40~mL of a mixture of CH3COOH and HCl (aqueous solution) is titrated against 0.1~M~NaOH solution conductometrically…

Year	2026	2025	2024	Total
Questions	13	19	8	40

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

At the equivalence point during titration:

Macid Vacid = Mbase Vbase

Core Logic

In a mixture of a strong acid ( $HCl$ $\mathrm{HCl}$) and a weak acid ( $CH₃COOH$ $\mathrm{CH}_3\mathrm{COOH}$):

$HCl$ $\mathrm{HCl}$ is a strong acid and is completely ionized. When $NaOH$ $\mathrm{NaOH}$ is added, highly mobile $H^+$ $\mathrm{H}^+$ ions are replaced by less mobile $Na^+$ $\mathrm{Na}^+$ ions, causing a sharp drop in conductance (segment AB).
At point B ( $2.0~mL$ $2.0\mathrm{~mL}$), $HCl$ $\mathrm{HCl}$ is completely neutralized.
Segment BC represents the neutralization of the weak acid $CH₃COOH$ $\mathrm{CH}_3\mathrm{COOH}$ to form highly conducting sodium acetate, causing a moderate rise in conductance up to point C ( $5.0~mL$ $5.0\mathrm{~mL}$).
Beyond point C, excess $OH^-$ $\mathrm{OH}^-$ ions cause a rapid rise in conductance (segment CD).

Step 1: Calculate concentration of $HCl$ $\mathrm{HCl}$

Volume of $NaOH$ $\mathrm{NaOH}$ used to neutralize $HCl$ $\mathrm{HCl}$ is $V₁ = 2.0~mL$ $V_1 = 2.0\mathrm{~mL}$:

MHCl \times 40~mL = 0.1~M \times 2.0~mL

MHCl = (0.2)/(40) = 0.005~M

Step 2: Calculate concentration of $CH₃COOH$ $\mathrm{CH}_3\mathrm{COOH}$

Volume of $NaOH$ $\mathrm{NaOH}$ used to neutralize $CH₃COOH$ $\mathrm{CH}_3\mathrm{COOH}$ is $V₂ = 5.0~mL - 2.0~mL = 3.0~mL$ :

MCH₃COOH \times 40~mL = 0.1~M \times 3.0~mL

MCH₃COOH = (0.3)/(40) = 0.0075~M

Pattern Recognition

Conductometric titration curves are analyzed sequentially: the strongest electrolyte is always neutralized first. A steep drop in conductance always signals the neutralization of a strong acid ( $H^+$ $\mathrm{H}^+$ depletion). A weak acid titration shows a gentle upward slope due to salt formation.

Chapter Mix

Class 12 Chemistry: Electrochemistry Class 11 Chemistry: Equilibrium

More Electrochemistry Previous-Year Questions — Page 8

Q82 jee_main_2024_29_jan_morning Faradays Laws of Electrolysis

The mass of zinc produced by the electrolysis of zinc sulphate solution with a steady current of 0.015 A for 15 minutes is

______ \times 10⁻⁴

g. (Atomic mass of zinc

= 65.4

amu)

Numerical Answer. Answer: 45.75 to 46

Solution

Related Formula

W = Z \cdot I \cdot t = (M)/(n \cdot F) \cdot I \cdot t

where, $W$ $W$ = mass deposited $Z$ $Z$ = electrochemical equivalent $I$ $I$ = current in amperes $t$ $t$ = time in seconds $M$ $M$ = molar mass $n$ $n$ = n-factor (electrons exchanged) $F$ $F$ = Faraday's constant ( $96500 C/mol$ $96500 \text{ C/mol}$)

Core Logic

The electrolysis of zinc sulphate ( $ZnSO₄$ $ZnSO_4$) involves the reduction of zinc ions at the cathode:

Zn⁺² + 2e^- arrow Zn

Here, the n-factor ( $n$ $n$) is 2.

Step 1: Calculation

Given values: $I = 0.015 A$ $I = 0.015\text{ A}$ $t = 15 minutes = 15 \times 60 seconds = 900 s$ $M = 65.4 g/mol$ $M = 65.4\text{ g/mol}$ $F \approx 96500 C$ $F \approx 96500\text{ C}$

Plugging the values into Faraday's First Law:

W = (65.4)/(2 \times 96500) \times 0.015 \times 15 \times 60

W = (65.4)/(193000) \times 13.5

W = 3.3886 \times 10⁻⁴ \times 13.5

W = 45.746 \times 10⁻⁴ g

Rounding to two decimal places (or nearest integer depending on convention), we get $45.75 \times 10⁻⁴ g$ $45.75 \times 10^{-4}\text{ g}$.

