Separate the measured cell conductance from the material conductivity and the per-mole response.
JEE Main · JEE Advanced · NEET UG44 min study guide9 indexed sections
BY THE END, YOU CAN
Relate resistance, conductance, conductivity and cell constant with consistent units.
Explain why conductivity and molar conductivity can change oppositely on dilution.
Apply limiting ionic contributions with correct stoichiometric coefficients.
Estimate weak-electrolyte dissociation, dissociation constants and sparingly soluble salt solubility.
Interpret strong-electrolyte extrapolations and conductometric titration trends within their limits.
BEFORE YOU START
Ionic solutions and analytical concentration.
Resistance, conductance and geometric dimensions.
Weak-acid equilibria and dilute-solution approximations.
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
CONNECT THE CONCEPT TO THE EXAM
Selected exam questions & worked solutions
These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.
Available in this lesson: JEE Advanced 2017. This is a limited lesson sample, not a complete exam-question bank.
1 sourced sample question available across the listed exams and years.
Independently derived and checked solution. Archived key finality is unverified; this is not presented as an official final answer.
A weak monoprotic acid has concentration 0.0015 mol/L and pH 4. Its measured conductance is 5×10⁻⁷ S in a cell with electrode separation 120 cm and area 1 cm². If limiting molar conductivity is Z×10² S cm² mol⁻¹, find Z. The paper’s molar-conductivity unit typo is corrected here.
Reveal answer & worked solution
Answer: Z=6 (independently derived).
Conductivity κ=G(l/A)=5×10⁻⁷×120=6×10⁻⁵ S/cm.
Molar conductivity Λ=1000κ/c=40 S cm²/mol. From pH 4, [H⁺]=10⁻⁴ mol/L, so dissociation fraction α=[H⁺]/c=1/15.
For the intended dilute weak-acid model, Λ=αΛ∞. Thus Λ∞=40×15=600 S cm²/mol, giving Z=6.
Source: JEE (Advanced) 2017, Paper 1, Code 1, Chemistry Q27; independently derived solution; key finality unverified. Solution written for this lesson.
01 / UNDERSTAND & APPLY
1. Both signs of ion contribute to the current
In a metallic wire, electronic motion carries current. In an electrolyte solution, positive and negative ions move in opposite directions under an electric field, but both motions contribute to the same conventional current direction. A moving negative charge in one direction is electrically equivalent to positive charge moving oppositely.
The number of mobile ions matters, but so do ionic charge, mobility, solvent viscosity, solvation and temperature. Counting ions in a chemical formula is therefore not enough to rank conductivities quantitatively. Two solutions with the same analytical molarity can have different degrees of dissociation and different ionic mobilities.
Hydrogen and hydroxide ions have unusually high aqueous mobilities because charge transport can involve rapid rearrangement of hydrogen-bonded water networks, in addition to ordinary movement of a solvated ion. This helps explain the characteristic conductance changes in acid–base titrations. It does not mean bare protons travel through water without interaction.
Electrical neutrality does not imply zero conductivity. A bulk solution can contain equal positive and negative charge while those charge carriers move under an applied field. Conversely, a solution of a neutral nonelectrolyte can contain many dissolved molecules yet contribute little ionic conductivity beyond the solvent and trace impurities.
02 / UNDERSTAND & APPLY
2. Distinguish four quantities before doing arithmetic
Resistance R measures opposition to current in a particular cell and has unit ohm, Ω. Conductance G=1/R has unit siemens, S. For an ideal uniform sample between parallel electrodes of area A separated by distance l, R=ρl/A, where ρ is resistivity. Changing cell geometry changes R and G even if the solution composition stays identical.
Conductivity κ=1/ρ is the material property at the stated composition and temperature. It is related to measured conductance by κ=G(l/A). The factor l/A is the cell constant, with dimensions inverse length. In SI units it is m⁻¹ and κ is S m⁻¹; in common chemistry units it is cm⁻¹ and κ is S cm⁻¹.
A larger electrode area gives larger conductance because more parallel ionic pathways contribute. A larger separation gives smaller conductance because the current path is longer. These geometric trends are a useful check on the formula: multiplying G by l/A removes the geometry rather than exaggerating it.
