40mathrm~mL$40\mathrm{~mL}$ of a mixture of mathrmCH_3mathrmCOOH$\mathrm{CH}_{3}\mathrm{COOH}$ and mathrmHCl$\mathrm{HCl}$ (aqueous solution) is titrated against 0.1mathrm~M~NaOH$0.1\mathrm{~M~NaOH}$ solution conductometrically. Which of the following statements is correct? Conductance vs Volume of NaOH added curve showing two equivalence points at 2.0 mL and 5.0 mL.
A.The concentration of mathrmCH_3mathrmCOOH$\mathrm{CH}_3\mathrm{COOH}$ in the original mixture is 0.005mathrm~M$0.005\mathrm{~M}$
B.The concentration of mathrmHCl$\mathrm{HCl}$ in the original mixture is 0.005mathrm~M$0.005\mathrm{~M}$
C.mathrmCH_3mathrmCOOH$\mathrm{CH}_3\mathrm{COOH}$ is neutralised first followed by neutralisation of mathrmHCl$\mathrm{HCl}$
D.Point 'C' indicates the complete neutralisation of mathrmHCl$\mathrm{HCl}$
Solution & Explanation
### Related Formula
At the equivalence point during titration:
M_textacid V_textacid = M_textbase V_textbase$$M_{\text{acid}} V_{\text{acid}} = M_{\text{base}} V_{\text{base}}$$
### Core Logic
In a mixture of a strong acid (mathrmHCl$\mathrm{HCl}$) and a weak acid (mathrmCH_3mathrmCOOH$\mathrm{CH}_3\mathrm{COOH}$):
1. mathrmHCl$\mathrm{HCl}$ is a strong acid and is completely ionized. When mathrmNaOH$\mathrm{NaOH}$ is added, highly mobile mathrmH^+$\mathrm{H}^+$ ions are replaced by less mobile mathrmNa^+$\mathrm{Na}^+$ ions, causing a sharp drop in conductance (segment AB).
2. At point B (2.0mathrm~mL$2.0\mathrm{~mL}$), mathrmHCl$\mathrm{HCl}$ is completely neutralized.
3. Segment BC represents the neutralization of the weak acid mathrmCH_3mathrmCOOH$\mathrm{CH}_3\mathrm{COOH}$ to form highly conducting sodium acetate, causing a moderate rise in conductance up to point C (5.0mathrm~mL$5.0\mathrm{~mL}$).
4. Beyond point C, excess mathrmOH^-$\mathrm{OH}^-$ ions cause a rapid rise in conductance (segment CD).
### Step 1: Calculate concentration of mathrmHCl$\mathrm{HCl}$
Volume of mathrmNaOH$\mathrm{NaOH}$ used to neutralize mathrmHCl$\mathrm{HCl}$ is V_1 = 2.0mathrm~mL$V_1 = 2.0\mathrm{~mL}$:
M_mathrmHCl times 40mathrm~mL = 0.1mathrm~M times 2.0mathrm~mL$$M_{\mathrm{HCl}} \times 40\mathrm{~mL} = 0.1\mathrm{~M} \times 2.0\mathrm{~mL}$$M_mathrmHCl = frac0.240 = 0.005mathrm~M$$M_{\mathrm{HCl}} = \frac{0.2}{40} = 0.005\mathrm{~M}$$
### Step 2: Calculate concentration of mathrmCH_3mathrmCOOH$\mathrm{CH}_3\mathrm{COOH}$
Volume of mathrmNaOH$\mathrm{NaOH}$ used to neutralize mathrmCH_3mathrmCOOH$\mathrm{CH}_3\mathrm{COOH}$ is V_2 = 5.0mathrm~mL - 2.0mathrm~mL = 3.0mathrm~mL$V_2 = 5.0\mathrm{~mL} - 2.0\mathrm{~mL} = 3.0\mathrm{~mL}$:
M_mathrmCH_3mathrmCOOH times 40mathrm~mL = 0.1mathrm~M times 3.0mathrm~mL$$M_{\mathrm{CH}_3\mathrm{COOH}} \times 40\mathrm{~mL} = 0.1\mathrm{~M} \times 3.0\mathrm{~mL}$$M_mathrmCH_3mathrmCOOH = frac0.340 = 0.0075mathrm~M$$M_{\mathrm{CH}_3\mathrm{COOH}} = \frac{0.3}{40} = 0.0075\mathrm{~M}$$
### 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 (mathrmH^+$\mathrm{H}^+$ depletion). A weak acid titration shows a gentle upward slope due to salt formation.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Electrochemistry
Class 11 Chemistry: Equilibrium
Keywords:#conductometric titration curve HCl CH3COOH#JEE Main 2025 Evening Q26#Electrochemistry conductometric titration#neutralization of strong and weak acid mixtures#Conductometric titration#Conductance curve#Equivalence points
More Electrochemistry Previous-Year Questions — Page 4
Q26jee_main_2025_24_jan_morningGalvanic Cells and Standard Cell Potential
For the given cell:
Fe^2+(aq) + Ag^+(aq) rightarrow Fe^3+(aq) + Ag(s)$$Fe^{2+}(aq) + Ag^{+}(aq) \rightarrow Fe^{3+}(aq) + Ag(s)$$
The standard cell potential of the above reaction is given by:
