The pH and conductance of a weak acid (HX) was found to be 5 and 4 times 10^-5$4 \times 10^{-5}$ S, respectively. The conductance was measured under standard condition using a cell where the electrode plates having a surface area of 1text cm^2$1\text{ cm}^2$ were at a distance of 15text cm$15\text{ cm}$ apart. The value of the limiting molar conductivity is ..... textS m^2text mol^-1$\text{S m}^2\text{ mol}^{-1}$. (nearest integer)
(Given: degree of dissociation of the weak acid (alpha) ll 1)$(\alpha) \ll 1)$
Numerical Answer Type:
Enter a numerical valueAnswer: 6 to 6+4 marks
Solution & Explanation
### Related Formula
kappa = G times fraclA$$\kappa = G \times \frac{l}{A}$$Lambda_m = frackappa times 1000C$$\Lambda_m = \frac{\kappa \times 1000}{C}$$alpha = fracLambda_mLambda_m^infty$$\alpha = \frac{\Lambda_m}{\Lambda_m^{\infty}}$$[H^+] = Calpha$[H^+] = C\alpha$
### Core Logic
Given: pH = 5$pH = 5$, so [H^+] = 10^-5text M$[H^+] = 10^{-5}\text{ M}$.
Since [H^+] = C cdot alpha = C cdot fracLambda_mLambda_m^infty$[H^+] = C \cdot \alpha = C \cdot \frac{\Lambda_m}{\Lambda_m^{\infty}}$, we have 10^-5 = C cdot fracLambda_mLambda_m^infty$10^{-5} = C \cdot \frac{\Lambda_m}{\Lambda_m^{\infty}}$.
First, calculate conductivity (kappa$\kappa$):
Conductance G = 4 times 10^-5text S$G = 4 \times 10^{-5}\text{ S}$.
Cell constant G^* = fraclA = frac15text cm1text cm^2 = 15text cm^-1$G^* = \frac{l}{A} = \frac{15\text{ cm}}{1\text{ cm}^2} = 15\text{ cm}^{-1}$.
kappa = G cdot G^* = (4 times 10^-5) times 15 = 6 times 10^-4text S cm^-1$$\kappa = G \cdot G^* = (4 \times 10^{-5}) \times 15 = 6 \times 10^{-4}\text{ S cm}^{-1}$$
Molar conductivity (Lambda_m$\Lambda_m$):
Lambda_m = frackappa times 1000C = frac6 times 10^-4 times 1000C = frac0.6C$$\Lambda_m = \frac{\kappa \times 1000}{C} = \frac{6 \times 10^{-4} \times 1000}{C} = \frac{0.6}{C}$$
Substitute Lambda_m$\Lambda_m$ into the proton concentration formula:
[H^+] = 10^-5 = C cdot frac0.6 / CLambda_m^infty$$[H^+] = 10^{-5} = C \cdot \frac{0.6 / C}{\Lambda_m^{\infty}}$$10^-5 = frac0.6Lambda_m^infty$$10^{-5} = \frac{0.6}{\Lambda_m^{\infty}}$$Lambda_m^infty = frac0.610^-5 = 60000text S cm^2text mol^-1$$\Lambda_m^{\infty} = \frac{0.6}{10^{-5}} = 60000\text{ S cm}^2\text{ mol}^{-1}$$
Convert units to textS m^2text mol^-1$\text{S m}^2\text{ mol}^{-1}$:
Since 1text m^2 = 10^4text cm^2$1\text{ m}^2 = 10^4\text{ cm}^2$,
Lambda_m^infty = 60000 times 10^-4text S m^2text mol^-1 = 6text S m^2text mol^-1$$\Lambda_m^{\infty} = 60000 \times 10^{-4}\text{ S m}^2\text{ mol}^{-1} = 6\text{ S m}^2\text{ mol}^{-1}$$
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Electrochemistry
Class 11 Chemistry: Equilibrium
Keywords:#limiting molar conductivity#weak acid conductance#JEE Main 2026 Morning Q72#Electrochemistry JEE Main 2026
More Electrochemistry Previous-Year Questions — Page 3
Q46jee_main_2025_28_jan_morningMolar Conductivity and Cell Resistance
Given below is the plot of the molar conductivity vs sqrttextconcentration$\sqrt{\text{concentration}}$ for KCl in aqueous solution.
The image features a standard linear plot tracing electrolytic molar conductance trends over root concentration variations.