Real conductivity probes may have nonideal field geometry, so their effective cell constant is commonly calibrated with a reference solution of known conductivity at controlled temperature. The ideal l/A expression explains the dimensions, while calibration captures the actual probe geometry. A cell constant is not the same quantity as a reaction equilibrium constant.
G=1/R; κ=G K_cell; K_cell=l/A
R in Ω, G in S; choose either cm-based or SI geometry consistently.
ORIGINAL WORKED EXAMPLE
Calibrate and reuse a cell
A reference solution has κ=1.40×10⁻³ S cm⁻¹ and gives R=500 Ω. The same cell gives R=800 Ω for an unknown at the same temperature. Find the cell constant and unknown conductivity.
For the reference, G=1/500=0.00200 S.
K_cell=κ/G=0.00140/0.00200=0.700 cm⁻¹.
For the unknown, G=1/800=0.00125 S.
κ_unknown=0.700×0.00125=8.75×10⁻⁴ S cm⁻¹.
Answer K_cell=0.700 cm⁻¹; κ_unknown=8.75×10⁻⁴ S cm⁻¹.
03 / UNDERSTAND & APPLY
3. Molar conductivity normalizes by analytical concentration
Molar conductivity Λ_m is conductivity divided by the analytical molar concentration, with the concentration expressed in units compatible with the length units in κ. In SI, κ in S m⁻¹ divided by c in mol m⁻³ gives Λ_m in S m² mol⁻¹. This dimensional relation is the safest starting definition.
When κ is in S cm⁻¹ and c is in mol L⁻¹, convert liters to cubic centimeters. A molarity c corresponds to c/1000 mol cm⁻³, so Λ_m=1000κ/c in S cm² mol⁻¹. The factor 1000 is a unit conversion, not an extra physical law.
Molar conductivity is not simply the conductance of one mole placed in an arbitrary vessel. It is a composition-normalized transport property that still depends on concentration through mobility and, for weak electrolytes, dissociation. Its limiting value Λ_m° is approached as electrolyte concentration tends toward zero under the specified solvent and temperature conditions.
Concentration must refer to the electrolyte whose molar conductivity is being defined. For 0.0100 M CaCl₂, the formula-unit analytical concentration is 0.0100 M even though ideal complete dissociation produces 0.0200 M chloride. Using total ion concentration in the denominator would define a different normalization.
Λ_m=κ/c in SI; Λ_m=1000κ/c in S cm² mol⁻¹
The second form uses κ in S cm⁻¹ and c in mol L⁻¹.
ORIGINAL WORKED EXAMPLE
Compute and convert molar conductivity
A 0.0200 mol L⁻¹ electrolyte has κ=2.40×10⁻³ S cm⁻¹. Find Λ_m in chemistry and SI units.
Λ_m=1000×0.00240/0.0200=120 S cm² mol⁻¹.
Since 1 cm²=10⁻⁴ m², multiply by 10⁻⁴ to obtain 0.0120 S m² mol⁻¹.
Alternatively κ=0.240 S m⁻¹ and c=20.0 mol m⁻³, giving 0.240/20.0=0.0120.
The independent unit route verifies the conversion.
Answer 120 S cm² mol⁻¹ = 0.0120 S m² mol⁻¹.
04 / UNDERSTAND & APPLY
4. Dilution lowers carriers per volume while improving the per-mole response
For many ordinary dilute electrolyte solutions, conductivity decreases upon dilution because fewer charge carriers occupy a given volume. Molar conductivity can nevertheless increase because it divides out the analytical concentration and reflects improved ionic mobility or increased dissociation. There is no contradiction: κ and Λ_m have different denominators.
A strong electrolyte is substantially dissociated in the ordinary dilute regime. Its molar conductivity still depends on concentration because ions interact with their ionic atmosphere and with solvent flow. Lowering concentration reduces these interionic transport effects. It is incorrect to explain every change in strong-electrolyte molar conductivity as newly breaking undissociated molecules apart.
Over a sufficiently dilute range at fixed temperature, a useful limiting form is Λ_m=Λ_m°−B√c, with B depending on the electrolyte, solvent, temperature and chosen concentration units. A plot of Λ_m against √c is approximately linear, and its zero-concentration intercept estimates Λ_m°. The equation should not be extended indiscriminately into concentrated solutions.
