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The molar conductivity of a weak electrolyte at a given concentration is 90 S cm² mol⁻¹. If the molar conductivity at infinite dilution $\Lambda_m^\infty = 400$ S cm² mol⁻¹, what is the degree of dissociation $\alpha$?
Options
1
$0.125$
2
$0.225$
3
$0.450$
4
$0.250$
Correct Answer
$\alpha = 0.225$
Solution
1

Degree of dissociation formula:

$$\alpha = \frac{\Lambda_m}{\Lambda_m^\infty}$$
2
$$\alpha = \frac{90}{400} = \boxed{0.225}$$
$\alpha = \Lambda_m / \Lambda_m^\infty = 90/400 = 0.225$
22.5% of weak electrolyte dissociated at this concentration
Theory: Electrochemistry
1. Molar Conductivity and Its Variation

Molar conductivity $\Lambda_m = \kappa \times 1000/c$ (where $\kappa$ = specific conductivity in S/cm, $c$ = concentration in mol/L). Units: S cm² mol⁻¹. For strong electrolytes: $\Lambda_m = \Lambda_m^\infty - A\sqrt{c}$ (Debye-Hückel-Onsager equation). As $c \to 0$, $\Lambda_m \to \Lambda_m^\infty$ (molar conductivity at infinite dilution). The decrease in $\Lambda_m$ with increasing concentration for strong electrolytes is due to interionic attractions that slow down ion movement. For weak electrolytes, $\Lambda_m$ increases sharply with dilution because degree of dissociation increases — at very low concentrations, nearly complete dissociation occurs. This is why weak electrolytes cannot be treated like strong electrolytes in conductivity measurements.

2. Kohlrausch Law of Independent Migration

Kohlrausch's Law states that at infinite dilution, each ion migrates independently of other ions. The molar conductivity at infinite dilution is the sum of contributions from individual ions: $\Lambda_m^\infty = \nu_+\lambda_+^\infty + \nu_-\lambda_-^\infty$ where $\nu_+$ and $\nu_-$ are the stoichiometric coefficients and $\lambda_+^\infty$, $\lambda_-^\infty$ are limiting molar conductivities of cation and anion. This law is extremely important because it allows calculation of $\Lambda_m^\infty$ for weak electrolytes (like acetic acid, NH₄OH) which cannot be determined by extrapolation. For example: $\Lambda_m^\infty(CH_3COOH) = \Lambda_m^\infty(CH_3COONa) + \Lambda_m^\infty(HCl) - \Lambda_m^\infty(NaCl)$. This uses strong electrolytes whose $\Lambda_m^\infty$ can be measured to find the weak electrolyte value indirectly.

3. Degree of Dissociation from Conductivity

For a weak electrolyte at concentration $c$: only a fraction $\alpha$ is dissociated. The molar conductivity $\Lambda_m$ at concentration $c$ is proportional to the fraction dissociated: $\alpha = \Lambda_m/\Lambda_m^\infty$. This is because ions are the charge carriers — more dissociation → more ions → higher conductivity. At infinite dilution, complete dissociation is assumed ($\alpha = 1$, $\Lambda_m = \Lambda_m^\infty$). The dissociation constant $K_a = c\alpha^2/(1-\alpha)$ (Ostwald dilution law). For small $\alpha$: $K_a \approx c\alpha^2$, so $\alpha \approx \sqrt{K_a/c}$ — degree of dissociation increases with dilution (as $c$ decreases, $\alpha$ increases). This is consistent with Le Chatelier's principle: dilution favours dissociation.

4. Specific Conductance and Cell Constant

Specific conductance (conductivity) $\kappa = 1/\rho$ (where $\rho$ = resistivity). Unit: S/m or S/cm. Conductance $G = \kappa A/l$ (where $A$ = electrode area, $l$ = distance between electrodes). Cell constant $G^* = l/A$ (unit: cm⁻¹). Measured conductance $G = \kappa/G^*$. So $\kappa = G \times G^*$. Cell constant is determined using a solution of known conductivity (usually KCl solutions with precisely known $\kappa$). Once $G^*$ is known, measuring $G$ directly gives $\kappa$ for any solution. A conductivity cell is essentially a container with two platinum electrodes — AC is used (not DC) to prevent electrolysis and polarisation of electrodes.