If, for the higher concentration of KCl solution, the resistance of the conductivity cell is 100Omega$100\Omega$ then the resistance of the same cell with the dilute solution is mathrmxOmega$\mathrm{x}\Omega$
The value of mathbfx$\mathbf{x}$ is (Nearest integer)
Numerical Answer.Answer: 150 to 150
Solution
### Related Formula
Conductivity relationship with cell parameters:
kappa = G cdot G^* = fracG^*R$$\kappa = G \cdot G^* = \frac{G^*}{R}$$lambda_m = frackappa times 1000C$$\lambda_m = \frac{\kappa \times 1000}{C}$$
where G^*$G^*$ represents the static cell constant.
### Step 1: Setting Up Ratios
Using concentration subscripts c$c$ (concentrated) and d$d$ (dilute):
frackappa_ckappa_d = fracR_dR_c$$\frac{\kappa_c}{\kappa_d} = \frac{R_d}{R_c}$$
Expressing conductivity through molar conductivity values:
kappa = fraclambda_m cdot C1000$$\kappa = \frac{\lambda_m \cdot C}{1000}$$frac(lambda_m cdot C)_c(lambda_m cdot C)_d = fracR_dR_c$$\frac{(\lambda_m \cdot C)_c}{(\lambda_m \cdot C)_d} = \frac{R_d}{R_c}$$
Substituting the graphical read coordinates (C_c = 0.15^2, C_d = 0.1^2$C_c = 0.15^2, C_d = 0.1^2$ with scaled lambda_m$\lambda_m$ parameters):
frac100 cdot (0.15)^2150 cdot (0.1)^2 = fracR_d100$$\frac{100 \cdot (0.15)^2}{150 \cdot (0.1)^2} = \frac{R_d}{100}$$R_d = 150\,Omega$$R_d = 150\,\Omega$$
### Pattern Recognition
Sees: Resistance correlation across specific graph coordinates.
Shortcut: Equate cell parameters through kappa propto frac1R$\kappa \propto \frac{1}{R}$ and solve for the target resistance directly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Electrochemistry
### Core Logic
Limiting molar conductivity relies heavily on the charge and hydrodynamic radius of the hydrated ion. Let us verify the standard experimental limiting molar ionic conductivities (lambda^circ$\lambda^{\circ}$) at 298text K$298\text{ K}$:
* H^+: 349.8text S cm^2text mol^-1$H^{+}: 349.8\text{ S cm}^2\text{ mol}^{-1}$ (exhibits Grotthuss proton-hopping conduction mechanism)
* Ca^2+: 119.0text S cm^2text mol^-1$Ca^{2+}: 119.0\text{ S cm}^2\text{ mol}^{-1}$
* Mg^2+: 106.1text S cm^2text mol^-1$Mg^{2+}: 106.1\text{ S cm}^2\text{ mol}^{-1}$
* K^+: 73.5text S cm^2text mol^-1$K^{+}: 73.5\text{ S cm}^2\text{ mol}^{-1}$
* Na^+: 50.1text S cm^2text mol^-1$Na^{+}: 50.1\text{ S cm}^2\text{ mol}^{-1}$
### Step 1: Trend Layout
Arranging these values in descending order yields:
H^+ > Ca^2+ > Mg^2+ > K^+ > Na^+$$H^{+} > Ca^{2+} > Mg^{2+} > K^{+} > Na^{+}$$
### Pattern Recognition
Shortcut: H^+$H^{+}$ always has the absolute highest value due to its unique proton-hopping transport system. For metal ions, a higher ionic charge boosts conductivity (M^2+ > M^+$M^{2+} > M^{+}$), and within a group, a smaller hydrated radius (larger bare ion) increases mobility (K^+ > Na^+$K^{+} > Na^{+}$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Electrochemistry
Q49jee_main_2025_04_april_eveningConductance of Electrolytic Solutions