Weak electrolytes show an additional strong concentration dependence from their changing dissociation fraction. Their ordinary Λ_m versus √c data generally cannot be treated as the same simple straight-line extrapolation. Independent ionic limiting values provide a more reliable route to Λ_m° in introductory calculations.
Λ_m≈Λ_m°−B√c
Dilute strong-electrolyte limiting relation at fixed temperature; B depends on the unit convention.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Hypothetical fixed-temperature dilute strong-electrolyte model. The limiting line is not a universal concentrated-solution law or a weak-electrolyte dissociation curve.
ORIGINAL WORKED EXAMPLE
Find a limiting intercept
A dilute strong electrolyte has Λ_m=140 at √c=0.0100 and Λ_m=130 at √c=0.0300, in a consistent concentration convention and S cm² mol⁻¹ conductivity units. Estimate Λ_m°.
The slope is (130−140)/(0.0300−0.0100)=−500.
The line is Λ_m=Λ_m°−500√c.
Using the first point gives Λ_m°=140+500×0.0100=145.
The second gives 130+500×0.0300=145, confirming consistency.
Answer Λ_m°=145 S cm² mol⁻¹ under the stated linear model.
05 / UNDERSTAND & APPLY
5. At infinite dilution, add ionic contributions with their counts
Kohlrausch’s law of independent ionic migration states that the limiting molar conductivity of an electrolyte is the sum of the limiting contributions of its ions, multiplied by the stoichiometric number of each ion per formula unit. Thus Λ_m°=ν_+λ_+°+ν_-λ_-° for a simple binary electrolyte.
For CaCl₂, use λ_Ca²⁺°+2λ_Cl⁻°, not merely one of each. If ionic molar conductivities are supplied, the ionic charge is already reflected in their defined contribution; do not insert another factor of two for calcium solely because its charge is +2. Equivalent ionic conductivities use a different normalization and must not be mixed with molar values.
The independence is a limiting statement at fixed solvent and temperature. It does not say that ions ignore one another in concentrated solutions. At finite concentration, ionic interactions and composition-dependent transport alter the simple additive limiting picture. Use Λ_m° values, not unrelated finite-concentration measurements, in cancellation combinations.
A weak-acid limiting value can be constructed from strong-electrolyte data. Adding Λ°(HCl) and Λ°(NaA), then subtracting Λ°(NaCl), cancels sodium and chloride contributions and leaves λ_H⁺°+λ_A⁻°=Λ°(HA). The algebra is ionic bookkeeping, not a physical instruction to mix those solutions and measure the resulting acid.
Λ_m°=Σν_i λ_i°
ν_i is the number of ions i per electrolyte formula unit.
Λ°(HA)=Λ°(HCl)+Λ°(NaA)−Λ°(NaCl)
All values at the same temperature and in the same solvent and unit convention.
ORIGINAL WORKED EXAMPLE
Construct a weak-acid limiting value
At one temperature, hypothetical limiting molar conductivities are HCl 425, NaA 95 and NaCl 125 S cm² mol⁻¹. Find Λ°(HA).
Write each as ionic sums: H+Cl, Na+A, and Na+Cl.
Add the first two and subtract the third; sodium and chloride cancel.
Λ°(HA)=425+95−125=395 S cm² mol⁻¹.
Only the limiting values can be combined this way without finite-concentration corrections.
Answer 395 S cm² mol⁻¹.
06 / UNDERSTAND & APPLY
6. Conductance can estimate dissociation under an explicit dilute model
For a weak 1:1 electrolyte HA⇌H⁺+A⁻, let c be the analytical concentration and α the fraction dissociated. If the ions at that concentration have approximately their limiting mobilities and background conductivity is properly handled, Λ_m≈αΛ_m°. Thus α≈Λ_m/Λ_m° is a useful dilute-model estimate.
The approximation is not an exact definition of dissociation for every electrolyte. At appreciable ionic strength, mobility changes can make Λ_m/Λ_m° less than one even for a strongly dissociated electrolyte. Therefore that ratio must not be interpreted mechanically as the dissociation fraction of a strong salt.
For a weak monoprotic acid in otherwise pure solvent, neglecting water autoionization and activity corrections, equilibrium concentrations are [H⁺]=[A⁻]=cα and [HA]=c(1−α). The concentration-form dissociation constant is K_a=cα²/(1−α). If α is small, the denominator can be approximated by one, but the full expression is easy to retain.