5. Equivalent Conductance vs Molar Conductance

Equivalent conductance $\Lambda_{eq} = \kappa \times 1000/N$ (where $N$ = normality). Molar conductance $\Lambda_m = \kappa \times 1000/M$ (where $M$ = molarity). Relation: $\Lambda_m = n \times \Lambda_{eq}$ where $n$ = n-factor (valence factor). For HCl: $n=1$, so $\Lambda_m = \Lambda_{eq}$. For H₂SO₄: $n=2$, so $\Lambda_m = 2\Lambda_{eq}$. For Al₂(SO₄)₃: $n=6$, so $\Lambda_m = 6\Lambda_{eq}$. Modern IUPAC recommends molar conductance. Equivalent conductance is older terminology but still appears in NEET questions. Both increase with dilution for weak electrolytes; for strong electrolytes, the variation is smaller and follows the Debye-Hückel-Onsager equation.

6. Electrochemical Cells — Galvanic and Electrolytic

Galvanic (voltaic) cell: converts chemical energy to electrical energy. Spontaneous redox reaction. Examples: Daniel cell (Zn/ZnSO₄ || CuSO₄/Cu), standard hydrogen electrode, dry cell, lead-acid battery. EMF of cell $E_{cell} = E_{cathode} - E_{anode}$ (reduction potentials). For spontaneous reaction: $E_{cell} > 0$ and $\Delta G < 0$. Nernst equation: $E_{cell} = E°_{cell} - (RT/nF)\ln Q = E°_{cell} - (0.0592/n)\log Q$ at 298 K. Electrolytic cell: uses electrical energy to drive non-spontaneous redox reaction. $E_{cell} < 0$, external EMF must be applied. Examples: electrolysis of water, electroplating, Hall-Héroult process (Al extraction), Downs cell (Na from molten NaCl).

7. Faradays Laws of Electrolysis

First Law: mass of substance deposited/liberated is proportional to quantity of electricity passed. $m = ZQ = ZIt$ where $Z$ = electrochemical equivalent, $Q$ = charge, $I$ = current, $t$ = time. Second Law: masses deposited by same quantity of electricity are proportional to their chemical equivalents (molar mass/n-factor). $m = MIt/(nF)$ where $F = 96485$ C/mol (Faraday constant) $\approx 96500$ C/mol. For depositing 1 mole of $M^{n+}$ ions: requires $nF$ coulombs. Example: depositing 1 mol Cu²⁺ needs $2F = 193000$ C. At 1 A current: time $= 193000$ s $\approx 53.6$ hours. Applications: electroplating (thin metal coating), electrorefining (purifying metals), electrochemical synthesis (NaOH by chlor-alkali process).

8. Standard Electrode Potentials and Applications

Standard hydrogen electrode (SHE): $E° = 0$ V (reference). Other electrode potentials measured relative to SHE. Standard reduction potentials: $F_2/F^-$: $+2.87$ V (strongest oxidising agent). $Li^+/Li$: $-3.04$ V (strongest reducing agent). $Cu^{2+}/Cu$: $+0.34$ V. $Zn^{2+}/Zn$: $-0.76$ V. Daniel cell: $E° = E°_{Cu} - E°_{Zn} = 0.34 - (-0.76) = 1.10$ V. Electrochemical series: predict: which metal displaces which from solution (more negative $E°$ displaces more positive), direction of cell reaction, relative reactivity of metals (used in activity series). Relation to thermodynamics: $\Delta G° = -nFE°$. $\Delta G° = -RT\ln K$. So $\ln K = nFE°/(RT) = nE°/0.0257$ at 298 K.

