2.3 Conductance of Electrolytic Solutions
Ions carry current through a solution. Measure the resistance, correct for the shape of the cell, and you get the conductivity $\kappa$; divide by concentration and you get the molar conductivity $\Lambda_{m}=\kappa/c$, whose behaviour on dilution separates strong electrolytes from weak ones.
Water-treatment plants, power stations and semiconductor factories all watch one number round the clock: the conductivity of their water. Ultrapure water hardly conducts at all; a trace of dissolved salt raises its conductivity at once, so a cheap probe raises the alarm long before any chemical test could. Doctors use the same physics when they measure body composition, and farmers when they check the saltiness of irrigation water. In every case the current is carried not by electrons, as in a copper wire, but by ions moving through the liquid.
This section explains how the conductance of an electrolyte solution is measured and how to turn a raw resistance reading into a property of the solution itself, the conductivity $\kappa$, and then into a property per mole of electrolyte, the molar conductivity $\Lambda_{m}$. How $\Lambda_{m}$ changes on dilution tells strong electrolytes such as $\mathrm{KCl}$ apart from weak ones such as ethanoic acid (acetic acid). Kohlrausch's law then lets us find the limiting molar conductivity of a weak electrolyte, its degree of dissociation and even its acid dissociation constant, all from conductance measurements.
Resistance, Conductivity and the Cell Constant
A solution between two electrodes obeys Ohm's law, $V=IR$, just like a metal wire. Its resistance depends on the solution, but also on the geometry: a longer column resists more, a wider one less.
For a uniform conductor of length $l$ and cross-sectional area $A$,
$$R=\rho\,\frac{l}{A},$$where the constant $\rho$ is the resistivity (specific resistance), unit $\Omega\,\mathrm{m}$ or, more commonly in chemistry, $\Omega\,\mathrm{cm}$. The reciprocal quantities are
- conductance $G=\dfrac{1}{R}$, unit siemens, $\mathrm{S}=\Omega^{-1}$;
- conductivity (specific conductance) $\kappa=\dfrac{1}{\rho}=G\,\dfrac{l}{A}$, unit $\mathrm{S\,m^{-1}}$ (SI) or $\mathrm{S\,cm^{-1}}$; $\ 1\ \mathrm{S\,m^{-1}}=0.01\ \mathrm{S\,cm^{-1}}$.
Conductivity is the conductance of a $1\ \mathrm{cm}$ cube of the solution (electrodes $1\ \mathrm{cm^{2}}$ in area and $1\ \mathrm{cm}$ apart). It depends only on the solution and the temperature, not on the cell.
Two kinds of conduction. In metals, graphite and some oxides, electrons carry the current (electronic conductance). It depends on the structure and number of valence electrons and falls slightly as temperature rises, because the vibrating lattice scatters the electrons. In electrolyte solutions and molten salts, ions carry the current (electrolytic or ionic conductance), and matter moves with the charge. Ionic conductance depends on (i) the nature of the electrolyte (how many ions it gives), (ii) the size and solvation of the ions, (iii) the nature and viscosity of the solvent, (iv) the concentration, and (v) the temperature. It rises with temperature, roughly 2% per kelvin, because the solvent becomes less viscous and the ions move faster.
A conductivity cell has two platinum electrodes coated with platinum black, fixed a distance $l$ apart, each of area $A$. The quantity $l/A$ depends only on the cell and is called the cell constant, $G^{*}$:
$$G^{*}=\frac{l}{A}=R\,\kappa,\qquad \kappa=\frac{G^{*}}{R}.$$Its unit is $\mathrm{cm^{-1}}$ (or $\mathrm{m^{-1}}$). Since $l$ and $A$ of a real cell are hard to measure accurately, $G^{*}$ is found by filling the cell with a standard solution of known conductivity, usually $\mathrm{KCl}$ (for example $\kappa=0.0129\ \mathrm{S\,cm^{-1}}$ for $0.100\ \mathrm{M}$ $\mathrm{KCl}$ at $298\ \mathrm{K}$), and measuring its resistance.