Chapter Mix

Class 12 Chemistry: Electrochemistry

Q80 jee_main_2024_30_january_evening Standard Electrode Potential

Reduction potential of ions are given below:

ClO₄^- E° = 1.19V

IO₄^- E° = 1.65V

BrO₄^- E° = 1.74V

The correct order of their oxidising power is:

A. $ClO₄^- > IO₄^- > BrO₄^-$
B. $BrO₄^- > IO₄^- > ClO₄^-$
C. $BrO₄^- > ClO₄^- > IO₄^-$
D. $IO₄^- > BrO₄^- > ClO₄^-$

Solution

Core Logic

The Standard Reduction Potential ( $E^°$ $E^\circ$) measures a species' tendency to undergo reduction (gain electrons). A higher, more positive $E^°$ $E^\circ$ value means the species has a stronger tendency to be reduced, which in turn makes it a stronger oxidizing agent.

Comparing the given $E^°$ $E^\circ$ values: $BrO₄^-$ $\mathrm{BrO}_4^-$: $1.74V$ $1.74\mathrm{V}$ $IO₄^-$ $\mathrm{IO}_4^-$: $1.65V$ $1.65\mathrm{V}$ $ClO₄^-$ $\mathrm{ClO}_4^-$: $1.19V$ $1.19\mathrm{V}$

The order of oxidizing power follows the magnitude of the reduction potential: $BrO₄^- > IO₄^- > ClO₄^-$

Pattern Recognition

Higher +ve Standard Reduction Potential (SRP) = Stronger Oxidising Agent.

Chapter Mix

Class 12 Chemistry: Electrochemistry Class 12 Chemistry: The p Block Elements

Q65 jee_main_2024_31_jan_morning Batteries

The metals that are employed in the battery industries are A. Fe B. Mn C. Ni D. Cr E. Cd Choose the correct answer from the options given below:

A. $B, C and E only$ $\text{B, C and E only}$
B. $A, B, C, D and E$ $\text{A, B, C, D and E}$
C. $A, B, C and D only$ $\text{A, B, C and D only}$
D. $B, D and E only$ $\text{B, D and E only}$

Solution

Core Logic

Mn, Ni, and Cd metals are predominantly used in battery industries.

Mn is used in dry cells (Leclanche cell).
Ni and Cd are used in Nickel-Cadmium (Ni-Cd) rechargeable batteries.

Chapter Mix

Class 12 Chemistry: Electrochemistry

Q67 jee_main_2024_31_jan_morning Electrolytic Conductance

Identify the factor from the following that does not affect electrolytic conductance of a solution.

A. $The nature of the electrolyte added.$
B. $The nature of the electrode used.$
C. $Concentration of the electrolyte.$
D. $The nature of solvent used.$ $\text{The nature of solvent used.}$

Solution

Core Logic

Conductivity of an electrolytic cell is affected by the concentration of the electrolyte, the nature of the electrolyte, and the nature of the solvent. It does not depend on the nature of the electrode used.

Chapter Mix

Class 12 Chemistry: Electrochemistry

Q90 jee_main_2024_31_jan_morning Faraday's Laws of Electrolysis

One Faraday of electricity liberates

x \times 10⁻¹

gram atom of copper from copper sulphate, x is

Numerical Answer. Answer: 5 to 5

Solution

Core Logic

The reduction reaction for copper is:

Cu²⁺ + 2e^- arrow Cu

This shows that 2 moles of electrons (2 Faraday) are required to deposit 1 mole (or 1 gram atom) of Cu.

Therefore, 1 Faraday of electricity will deposit:

(1)/(2) = 0.5 moles of Cu

Step 1: Finding x

0.5 mole = 0.5 gram atom = 5 \times 10⁻¹ gram atom

Hence, $x = 5$ $x = 5$.

Chapter Mix

Class 12 Chemistry: Electrochemistry

Solution & Explanation

Related Formula

Core Logic

Step 1: Calculate concentration of $HCl$ $\mathrm{HCl}$

Step 2: Calculate concentration of $CH₃COOH$ $\mathrm{CH}_3\mathrm{COOH}$

Pattern Recognition

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Solution & Explanation

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Core Logic

Step 1: Calculate concentration of HCl$\mathrm{HCl}$

Step 2: Calculate concentration of CH₃COOH$\mathrm{CH}_3\mathrm{COOH}$

Pattern Recognition

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More Electrochemistry Previous-Year Questions — Page 8

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Step 1: Calculate concentration of $HCl$ $\mathrm{HCl}$

Step 2: Calculate concentration of $CH₃COOH$ $\mathrm{CH}_3\mathrm{COOH}$