The molar conductance of an infinitely dilute solution of ammonium chloride was found to be 185mathrm~S~cm^2mathrmmol^-1$185\mathrm{~S~cm}^2\mathrm{mol}^{-1}$ and the ionic conductance of hydroxyl and chloride ions are 170 and 70mathrm~S~cm^2mathrmmol^-1$70\mathrm{~S~cm}^2\mathrm{mol}^{-1}$ , respectively. If molar conductance of 0.02mathrm~M$0.02\mathrm{~M}$ solution of ammonium hydroxide is 85.5mathrm~S~cm^2mathrmmol^-1$85.5\mathrm{~S~cm}^2\mathrm{mol}^{-1}$ , its degree of dissociation is given by mathbfxtimes 10^-1$\mathbf{x}\times 10^{-1}$ . The value of mathbfx$\mathbf{x}$ is _______. (Nearest integer)
Numerical Answer.Answer: 2.9 to 3.1
Solution
### Related Formula
Lambda_m^circ(NH_4OH) = lambda^circ(NH_4^+) + lambda^circ(OH^-) quad text(Kohlrausch's Law)$$\Lambda_m^\circ(NH_4OH) = \lambda^\circ(NH_4^+) + \lambda^\circ(OH^-) \quad \text{(Kohlrausch's Law)}$$alpha = fracLambda_m^cLambda_m^circ$$\alpha = \frac{\Lambda_m^c}{\Lambda_m^\circ}$$
### Core Logic
1. Find the limiting molar conductance of NH_4^+$NH_4^+$ using the NH_4Cl$NH_4Cl$ data:
Lambda_m^circ(NH_4Cl) = lambda^circ(NH_4^+) + lambda^circ(Cl^-) = 185$$\Lambda_m^\circ(NH_4Cl) = \lambda^\circ(NH_4^+) + \lambda^\circ(Cl^-) = 185$$lambda^circ(NH_4^+) = 185 - 70 = 115 mathrm~S cdot cm^2 cdot mol^-1$$\lambda^\circ(NH_4^+) = 185 - 70 = 115 \mathrm{~S \cdot cm^2 \cdot mol^{-1}}$$
2. Calculate Lambda_m^circ$\Lambda_m^\circ$ for the weak electrolyte ammonium hydroxide (NH_4OH$NH_4OH$):
Lambda_m^circ(NH_4OH) = 115 + 170 = 285 mathrm~S cdot cm^2 cdot mol^-1$$\Lambda_m^\circ(NH_4OH) = 115 + 170 = 285 \mathrm{~S \cdot cm^2 \cdot mol^{-1}}$$
3. Evaluate the degree of dissociation (alpha$\alpha$):
alpha = frac85.5285 = 0.3 = 3 times 10^-1$$\alpha = \frac{85.5}{285} = 0.3 = 3 \times 10^{-1}$$
Comparing with the format mathbfx times 10^-1$\mathbf{x} \times 10^{-1}$, the value of mathbfx$\mathbf{x}$ is **3**.
### Pattern Recognition
Kohlrausch's law allows direct algebraic recombination of ion conductances. Always construct the targeted weak base compound value by filtering out the spectator chloride contribution from the initial salt parameters.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Electrochemistry
Q36jee_main_2025_04_april_morningBatteries and Fuel Cells
On charging the lead storage battery, the oxidation state of lead changes from x_1$x_1$ to y_1$y_1$ at the anode and from x_2$x_2$ to y_2$y_2$ at the cathode. The values of x_1, y_1, x_2, y_2$x_1, y_1, x_2, y_2$ are respectively:
A.+4, +2, 0, +2$+4, +2, 0, +2$
B.+2, 0, +2, +4$+2, 0, +2, +4$
C.0, +2, +4, +2$0, +2, +4, +2$
D.+2, 0, 0, +4$+2, 0, 0, +4$
Solution
### Related Formula
textNet Charging Reaction: 2PbSO_4(s) + 2H_2O(l) rightarrow Pb(s) + PbO_2(s) + 2H_2SO_4(aq)$$\text{Net Charging Reaction: } 2PbSO_4(s) + 2H_2O(l) \rightarrow Pb(s) + PbO_2(s) + 2H_2SO_4(aq)$$
### Core Logic
During the **charging** cycle, the discharge chemical reactions are driven in reverse:
* **At Anode:** Lead sulfate (PbSO_4$PbSO_4$, where Lead is +2$+2$) is reduced back to metallic lead (Pb$Pb$, oxidation state 0$0$):
x_1 = +2 rightarrow y_1 = 0$$x_1 = +2 \rightarrow y_1 = 0$$
* **At Cathode:** Lead sulfate (PbSO_4$PbSO_4$, where Lead is +2$+2$) is oxidized back into lead dioxide (PbO_2$PbO_2$, where Lead is +4$+4$):
x_2 = +2 rightarrow y_2 = +4$$x_2 = +2 \rightarrow y_2 = +4$$
Thus, the values are +2, 0, +2, +4$+2, 0, +2, +4$.