Measuring the resistance. A solution cannot be measured with an ordinary ohmmeter. A direct current would electrolyse the solution, change its composition and coat the electrodes with gas, so the reading would drift. Instead the cell forms one arm of a Wheatstone bridge driven by an alternating current (about $1\ \mathrm{kHz}$), which reverses before any net electrolysis can occur. A headphone or an electronic null detector replaces the galvanometer. At balance, the unknown resistance follows from the three known ones. Modern conductivity meters do the same job electronically and display $\kappa$ directly, once the cell constant has been entered.
Molar Conductivity and How It Changes with Concentration
Conductivity is not a fair way to compare electrolytes: a more concentrated solution simply has more ions per $\mathrm{cm^{3}}$. To compare like with like we divide by the concentration, and count the conductance per mole of electrolyte.
The molar conductivity $\Lambda_{m}$ of a solution of concentration $c$ is
$$\Lambda_{m}=\frac{\kappa}{c}.$$In SI units ($\kappa$ in $\mathrm{S\,m^{-1}}$, $c$ in $\mathrm{mol\,m^{-3}}$) it is in $\mathrm{S\,m^{2}\,mol^{-1}}$. In the units most used in chemistry ($\kappa$ in $\mathrm{S\,cm^{-1}}$, $c$ in $\mathrm{mol\,L^{-1}}$, and $1\ \mathrm{L}=1000\ \mathrm{cm^{3}}$),
$$\Lambda_{m}\ (\mathrm{S\,cm^{2}\,mol^{-1}})=\frac{\kappa\ (\mathrm{S\,cm^{-1}})\times1000}{c\ (\mathrm{mol\,L^{-1}})},\qquad 1\ \mathrm{S\,m^{2}\,mol^{-1}}=10^{4}\ \mathrm{S\,cm^{2}\,mol^{-1}}.$$Picture. Take a volume $V$ of solution containing exactly $1\ \mathrm{mol}$ of electrolyte, and spread it between two large parallel electrodes $1\ \mathrm{cm}$ apart. The conductance of that slab is $\Lambda_{m}$. Diluting the solution makes $V$ larger, so the electrodes must be larger too, but they always hold the same one mole of electrolyte.
Dilution: κ falls, Λm rises. On dilution the number of ions per $\mathrm{cm^{3}}$ falls, so $\kappa$ decreases for every electrolyte. But $\Lambda_{m}$ always increases, for two different reasons:
- Strong electrolytes ($\mathrm{KCl}$, $\mathrm{NaCl}$, $\mathrm{HCl}$, $\mathrm{MgSO_{4}}$) are fully ionised at all concentrations. In a concentrated solution each ion is surrounded by an "atmosphere" of oppositely charged ions that drags on it. On dilution the ions are farther apart, the drag weakens and each ion moves more freely. $\Lambda_{m}$ rises gently.
- Weak electrolytes ($\mathrm{CH_{3}COOH}$, $\mathrm{NH_{4}OH}$, water) are only partly ionised, and their degree of dissociation $\alpha$ rises sharply on dilution (Ostwald's dilution law, Class 11). The number of ions per mole of electrolyte grows, so $\Lambda_{m}$ rises steeply, especially at very low concentration.
The value that $\Lambda_{m}$ approaches as $c\rightarrow0$ is the limiting molar conductivity $\Lambda_{m}^{\circ}$, the molar conductivity at infinite dilution, when the ions no longer affect one another.
For a strong electrolyte in dilute solution,
$$\Lambda_{m}=\Lambda_{m}^{\circ}-A\sqrt{c}.$$So a plot of $\Lambda_{m}$ against $\sqrt{c}$ is a straight line of slope $-A$. Extending it to $\sqrt{c}=0$ gives $\Lambda_{m}^{\circ}$ as the intercept. The constant $A$ depends on the solvent, the temperature and the type of electrolyte, i.e. the charges of its ions: all 1:1 electrolytes ($\mathrm{NaCl}$, $\mathrm{KBr}$, $\mathrm{HCl}$…) have similar $A$, 2:1 electrolytes ($\mathrm{CaCl_{2}}$) a larger one, and 2:2 electrolytes ($\mathrm{MgSO_{4}}$) larger still. The relation holds well only below about $0.01\ \mathrm{mol\,L^{-1}}$. For weak electrolytes the plot is a steep curve, not a line, and cannot be extrapolated.