At extremely low acid concentration, water’s own ions and trace contamination become important; at higher concentration, ionic activity and mobility corrections matter. A conductivity-derived K_a is meaningful only inside the model assumptions and with adequate background correction. Adding a common ion changes the equilibrium and invalidates the simple [H⁺]=[A⁻] setup.
α≈Λ_m/Λ_m°; K_a≈cα²/(1−α)
Weak 1:1 acid, dilute mobility model, no common ion, negligible water and activity corrections.
ORIGINAL WORKED EXAMPLE
Estimate a dissociation constant
A weak 1:1 acid at c=0.0100 M has corrected Λ_m=20.0 and Λ_m°=400 S cm² mol⁻¹. Estimate α and K_a under the stated ideal dilute model.
α=20.0/400=0.0500, or 5.00%.
K_a=cα²/(1−α)=0.0100×0.0500²/0.9500.
The result is 2.632×10⁻⁵ in the concentration-form convention.
Using 1−α≈1 would give 2.50×10⁻⁵, about 5% below the retained-denominator result.
Answer α=0.0500; K_a≈2.63×10⁻⁵ with the stated concentration convention.
07 / UNDERSTAND & APPLY
7. A small conductivity can reveal a small dissolved amount
For a sparingly soluble salt whose saturated solution is sufficiently dilute and essentially fully dissociated, approximate its molar conductivity by the limiting value. After subtracting the solvent background, κ_salt≈Λ_m°s/1000 when κ is in S cm⁻¹, Λ_m° in S cm² mol⁻¹ and molar solubility s in mol L⁻¹.
This gives s≈1000κ_salt/Λ_m°. The formula assumes the dissolved salt is the dominant additional conducting species and that hydrolysis, complexation and other impurities do not substantially change the speciation. A conductivity measurement alone cannot identify which ions produced the signal.
Water background is particularly important for low-solubility salts. Dissolved carbon dioxide, contamination and temperature variation can be comparable to the desired conductivity increment. Subtracting a nominal universal water value without measuring a relevant blank can give a misleading solubility estimate.
Once s is found, use the salt’s dissolution stoichiometry to build K_sp. For AB→A⁺+B⁻, K_sp≈s². For AB₂→A²⁺+2B⁻, K_sp≈s(2s)²=4s³. The conductivity conversion and solubility-product conversion require separate stoichiometric checks.
ORIGINAL WORKED EXAMPLE
Infer solubility from a conductivity increment
A saturated 1:1 salt solution has κ=2.50×10⁻⁶ S cm⁻¹; the solvent blank is 0.50×10⁻⁶ S cm⁻¹. Given Λ_m°=100 S cm² mol⁻¹, estimate s and K_sp, neglecting other equilibria and activities.
Corrected conductivity is (2.50−0.50)×10⁻⁶=2.00×10⁻⁶ S cm⁻¹.
s=1000×2.00×10⁻⁶/100=2.00×10⁻⁵ mol L⁻¹.
For a 1:1 salt, each ion concentration equals s.
K_sp≈s²=4.00×10⁻¹⁰ in the classroom concentration convention.
Answer s≈2.00×10⁻⁵ M; K_sp≈4.00×10⁻¹⁰.
08 / UNDERSTAND & APPLY
8. A conductivity curve reflects which ions are being replaced
Conductivity measurements commonly use alternating excitation to reduce electrode polarization and net electrolysis effects. The measured response can depend on frequency and electrode interfaces, so the simple bulk-resistance model is an approximation chosen for suitable experimental conditions. Maintain temperature because ionic mobility and solvent properties change with it.
In a strong-acid/strong-base conductometric titration such as HCl with NaOH, highly mobile H⁺ is consumed while Na⁺ remains in solution. Conductance typically falls before equivalence. Beyond equivalence, excess OH⁻ produces a rising branch. The intersection of approximately linear corrected branches can estimate the equivalence volume.
The minimum is not zero: spectator ions such as Na⁺ and Cl⁻ remain mobile at equivalence. Dilution from titrant addition also affects raw conductance. If a simple volume correction is appropriate, multiply measured conductance by (V_initial+V_added)/V_initial to compare the changing ionic composition on a common volume basis.