Ag^+ + e^- rightarrow Ag quad E^0 = xtext V$$Ag^{+} + e^{-} \rightarrow Ag \quad E^{0} = x\text{ V}$$Fe^2+ + 2e^- rightarrow Fe quad E^0 = ytext V$$Fe^{2+} + 2e^{-} \rightarrow Fe \quad E^{0} = y\text{ V}$$Fe^3+ + 3e^- rightarrow Fe quad E^0 = ztext V$$Fe^{3+} + 3e^{-} \rightarrow Fe \quad E^{0} = z\text{ V}$$
A.x + y - z$x + y - z$
B.x + 2y - 3z$x + 2y - 3z$
C.y - 2x$y - 2x$
D.x + 2y$x + 2y$
Solution
### Related Formula
Delta G^0 = -nFE^0$$\Delta G^{0} = -nFE^{0}$$
### Core Logic
Using Gibbs free energy changes for individual steps to find the target reduction potential:
1. Ag^+ + e^- rightarrow Ag quad Delta G_1^0 = -1Fx$$Ag^{+} + e^{-} \rightarrow Ag \quad \Delta G_1^0 = -1Fx$$
2. Fe^2+ + 2e^- rightarrow Fe quad Delta G_2^0 = -2Fy$$Fe^{2+} + 2e^{-} \rightarrow Fe \quad \Delta G_2^0 = -2Fy$$
3. Fe^3+ + 3e^- rightarrow Fe quad Delta G_3^0 = -3Fz$$Fe^{3+} + 3e^{-} \rightarrow Fe \quad \Delta G_3^0 = -3Fz$$
For the conversion of Fe^2+ rightarrow Fe^3+ + e^-$$Fe^{2+} \rightarrow Fe^{3+} + e^{-}$$, we compute the free energy change as:
Delta G^0 = Delta G_2^0 - Delta G_3^0 = -2Fy - (-3Fz) = 3Fz - 2Fy$$\Delta G^0 = \Delta G_2^0 - \Delta G_3^0 = -2Fy - (-3Fz) = 3Fz - 2Fy$$
Thus, E^0_Fe^2+/Fe^3+ = 2y - 3z$$E^{0}_{Fe^{2+}/Fe^{3+}} = 2y - 3z$$
Combining with silver reduction:
E^0_cell = E^0_Ag^+/Ag + E^0_Fe^2+/Fe^3+ = x + 2y - 3z$$E^{0}_{cell} = E^{0}_{Ag^{+}/Ag} + E^{0}_{Fe^{2+}/Fe^{3+}} = x + 2y - 3z$$
### Step 1: Final Calculation
The overall potential equals x + 2y - 3z$x + 2y - 3z$. Galvanic Cells and Standard Cell Potential diagram for Q26 - JEE Main 2025 Morning
### Pattern Recognition
Direct application of Delta G^0$\Delta G^0$ summation. Remember that standard cell potentials cannot be added directly unless the number of electrons involved is identical.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Electrochemistry
Q48jee_main_2025_28_jan_eveningFaraday's Laws of Electrolysis
Electrolysis of 600mathrm~mL$600\mathrm{~mL}$ aqueous solution of NaCl for 5mathrm\ min$5\mathrm{\ \min}$ changes the mathrmpH$\mathrm{pH}$ of the solution to 12. The current in Amperes used for the given electrolysis is ______ (Nearest integer).
Numerical Answer.Answer: 2 to 2
Solution
### Related Formula
Faraday's law of electrolysis equation:
textMoles of electrons (equivalents) = fracI cdot tF$$\text{Moles of electrons (equivalents)} = \frac{I \cdot t}{F}$$
Water ion product relation:
textpH + textpOH = 14$$\text{pH} + \text{pOH} = 14$$
### Core Logic
During the electrolysis of brine (NaCl(aq)$NaCl(aq)$), hydroxide ions (OH^-$OH^-$) are generated at the cathode:
2H_2O + 2e^- rightarrow H_2 + 2OH^-$$2H_2O + 2e^- \rightarrow H_2 + 2OH^-$$
Given metrics:
- Final textpH = 12 implies textpOH = 14 - 12 = 2$\text{pH} = 12 \implies \text{pOH} = 14 - 12 = 2$
- [OH^-] = 10^-2mathrm\ M$[OH^-] = 10^{-2}\mathrm{\ M}$
- textVolume = 600mathrm\ mL = 0.6mathrm\ L$\text{Volume} = 600\mathrm{\ mL} = 0.6\mathrm{\ L}$
- textTime = 5mathrm\ min = 300mathrm\ s$\text{Time} = 5\mathrm{\ \min} = 300\mathrm{\ s}$
### Step 1: Calculate Moles of Hydroxide Produced
Find the absolute moles of OH^-$OH^-$ ions generated:
textMoles = textMolarity times textVolume (L) = 10^-2 times 0.6 = 6 times 10^-3text moles$$\text{Moles} = \text{Molarity} \times \text{Volume (L)} = 10^{-2} \times 0.6 = 6 \times 10^{-3}\text{ moles}$$
### Step 2: Relate to Electrical Current
Since 1$1$ mole of electrons produces 1$1$ mole of OH^-$OH^-$, the moles of charge equals 6 times 10^-3$6 \times 10^{-3}$. Applying Faraday's equation:
6 times 10^-3 = fracI times 30096500$$6 \times 10^{-3} = \frac{I \times 300}{96500}$$I = frac6 times 10^-3 times 96500300 = 1.93mathrm\ A$$I = \frac{6 \times 10^{-3} \times 96500}{300} = 1.93\mathrm{\ A}$$
Rounding to the nearest integer gives 2$2$.