Frequently Asked Questions
1. Why is alpha = Lambda_m / Lambda_m_infinity?
At infinite dilution, we assume 100% dissociation ($\alpha = 1$) and $\Lambda_m = \Lambda_m^\infty$. At concentration $c$, only a fraction $\alpha$ of molecules are dissociated. Since conductivity is due to ions only (undissociated molecules don't carry current): $\Lambda_m \propto \alpha$. At infinite dilution: $\Lambda_m^\infty \propto 1$. So $\Lambda_m/\Lambda_m^\infty = \alpha/1 = \alpha$. This assumes: (1) Ion mobilities are the same at both concentrations (valid at low concentrations). (2) Interionic effects are negligible. For more accurate treatment, Debye-Hückel corrections are applied.
2. What is Ostwald dilution law?
Ostwald dilution law relates the degree of dissociation $\alpha$ to the dissociation constant $K_a$ and concentration $c$: for weak acid $HA \rightleftharpoons H^+ + A^-$, $K_a = c\alpha^2/(1-\alpha)$. For small $\alpha$ ($\alpha \ll 1$): $K_a \approx c\alpha^2$, so $\alpha \approx \sqrt{K_a/c}$. This shows that $\alpha$ increases as $c$ decreases (dilution increases degree of dissociation). Substituting $\alpha = \Lambda_m/\Lambda_m^\infty$: $K_a = c(\Lambda_m/\Lambda_m^\infty)^2/(1 - \Lambda_m/\Lambda_m^\infty)$. This allows calculation of $K_a$ from conductivity measurements alone — a powerful experimental method for finding dissociation constants of weak electrolytes.
3. How is Lambda_m_infinity found for weak electrolytes?
Weak electrolytes (e.g., CH₃COOH, NH₄OH) cannot be measured directly because their $\Lambda_m$ vs $\sqrt{c}$ graph is not linear and does not extrapolate reliably to $c = 0$. Kohlrausch law is used instead. Example for acetic acid: $\Lambda_m^\infty(CH_3COOH) = \lambda^\infty(CH_3COO^-) + \lambda^\infty(H^+) = [\Lambda_m^\infty(CH_3COONa) - \lambda^\infty(Na^+)] + [\lambda^\infty(H^+)] = \Lambda_m^\infty(CH_3COONa) + \Lambda_m^\infty(HCl) - \Lambda_m^\infty(NaCl)$. All three values on the right are for strong electrolytes, easily measured by extrapolation. This indirect method is the standard approach in all textbooks.
4. What is the difference between conductance and conductivity?
Conductance $G = 1/R$ (siemens, S). It depends on both the material AND the dimensions of the conductor. Conductivity (specific conductance) $\kappa = G \times l/A = G \times G^*$ (S/m or S/cm). It is an intrinsic property of the material only — independent of dimensions. Molar conductivity $\Lambda_m = \kappa/c$ (or $\kappa \times 1000/c$ in CGS): relates conductivity to concentration. This is the most useful parameter for comparing different electrolytes at different concentrations. The progression $G \to \kappa \to \Lambda_m$ represents increasing normalisation: from geometry-dependent to material-dependent to concentration-normalised.
5. What are typical values of limiting molar conductivities?
Some important limiting molar conductivities at 298 K (S cm² mol⁻¹): H⁺: 349.6 (highest — moves by Grotthuss mechanism through H-bonding network). OH⁻: 197.6 (second highest — similar mechanism). K⁺: 73.5. Na⁺: 50.1. Li⁺: 38.7 (smallest alkali — most hydrated, moves slowest). Cl⁻: 76.4. SO₄²⁻: 160.0 (for the ion, per equivalent = 80.0). Mg²⁺: 106.0 (per mole). These values are additive (Kohlrausch law) to give $\Lambda_m^\infty$ for any electrolyte. Note: H⁺ has anomalously high conductivity (5× Na⁺) because proton transfer through water is faster than ion migration.
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