For a weak acid titrated with strong base, formation of its ionic salt generally raises conductance before equivalence, followed by a steeper increase from excess hydroxide. Initial details can depend on acid concentration and dilution, so the schematic trend is not a universal exact straight line. Identify the ion replacement mechanism rather than memorizing only a V-shaped drawing.
Using conductance and conductivity as synonyms.
Forgetting the factor 1000 when combining S cm⁻¹ and mol L⁻¹.
Adding extra ionic charge factors to already molar ionic conductivities.
Applying limiting ionic additivity to arbitrary concentrated-solution data.
Interpreting Λ/Λ° as an exact dissociation fraction for strong electrolytes.
Ignoring water background in very dilute solutions.
Expecting zero conductivity at acid–base equivalence.
ORIGINAL WORKED EXAMPLE
Why the endpoint still conducts
An idealized HCl–NaOH titration reaches equivalence. Does the solution’s conductivity become zero?
H⁺ and OH⁻ have largely neutralized to water.
Na⁺ and Cl⁻ remain as mobile ions in the solution.
Their mobilities are lower than that of H⁺ or OH⁻ under ordinary aqueous conditions, but they are not zero.
Thus the curve has a minimum or branch intersection, not a zero-conductivity requirement.
Answer No. Spectator ions still carry current at equivalence.
09 / UNDERSTAND & APPLY
9. Eight solved checks
Keep units visible and specify whether each concentration is analytical electrolyte concentration or a free-ion concentration.
ORIGINAL PRACTICE · 1
A cell has resistance 250 Ω. Find conductance.
Reveal answer & reasoning
0.00400 S.
G=1/R=1/250=0.00400 S.
ORIGINAL PRACTICE · 2
The same solution is measured with twice the electrode separation and unchanged area in an ideal cell. What happens to G and κ?
Reveal answer & reasoning
G halves; κ is unchanged.
R is proportional to l, so doubling l doubles R and halves G. Conductivity removes geometry through κ=Gl/A.
ORIGINAL PRACTICE · 3
κ=1.50×10⁻³ S cm⁻¹ at c=0.0100 M. Find Λ_m.
Reveal answer & reasoning
150 S cm² mol⁻¹.
Λ_m=1000κ/c=1000×0.00150/0.0100=150.
ORIGINAL PRACTICE · 4
Given ionic molar values λ_Ca²⁺°=120 and λ_Cl⁻°=75 S cm² mol⁻¹, find Λ°(CaCl₂).
Reveal answer & reasoning
270 S cm² mol⁻¹.
Use one calcium contribution and two chlorides: 120+2×75=270. No additional calcium charge multiplier is inserted.
ORIGINAL PRACTICE · 5
A weak 1:1 electrolyte has Λ_m=12 and Λ_m°=300 in the same units. Estimate α in the limiting-mobility model.
Reveal answer & reasoning
0.0400, or 4.00%.
α≈Λ_m/Λ_m°=12/300. The result depends on the stated weak-electrolyte model.
ORIGINAL PRACTICE · 6
For c=0.0200 M and α=0.0200, estimate the weak-acid concentration-form K_a.
Reveal answer & reasoning
8.16×10⁻⁶.
K_a=cα²/(1−α)=0.0200×0.000400/0.9800=8.163×10⁻⁶.
ORIGINAL PRACTICE · 7
A dilute strong-electrolyte fit is Λ_m=160−400√c in consistent units. What is Λ_m°?
Reveal answer & reasoning
160 in the stated molar-conductivity units.
At the limiting concentration c→0 the square-root term vanishes, leaving the intercept.
ORIGINAL PRACTICE · 8
Can conductivity identify a unique dissolved salt from its value alone?
Reveal answer & reasoning
No.
Different ion combinations and concentrations can produce the same total conductivity. Identification needs additional chemical information or independent measurements.
KEEP THE BIG IDEAS
Quick revision
Conductance depends on cell geometry; conductivity describes the solution at its stated conditions.
Molar conductivity divides conductivity by analytical concentration using consistent units.
Dilution can lower κ while raising Λ_m.
Limiting molar conductivity is the stoichiometric sum of limiting ionic contributions.
Weak-electrolyte dissociation and solubility calculations require explicit dilute-model and background assumptions.
Conductometric endpoints reflect changing mobile ions, not the disappearance of all ions.
The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.