer!

Animated Solution for Chemistry - Basic Concepts in Chemistry: 10.0 mL of 0.05 M $KMnO_{4}$ solution was consumed in a titration with 10.0 mL of given oxalic acid dihydrate solution. The strength of given oxalic acid solution is ...... $\times 1 0^{- 2} g/L$ . (Round off to the nearest integer)

Enter Numerical Value:

Visualized Solution

\text{Titration Setup}

\text{Given Data}

\text{Law of Equivalence}

\text{n-factors}

\text{Calculating Molarity}

\text{Molar Mass of Oxalic Acid Dihydrate}

\text{Strength of Solution}

\text{What if?}

The Sigma Insight: Stoichiometric and Volumetric Calculations

The Purple Indicator

The Master Equation

Decoding the n-factors

Calculating the Molarity

The Dihydrate Trap and Final Strength

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