### Pattern Recognition
Always convert a given textpH$\text{pH}$ value into [OH^-]$[OH^-]$ concentration when dealing with cathodic water reduction. Tracking the relationship where 1\ e^- equiv 1\ OH^-$1\ e^- \equiv 1\ OH^-$ provides a direct shortcut to link textpH$\text{pH}$ changes to current flow.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Electrochemistry
Qjee_main_2025_29_jan_morningStandard Reduction Potential and Oxidising Power
The standard reduction potential values of some of the p-block ions are given below. Predict the one with the strongest oxidising capacity.
### Related Formula
textOxidising Capacity propto textStandard Reduction Potential (E^ominus)$$\text{Oxidising Capacity} \propto \text{Standard Reduction Potential } (E^{\ominus})$$
### Core Logic
A higher positive value of standard reduction potential (E^ominus$E^{\ominus}$) indicates a stronger tendency to undergo reduction, hence behaving as a stronger oxidising agent. Comparing the given values:
* mathrmE_mathrmSn^4+/mathrmSn^2+^ominus = +1.15mathrmV$\mathrm{E}_{\mathrm{Sn}^{4+}/\mathrm{Sn}^{2+}}^{\ominus} = +1.15\mathrm{V}$
* mathrmE_mathrmTl^3+/mathrmTl^ominus = +1.26mathrmV$\mathrm{E}_{\mathrm{T}l^{3+}/\mathrm{T}l}^{\ominus} = +1.26\mathrm{V}$
* mathrmE_mathrmAl^3+/mathrmAl^ominus = -1.66mathrmV$\mathrm{E}_{\mathrm{Al}^{3+}/\mathrm{Al}}^{\ominus} = -1.66\mathrm{V}$
* mathrmE_mathrmPb^4+/mathrmPb^2+^ominus = +1.67mathrmV$\mathrm{E}_{\mathrm{Pb}^{4+}/\mathrm{Pb}^{2+}}^{\ominus} = +1.67\mathrm{V}$
Since +1.67mathrmV$+1.67\mathrm{V}$ is the highest value, mathrmPb^4+$\mathrm{Pb}^{4+}$ possesses the strongest oxidising capacity.
### Pattern Recognition
Strongest oxidising agent = Most positive reduction potential. Weakest oxidising agent = Most negative reduction potential.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Electrochemistry
Class 12 Chemistry: p-Block Elements
Qjee_main_2025_29_jan_morningVariation of Molar Conductivity with Concentration
The molar conductivity of a weak electrolyte when plotted against the square root of its concentration, which of the following is expected to be observed?
A. A small decrease in molar conductivity is observed at infinite dilution.
B. A small increase in molar conductivity is observed at infinite dilution.
C. Molar conductivity increases sharply with increase in concentration.
D. Molar conductivity decreases sharply with increase in concentration.
Solution
### Related Formula
For weak electrolytes, the degree of dissociation alpha$\alpha$ increases sharply near infinite dilution according to Ostwald's Dilution Law:
alpha = sqrtfracK_aC$$\alpha = \sqrt{\frac{K_a}{C}}$$
### Core Logic
When a weak electrolyte is diluted (concentration C rightarrow 0$C \rightarrow 0$), its molar conductivity increases steeply. Conversely, when plotted against sqrtC$\sqrt{C}$, as concentration increases, the degree of dissociation drops rapidly, causing a sharp decrease in molar conductivity. This matches the curve given below:
Variation of Molar Conductivity with Concentration diagram for Q28 - JEE Main 2025 Morning
### Pattern Recognition
Weak electrolyte plots feature a steep asymptotic exponential-like rise towards the y-axis as C rightarrow 0$C \rightarrow 0$, meaning a sharp decrease occurs with increasing concentration.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Electrochemistry
Qjee_main_2025_29_jan_morningNernst Equation
For a mathrmMg mid mathrmMg^2+$\mathrm{Mg} \mid \mathrm{Mg}^{2+}$ (aq) parallel mathrmAg^+(mathrmaq) mid mathrmAg$\parallel \mathrm{Ag}^{+}(\mathrm{aq}) \mid \mathrm{Ag}$ the correct Nernst Equation is :
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