Kohlrausch's Law and Its Applications
In 1875 Friedrich Kohlrausch compared the limiting molar conductivities of many pairs of salts and found a striking pattern. $\Lambda_{m}^{\circ}(\mathrm{KCl})-\Lambda_{m}^{\circ}(\mathrm{NaCl})$ is almost exactly the same as $\Lambda_{m}^{\circ}(\mathrm{KBr})-\Lambda_{m}^{\circ}(\mathrm{NaBr})$, about $23\ \mathrm{S\,cm^{2}\,mol^{-1}}$, whatever the common anion. At infinite dilution each ion must therefore carry current independently of its partner.
At infinite dilution, the limiting molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its cation and anion, each multiplied by the number of such ions in one formula unit. For an electrolyte that gives $\nu_{+}$ cations and $\nu_{-}$ anions,
$$\Lambda_{m}^{\circ}=\nu_{+}\lambda_{+}^{\circ}+\nu_{-}\lambda_{-}^{\circ}.$$Here $\lambda^{\circ}$ is the limiting molar conductivity of an ion, a property of that ion alone in the solvent at that temperature. Example: $\Lambda_{m}^{\circ}(\mathrm{CaCl_{2}})=\lambda^{\circ}(\mathrm{Ca^{2+}})+2\lambda^{\circ}(\mathrm{Cl^{-}})$. The law applies to strong and weak electrolytes alike, because at infinite dilution even a weak electrolyte is completely ionised.
| Cation | $\lambda^{\circ}$ / $\mathrm{S\,cm^{2}\,mol^{-1}}$ | Anion | $\lambda^{\circ}$ / $\mathrm{S\,cm^{2}\,mol^{-1}}$ |
|---|---|---|---|
| $\mathrm{H^{+}}$ | 349.6 | $\mathrm{OH^{-}}$ | 199.1 |
| $\mathrm{K^{+}}$ | 73.5 | $\mathrm{Br^{-}}$ | 78.1 |
| $\mathrm{Na^{+}}$ | 50.1 | $\mathrm{Cl^{-}}$ | 76.3 |
| $\mathrm{Ca^{2+}}$ | 119.0 | $\mathrm{SO_{4}^{2-}}$ | 160.0 |
| $\mathrm{Mg^{2+}}$ | 106.0 | $\mathrm{CH_{3}COO^{-}}$ | 40.9 |
Values at $298\ \mathrm{K}$, per mole of the ion as written. Notice the exceptionally large values for $\mathrm{H^{+}}$ and $\mathrm{OH^{-}}$. These ions do not have to travel through the water themselves: a proton can hop along a chain of hydrogen-bonded water molecules, so charge moves far faster than any ion could swim. Among the alkali-metal ions, small $\mathrm{Na^{+}}$ is slower than larger $\mathrm{K^{+}}$, because it holds a bigger shell of water molecules (it is more heavily hydrated) and must drag them along.
- $\Lambda_{m}^{\circ}$ of a weak electrolyte from strong ones. Choose strong electrolytes whose ions add and cancel to leave the weak one: $$\Lambda_{m}^{\circ}(\mathrm{CH_{3}COOH})=\Lambda_{m}^{\circ}(\mathrm{CH_{3}COONa})+\Lambda_{m}^{\circ}(\mathrm{HCl})-\Lambda_{m}^{\circ}(\mathrm{NaCl}).$$ The $\mathrm{Na^{+}}$ and $\mathrm{Cl^{-}}$ terms cancel, leaving $\lambda^{\circ}(\mathrm{H^{+}})+\lambda^{\circ}(\mathrm{CH_{3}COO^{-}})$.
- Degree of dissociation of a weak electrolyte at concentration $c$: $$\alpha=\frac{\Lambda_{m}}{\Lambda_{m}^{\circ}}.$$ (The ions of a weak electrolyte are so few that their mutual drag is negligible, so $\Lambda_{m}$ is simply the fraction $\alpha$ of the fully ionised value.)
