When 10 mL of an aqueous solution of Fe2+ ions was titrated in the presence of dil. H2SO4 using diph

Q: Why do we use equivalents instead of moles for this calculation?

Using equivalents allows us to bypass the tedious process of writing and balancing the full complex redox equation. By definition, 1 equivalent of any substance reacts with exactly 1 equivalent of another.

Q: How did we get an n-factor of 6 for dichromate? I thought the change was only 3.

You are right that each Chromium atom goes from a +6 to a +3 oxidation state, which is a change of 3. However, a single dichromate ion ($Cr_2O_7^{2-}$) contains TWO Chromium atoms. Therefore, the total electron transfer per molecule is $3 \times 2 = 6$.

Q: What is the role of $H_2SO_4$ in this titration?

Sulfuric acid provides the acidic medium ($H^+$ ions) necessary for the dichromate ion to act as a strong oxidizing agent and successfully reduce to $Cr^{3+}$.

Q: Why is diphenylamine used as an indicator?

Diphenylamine is a redox indicator. It changes color when the oxidation potential of the solution jumps sharply, which happens exactly at the endpoint when all the $Fe^{2+}$ has been consumed.

Q: Can we use $HCl$ instead of $H_2SO_4$ to make the medium acidic?

No, we cannot. Potassium dichromate is a strong enough oxidizing agent to oxidize the chloride ions ($Cl^-$) in $HCl$ into chlorine gas ($Cl_2$). This would consume extra dichromate, leading to an inaccurate, higher volume reading for the titration.

Question

When $10	ext{ mL}$ of an aqueous solution of $Fe^{2+}$ ions was titrated in the presence of dil. $H_2SO_4$ using diphenylamine indicator, $15	ext{ mL}$ of $0.02	ext{ M}$ solution of $K_2Cr_2O_7$ was required to get the end point. The molarity of the solution containing $Fe^{2+}$ ions is $x 	imes 10^{-2}	ext{ M}$. The value of $x$ is …… . (Nearest integer)

Sigma Percentile · Accepted Answer

### The Setup: A Classic Redox Battle

Imagine you are standing in a chemistry lab, looking at a classic titration setup. In your conical flask, you have $10	ext{ mL}$ of an unknown concentration of iron(II) ions ($Fe^{2+}$). Suspended above it is a burette filled with a standard solution: $15	ext{ mL}$ of $0.02	ext{ M}$ potassium dichromate ($K_2Cr_2O_7$).

Our mission is to find the exact molarity of the iron solution. To do this without writing out and balancing a massive, complex redox equation, we rely on the most powerful tool in volumetric analysis: **The Law of Equivalence**.

### The Law of Equivalence

The Law of Equivalence states that at the endpoint of a titration, the number of equivalents (or milliequivalents) of the reducing agent must perfectly equal the number of equivalents of the oxidizing agent.

Mathematically, this is expressed as:

$$n_1 M_1 V_1 = n_2 M_2 V_2$$

Here, $n$ represents the **n-factor** (the number of electrons lost or gained per molecule), $M$ is the molarity, and $V$ is the volume.

### Decoding the n-factors

Before we can plug numbers into our master equation, we need to determine the n-factors for both reactants.

First, let's look at the reducing agent, iron. Iron oxidizes from a $+2$ state to a $+3$ state:

$$Fe^{2+} \longrightarrow Fe^{3+} + e^-$$

Since exactly one electron is lost per iron atom, its n-factor is simply **$n_1 = 1$**.

Now, let's analyze the oxidizing agent, the dichromate ion ($Cr_2O_7^{2-}$). In an acidic medium, chromium goes from a $+6$ oxidation state down to $+3$:

$$Cr_2O_7^{2-} \longrightarrow 2Cr^{3+}$$

This is a change of $3$ electrons per chromium atom. However, because there are **two** chromium atoms in a single dichromate ion, the total change in oxidation state is $3 	imes 2 = 6$. Therefore, its n-factor is **$n_2 = 6$**.

*(Note: Forgetting to multiply by 2 here is one of the most common silly mistakes students make!)*

### The Final Calculation

Now we have all the pieces of the puzzle. Let's substitute our known values into the equivalence equation:

$$1 	imes M_1 	imes 10 = 6 	imes 0.02 	imes 15$$

Let's simplify the right side of the equation. $6 	imes 15$ is $90$, and $90 	imes 0.02$ gives us $1.8$.

$$10 M_1 = 1.8$$

Dividing both sides by $10$, we find the molarity of the iron solution:

$$M_1 = 0.18	ext{ M}$$

The question specifically asks for the answer in the format of $x 	imes 10^{-2}	ext{ M}$. So, we rewrite $0.18$ as:

$$M_1 = 18 	imes 10^{-2}	ext{ M}$$

Comparing this with our target format, we can clearly see that the value of $x$ is exactly **$18$**.

Enter Numerical Value:

Visualized Solution

$T i t r a t i o n S e t u p$

$L a w o f E q u i v a l e n ce$

$n -factor of F e^{2 +}$

$n -factor of C r_{2} O_{7}^{2 -}$

$S u b s t i t u t i o n$

$C a l c u l a t i o n$

$F ina l A n s w er$

$C o n ce pt u a l C h ec k$

The Sigma Insight: Stoichiometric and Volumetric Calculations

The Setup

The Law of Equivalence

Decoding the n-factors

The Final Calculation

Similar Questions

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).

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 $Fe = 56 g mol^{- 1}$ )

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)

A $100 mL$ solution was made by adding $1.43 g$ of $Na_{2} CO_{3} \cdot x H_{2} O$ . The normality of the solution is $0.1 N$ . The value of $x$ is ......... .

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Titration Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Law of Equivalence

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n-factor of Fe2+

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n-factor of Cr2​O72−​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substitution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conceptual Check