To measure the quantity of MnCl2 dissolved in an aqueous solution, it was completely converted to KM

Q: Why do we use equivalents instead of moles for the titration?

Equivalents account for the number of electrons transferred per molecule (the n-factor). This allows us to equate the reacting species directly without needing a balanced chemical equation.

Q: How do we know the n-factor of KMnO4 is 5?

In an acidic medium, the permanganate ion ($MnO_4^-$) where Mn is in the +7 oxidation state, is reduced to $Mn^{2+}$. The change in oxidation state is $7 - 2 = 5$, so the n-factor is 5.

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

Oxalic acid ($H_2C_2O_4$) oxidizes to $CO_2$. The carbon goes from +3 to +4. Since there are two carbon atoms per molecule, the total change is $2 \times (4 - 3) = 2$.

Q: Does the addition of K2S2O8 affect the titration?

The K2S2O8 is used to completely convert $Mn^{2+}$ to $MnO_4^-$. Any excess must be destroyed (usually by boiling) before titration, though the problem assumes ideal conditions where only $MnO_4^-$ reacts with oxalic acid.

Q: Why did we equate moles of MnCl2 to moles of KMnO4?

We apply the Principle of Atomic Conservation (POAC) on Manganese. All the Mn atoms from $MnCl_2$ end up in $KMnO_4$. Since both molecules contain exactly one Mn atom, their moles must be equal.

Question

To measure the quantity of $	ext{MnCl}_2$ dissolved in an aqueous solution, it was completely converted to $	ext{KMnO}_4$ using the reaction,
$	ext{MnCl}_2 + 	ext{K}_2	ext{S}_2	ext{O}_8 + 	ext{H}_2	ext{O} ightarrow 	ext{KMnO}_4 + 	ext{H}_2	ext{SO}_4 + 	ext{HCl}$ (equation not balanced).
Few drops of concentrated $	ext{HCl}$ were added to this solution and gently warmed. Further, oxalic acid ($225	ext{ mg}$) was added in portions till the colour of the permanganate ion disappeared. The quantity of $	ext{MnCl}_2$ (in $	ext{mg}$) present in the initial solution is ______.
(Atomic weights in $	ext{g mol}^{-1}$ : $	ext{Mn} = 55$, $	ext{Cl} = 35.5$)

Sigma Percentile · Accepted Answer

### The Mystery in the Beaker

Imagine you are a chemist handed a beaker containing a pale, almost colorless solution of manganese chloride ($MnCl_2$). Your mission is to find out exactly how much $MnCl_2$ is dissolved in it.

To do this, we can't just weigh it. We have to use the elegant power of redox titrations.

### Phase 1: The Oxidation Step

First, we need to transform our manganese into something we can easily measure. We add potassium persulfate ($K_2S_2O_8$), a very strong oxidizing agent.

This chemical beast rips electrons away from the $Mn^{2+}$ ions, forcing them into a highly oxidized $+7$ state.

$$Mn^{2+} \xrightarrow{	ext{Oxidation}} MnO_4^-$$

Suddenly, the solution turns a brilliant, deep purple. We have successfully converted all our $MnCl_2$ into potassium permanganate ($KMnO_4$).

### Phase 2: The Reduction Step

Now comes the actual measurement. We take this purple solution and titrate it with oxalic acid ($H_2C_2O_4$).

Oxalic acid is a reducing agent. As we add it drop by drop, it donates electrons back to the permanganate ions.

$$MnO_4^- + H_2C_2O_4 \xrightarrow{H^+} Mn^{2+} + CO_2$$

The moment the last drop of permanganate is reduced back to $Mn^{2+}$, the purple color completely vanishes. This is our endpoint!

### The Master Equation: Law of Equivalence

In any titration, the fundamental rule is the Law of Equivalence. It states that the equivalents of the oxidizing agent must equal the equivalents of the reducing agent.

$$	ext{Equivalents of } KMnO_4 = 	ext{Equivalents of } H_2C_2O_4$$

To calculate equivalents, we need the crucial **n-factor** for both chemicals.

For $KMnO_4$ in an acidic medium, manganese goes from $+7$ to $+2$. That is a change of $5$ electrons, so its n-factor is **$5$**.