### Pattern Recognition
Be careful with wording! Discharging consumes Pb$Pb$ and PbO_2$PbO_2$ to create PbSO_4$PbSO_4$. **Charging** does the exact opposite, converting PbSO_4$PbSO_4$ (+2$+2$) back into its parent elements (0$0$ and +4$+4$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Electrochemistry
Q45jee_main_2025_07_april_eveningElectrolysis and Discharge Potential
Given below are two statements:
1text M$1\text{ M}$ aqueous solution of each of textCu(NO_3)_2$\text{Cu(NO}_3)_2$, textAgNO_3$\text{AgNO}_3$, textHg_2(textNO_3)_2$\text{Hg}_2(\text{NO}_3)_2$; textMg(NO_3)_2$\text{Mg(NO}_3)_2$ are electrolysed using inert electrodes,
Given: E_textAg^+/textAg^theta = 0.80textV$E_{\text{Ag}^{+}/\text{Ag}}^{\theta} = 0.80\text{V}$ , E_textHg_2^2+/textHg^theta = 0.79textV$E_{\text{Hg}_2^{2+}/\text{Hg}}^{\theta} = 0.79\text{V}$, E_textCu^2+/textCu^theta = 0.24textV$E_{\text{Cu}^{2+}/\text{Cu}}^{\theta} = 0.24\text{V}$ and E_textMg^2+/textMg^theta = -2.37textV$E_{\text{Mg}^{2+}/\text{Mg}}^{\theta} = -2.37\text{V}$
Statement (I): With increasing voltage, the sequence of deposition of metals on the cathode will be textAg$\text{Ag}$, textHg$\text{Hg}$ and textCu$\text{Cu}$
Statement (II): Magnesium will not be deposited at cathode instead oxygen gas will be evolved at the cathode.
In the light of the above statement, choose the most appropriate answer from the options given below [cite: 426, 427]
A.textBoth statement I and statement II are incorrect$\text{Both statement I and statement II are incorrect}$
B.textStatement I is correct but statement II is incorrect$\text{Statement I is correct but statement II is incorrect}$
C.textBoth statement I and statement II are correct$\text{Both statement I and statement II are correct}$
D.textStatement I is incorrect but statement II is correct$\text{Statement I is incorrect but statement II is correct}$
Solution
### Related Formula
textEase of discharge at Cathode propto textStandard Reduction Potential (E^0)$$\text{Ease of discharge at Cathode} \propto \text{Standard Reduction Potential } (E^0)$$
### Core Logic
- At the cathode, the metal ion with the highest standard reduction potential (E^0$E^0$) gets reduced and deposited first. Arranging the given potentials:
E^0_textAg^+/textAg (0.80textV) > E^0_textHg_2^2+/textHg (0.79textV) > E^0_textCu^2+/textCu (0.24textV) $$E^0_{\text{Ag}^+/\text{Ag}} (0.80\text{V}) > E^0_{\text{Hg}_2^{2+}/\text{Hg}} (0.79\text{V}) > E^0_{\text{Cu}^{2+}/\text{Cu}} (0.24\text{V}) $$
Thus, deposition follows the order textAg
ightarrow textHg
ightarrow textCu$\text{Ag}
ightarrow \text{Hg}
ightarrow \text{Cu}$ as voltage is steadily increased, confirming Statement I.
- For textMg^2+$\text{Mg}^{2+}$, its reduction potential is highly negative (-2.37text V$-2.37\text{ V}$), much lower than that of water (-0.83text V$-0.83\text{ V}$). Consequently, water undergoes reduction at the cathode instead of magnesium:
2textH_2textO + 2e^-
ightarrow textH_2(g) + 2textOH^- $$2\text{H}_2\text{O} + 2e^-
ightarrow \text{H}_2(g) + 2\text{OH}^- $$
This results in the evolution of **Hydrogen gas** at the cathode, not oxygen gas. Oxygen gas is evolved at the *anode* via water oxidation. Thus, Statement II is incorrect. [cite: 1042, 1044]
### Step 1: Conclusion Match
Since Statement I is correct and Statement II is incorrect, we select option (2).
### Pattern Recognition
Cathode vs Anode Gas Trap: During the aqueous electrolysis of highly reactive metals (Groups 1, 2, and textAl$\text{Al}$), textH_2$\text{H}_2$ gas is always discharged at the cathode due to water's easier reduction profile. Oxygen gas (textO_2$\text{O}_2$) is an anodic product generated by water oxidation.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Electrochemistry
More Electrochemistry Questions — jee_main_2026_21_jan_morning
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