- Dissociation constant of a weak acid $\mathrm{HA(aq)\rightleftharpoons H^{+}(aq)+A^{-}(aq)}$: $$K_{a}=\frac{c\,\alpha^{2}}{1-\alpha}.$$
1. The factor 1000. $\Lambda_{m}=\kappa\times1000/c$ only when $\kappa$ is in $\mathrm{S\,cm^{-1}}$ and $c$ in $\mathrm{mol\,L^{-1}}$. In SI, convert $c$ to $\mathrm{mol\,m^{-3}}$ (multiply by 1000) and use $\kappa$ in $\mathrm{S\,m^{-1}}$.
2. Opposite trends. On dilution $\kappa$ goes down and $\Lambda_{m}$ goes up. Questions often test exactly this pair.
3. Extrapolating a weak electrolyte. The $\Lambda_{m}$–$\sqrt{c}$ curve of ethanoic acid shoots up near $c=0$; it cannot be extended to the axis. Use Kohlrausch's law instead.
4. Forgetting $\nu$. $\Lambda_{m}^{\circ}(\mathrm{CaCl_{2}})$ needs $2\lambda^{\circ}(\mathrm{Cl^{-}})$; $\Lambda_{m}^{\circ}(\mathrm{Ba(OH)_{2}})$ needs $2\lambda^{\circ}(\mathrm{OH^{-}})$.
5. Cell constant upside down. $G^{*}=l/A$ (length over area) $=R\kappa$; so $\kappa=G^{*}/R$, not $R/G^{*}$.
Exercises
- 1 Define conductivity and give its SI unit. Express $0.0129\ \mathrm{S\,cm^{-1}}$ in $\mathrm{S\,m^{-1}}$.
- 2 The electrodes of a cell have an area of $1.50\ \mathrm{cm^{2}}$ and are $0.750\ \mathrm{cm}$ apart. Find the cell constant.
- 3 A cell with $G^{*}=1.25\ \mathrm{cm^{-1}}$ contains a solution of resistance $250\ \Omega$. Find its conductance and conductivity.
- 4 A cell filled with $0.100\ \mathrm{M}$ $\mathrm{KCl}$ has a resistance of $310\ \Omega$. Find its cell constant.
- 5 The cell of part 4, filled with a $0.020\ \mathrm{M}$ solution of an electrolyte, has a resistance of $1450\ \Omega$. Find $\kappa$ and $\Lambda_{m}$ of that solution.
- 6 Why is alternating current, not direct current, used to measure the resistance of an electrolyte solution?
- 7 How does the conductivity change with rising temperature (a) for copper wire, (b) for aqueous $\mathrm{KCl}$? Explain each.
- 8 A solution has a resistivity of $80.0\ \Omega\,\mathrm{cm}$. Find its conductivity in $\mathrm{S\,cm^{-1}}$ and in $\mathrm{S\,m^{-1}}$.
- 1 $0.050\ \mathrm{M}$ $\mathrm{KCl}$ has $\kappa=6.67\times10^{-3}\ \mathrm{S\,cm^{-1}}$. Find $\Lambda_{m}$.
- 2 $0.0020\ \mathrm{M}$ $\mathrm{NaCl}$ has $\kappa=2.45\times10^{-4}\ \mathrm{S\,cm^{-1}}$. Find $\Lambda_{m}$.
- 3 Express $\Lambda_{m}=125\ \mathrm{S\,cm^{2}\,mol^{-1}}$ in SI units.
- 4 A $0.0050\ \mathrm{M}$ solution has $\Lambda_{m}=120.0\ \mathrm{S\,cm^{2}\,mol^{-1}}$. Find its conductivity.
- 5 Explain why, on dilution, the conductivity of a solution decreases while its molar conductivity increases.
- 6 A strong electrolyte has $\Lambda_{m}=140.5$ at $c=0.0025\ \mathrm{M}$ and $136.0\ \mathrm{S\,cm^{2}\,mol^{-1}}$ at $c=0.010\ \mathrm{M}$. Assuming $\Lambda_{m}=\Lambda_{m}^{\circ}-A\sqrt{c}$, find $A$ and $\Lambda_{m}^{\circ}$.
- 7 For $\mathrm{NaCl}$, $\Lambda_{m}^{\circ}=126.4\ \mathrm{S\,cm^{2}\,mol^{-1}}$ and $A=80$. Predict $\Lambda_{m}$ at $c=0.0090\ \mathrm{M}$.