For oxalic acid, each carbon atom goes from $+3$ to $+4$. Since there are two carbon atoms, the total electron change is $2$, making its n-factor **$2$**.

### Crunching the Numbers

We are given $225 	ext{ mg}$ of oxalic acid. Let's convert this to moles. The molar mass of anhydrous oxalic acid is $90 	ext{ g/mol}$.

$$	ext{Moles of } H_2C_2O_4 = \frac{225 	imes 10^{-3} 	ext{ g}}{90 	ext{ g/mol}} = 2.5 	imes 10^{-3} 	ext{ mol}$$

Now, we find its equivalents by multiplying by the n-factor.

$$	ext{Equivalents of } H_2C_2O_4 = 2.5 	imes 10^{-3} 	imes 2 = 5 	imes 10^{-3} 	ext{ eq}$$

By our Law of Equivalence, the equivalents of $KMnO_4$ must also be $5 	imes 10^{-3}$.

To find the moles of $KMnO_4$, we divide its equivalents by its n-factor of $5$.

$$	ext{Moles of } KMnO_4 = \frac{5 	imes 10^{-3}}{5} = 10^{-3} 	ext{ mol}$$

### The Final Calculation

Remember our very first step? We converted all the initial $MnCl_2$ into $KMnO_4$.

By the Principle of Atomic Conservation (POAC) on manganese, the moles of $MnCl_2$ must exactly equal the moles of $KMnO_4$ produced.

$$	ext{Moles of } MnCl_2 = 10^{-3} 	ext{ mol}$$

Finally, we multiply by the molar mass of $MnCl_2$, which is $55 + 2(35.5) = 126 	ext{ g/mol}$.

$$	ext{Mass of } MnCl_2 = 10^{-3} 	ext{ mol} 	imes 126 	ext{ g/mol} = 0.126 	ext{ g}$$

Converting this back to milligrams, we get our final, beautiful answer: **$126 	ext{ mg}$**.

Enter Numerical Value:

Visualized Solution

Initial State of $M n C l_{2}$

Reaction with $K_{2} S_{2} O_{8}$

Titration with $H_{2} C_{2} O_{4}$

Law of Equivalence

Calculating n-factors

Moles of $H_{2} C_{2} O_{4}$

Equivalents of $H_{2} C_{2} O_{4}$

Moles of $K M n O_{4}$

Mass of $M n C l_{2}$

Final Answer

The Way Forward

The Sigma Insight: Stoichiometric and Volumetric Calculations

The Mystery in the Beaker

Phase 1

Phase 2

The Master Equation

Crunching the Numbers

The Final Calculation

Similar Questions

Comprehension Passage

The value of x is ______.

The value of y is ______.

The volume, in mL, of $0.02 M K_{2} C r_{2} O_{7}$ solution required to react with $0.288 g$ of ferrous oxalate in acidic medium is …… . (Molar mass of $Fe = 56 g mol^{- 1}$ )

A $20.0 mL$ solution containing $0.2 g$ impure $H_{2} O_{2}$ reacts completely with $0.316 g$ of $KMnO_{4}$ in acid solution. The purity of $H_{2} O_{2}$ (in $%$ ) is ............. (molecular weight of $H_{2} O_{2} = 34$ ; molecular weight of $KMnO_{4} = 158$ ).

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)

The volume (in $mL$ ) of $0.125 M$ $AgNO_{3}$ required to quantitatively precipitate chloride ions in $0.3 g$ of $[Co (NH_{3})_{6}] Cl_{3}$ is ......... . $M_{[Co (NH_{3})_{6}] Cl_{3}} = 267.46 g/mol$ $M_{AgNO_{3}} = 169.87 g/mol$

15 mL of aqueous solution of $Fe^{2 +}$ in acidic medium completely reacted with 20 mL of 0.03 M aqueous $Cr_{2} O_{7}^{2 -}$ . The molarity of the $Fe^{2 +}$ solution is ...... $\times 1 0^{- 2}$ M (Round off to the nearest integer).

The mass of potassium dichromate crystals required to oxidise $750 cm^{3}$ of $0.6 M$ Mohr's salt solution is (molar mass $= 392$ )