- 8 Why can $\Lambda_{m}^{\circ}$ of ethanoic acid not be found by extrapolating a graph of $\Lambda_{m}$ against $\sqrt{c}$?
- 1 Find $\Lambda_{m}^{\circ}$ of $\mathrm{KBr}$.
- 2 Find $\Lambda_{m}^{\circ}$ of $\mathrm{MgCl_{2}}$.
- 3 Find $\Lambda_{m}^{\circ}$ of $\mathrm{K_{2}SO_{4}}$.
- 4 Find $\Lambda_{m}^{\circ}$ of $\mathrm{HBr}$ and of $\mathrm{NaOH}$.
- 5 State Kohlrausch's law and explain why it applies even to weak electrolytes.
- 6 Given $\Lambda_{m}^{\circ}(\mathrm{BaCl_{2}})=280.0$, $\Lambda_{m}^{\circ}(\mathrm{NaOH})=249.2$ and $\Lambda_{m}^{\circ}(\mathrm{NaCl})=126.4$, find $\Lambda_{m}^{\circ}(\mathrm{Ba(OH)_{2}})$.
- 7 $0.0050\ \mathrm{M}$ ethanoic acid has $\Lambda_{m}=22.6\ \mathrm{S\,cm^{2}\,mol^{-1}}$. Find $\alpha$ and $K_{a}$.
- 8 $0.025\ \mathrm{M}$ methanoic acid (formic acid, $\mathrm{HCOOH}$) has $\kappa=8.0\times10^{-4}\ \mathrm{S\,cm^{-1}}$. With $\lambda^{\circ}(\mathrm{HCOO^{-}})=54.6$, find $\Lambda_{m}^{\circ}$, $\Lambda_{m}$, $\alpha$ and $K_{a}$.
- 1 Ultrapure water. Water for chip manufacture has $\kappa=5.5\times10^{-8}\ \mathrm{S\,cm^{-1}}$ at $298\ \mathrm{K}$. Using $\lambda^{\circ}(\mathrm{H^{+}})$ and $\lambda^{\circ}(\mathrm{OH^{-}})$, find $[\mathrm{H^{+}}]$ in this water and hence the ionic product $K_{w}$.
- 2 Oceanography. Seawater has $\kappa\approx5.0\ \mathrm{S\,m^{-1}}$. What resistance would it show in a cell with $G^{*}=1.00\ \mathrm{cm^{-1}}$?
- 3 Which conducts electricity better, $0.10\ \mathrm{M}$ glucose or $0.10\ \mathrm{M}$ $\mathrm{NaCl}$, and why? Which conducts better, $0.10\ \mathrm{M}$ $\mathrm{HCl}$ or $0.10\ \mathrm{M}$ $\mathrm{NaCl}$, and why?
- 4 Sketch on one set of axes $\Lambda_{m}$ against $\sqrt{c}$ for $\mathrm{KCl}$ and for ethanoic acid, and explain the shape of each curve.
E1. Board · 3 marks (a) Define molar conductivity. (b) How does it vary with concentration for a strong and for a weak electrolyte? (c) The conductivity of $0.20\ \mathrm{M}$ $\mathrm{KCl}$ at $298\ \mathrm{K}$ is $0.0248\ \mathrm{S\,cm^{-1}}$. Calculate its molar conductivity.
E2. JEE Main Numerical value. A conductivity cell filled with $0.10\ \mathrm{M}$ $\mathrm{KCl}$ ($\kappa=1.29\ \mathrm{S\,m^{-1}}$) has a resistance of $150\ \Omega$ at $298\ \mathrm{K}$. Filled with $0.025\ \mathrm{M}$ solution of an electrolyte it has a resistance of $700\ \Omega$. The molar conductivity of that solution is $x\times10^{-4}\ \mathrm{S\,m^{2}\,mol^{-1}}$. The value of $x$ to the nearest integer is ______.
E3. NEET The limiting molar conductivities of $\mathrm{NH_{4}Cl}$, $\mathrm{NaOH}$ and $\mathrm{NaCl}$ are $a$, $b$ and $c$ respectively. The limiting molar conductivity of $\mathrm{NH_{4}OH}$ is (A) $a+b-c$ (B) $a-b+c$ (C) $a+b+c$ (D) $b+c-a$
E4. JEE Advanced A $0.010\ \mathrm{M}$ solution of a weak monobasic acid HA has a molar conductivity of $16.0\ \mathrm{S\,cm^{2}\,mol^{-1}}$ at $298\ \mathrm{K}$. Given $\lambda^{\circ}(\mathrm{H^{+}})=349.6$ and $\lambda^{\circ}(\mathrm{A^{-}})=50.4\ \mathrm{S\,cm^{2}\,mol^{-1}}$, find the degree of dissociation, $K_{a}$ and the pH of the solution.
E5. JEE Main NEET When a solution of a strong electrolyte is diluted, (A) both $\kappa$ and $\Lambda_{m}$ increase (B) $\kappa$ decreases and $\Lambda_{m}$ increases (C) both decrease (D) $\kappa$ increases and $\Lambda_{m}$ decreases
E6. Board NEET Assertion–reason. Assertion (A): The molar conductivity of ethanoic acid rises steeply as its solution is made very dilute. Reason (R): The degree of dissociation of a weak electrolyte increases on dilution. (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true.
Research question. Conductivity meters are used everywhere to judge how much salt water contains. How good is a single conductivity reading as a measure of total dissolved solids (TDS), and when does it mislead?
Why it matters. India's drinking-water standard prefers TDS below $500\ \mathrm{mg\,L^{-1}}$; farmers in Punjab, California and Australia worry about salty irrigation water; aquarium keepers and dialysis units watch conductivity daily. Field instruments convert electrical conductivity (EC, in $\mu\mathrm{S\,cm^{-1}}$) to TDS with a rule of thumb, $\text{TDS}\ (\mathrm{mg\,L^{-1}})\approx0.65\times\text{EC}\ (\mu\mathrm{S\,cm^{-1}})$.
What to do.
- Typical readings at $25\ ^{\circ}\mathrm{C}$: reverse-osmosis water $20$, tap water $450$, a borewell $1500$ and seawater $50\,000\ \mu\mathrm{S\,cm^{-1}}$. Convert each to TDS with the rule of thumb and decide which meet the $500\ \mathrm{mg\,L^{-1}}$ guideline.
- Now assume the tap water contains only $\mathrm{NaCl}$ with $\Lambda_{m}\approx120\ \mathrm{S\,cm^{2}\,mol^{-1}}$. Use $c=\kappa\times1000/\Lambda_{m}$ to find its molarity and mass concentration of $\mathrm{NaCl}$. Compare with the rule-of-thumb TDS.
- Explain why a water rich in calcium hydrogencarbonate, or slightly acidic water, would give a different EC for the same TDS (use the $\lambda^{\circ}$ table).
- Why is a sugar solution invisible to a conductivity meter, however concentrated?
Tools: Calculator; Interactive 1; a cheap pen-type TDS meter if your school has one.
Hand in: A table of EC, TDS by the rule of thumb and (for tap water) TDS from the $\mathrm{NaCl}$ model, and three sentences on the limits of conductivity as a purity test.
Going further. Ocean scientists define salinity from conductivity measured very precisely at a fixed temperature. Read how the "practical salinity scale" works and why temperature must be controlled to a few thousandths of a degree.
🧪 Self-check quiz
Q1. A common unit of molar conductivity is
Q2. A cell filled with a solution of $\kappa=0.0050\ \mathrm{S\,cm^{-1}}$ has a resistance of $200\ \Omega$. Its cell constant is
Q3. With $\lambda^{\circ}(\mathrm{Ca^{2+}})=119.0$ and $\lambda^{\circ}(\mathrm{Cl^{-}})=76.3\ \mathrm{S\,cm^{2}\,mol^{-1}}$, $\Lambda_{m}^{\circ}(\mathrm{CaCl_{2}})$ is
Q4. The limiting molar conductivity of a weak electrolyte such as ethanoic acid is best obtained
Q5. A solution of ethanoic acid has $\Lambda_{m}=39.05\ \mathrm{S\,cm^{2}\,mol^{-1}}$, and $\Lambda_{m}^{\circ}=390.5\ \mathrm{S\,cm^{2}\,mol^{-1}}$. Its degree of dissociation is