Strong and Weak Electrolytes

Dissolve table salt in water and it lights a bulb brightly. Dissolve the same mass of acetic acid and the bulb glows faintly. Both are electrolytes. Both make water conduct electricity. Yet they behave completely differently as concentration changes. The difference comes down to one question: how much of the solute actually splits into ions? Strong electrolytes answer "essentially all of it." Weak electrolytes answer "only a small, concentration-dependent fraction." This single fact — the degree of dissociation — governs conductivity, colligative properties, pH, and reaction equilibria throughout solution chemistry. This article builds the strong/weak distinction from first principles. You will see how conductivity data reveals dissociation behavior. Then you will meet the degree of dissociation as a quantitative variable. Finally you will see how this sets up the Ostwald dilution law and molar conductivity, covered next.

Key takeaways

Strong electrolytes (NaCl, HCl, KOH) dissociate essentially completely in dilute aqueous solution; weak electrolytes (CH3COOH, NH3, HF) dissociate only partially.
The degree of dissociation $\alpha$ is the fraction of dissolved solute present as ions; $\alpha \approx 1$ for strong electrolytes, $\alpha \ll 1$ typically for weak ones at ordinary concentrations.
Molar conductivity $\Lambda_m$ falls only gently with concentration for strong electrolytes (interionic effects) but rises steeply toward infinite dilution for weak electrolytes (more dissociation).
Weak electrolyte equilibria obey a mass-action law with a dissociation constant $K_a$ or $K_b$ , linking $\alpha$ to concentration.
Classification is not always black-and-white: some salts (MgSO4-type) show intermediate "ion association" behavior.
Key idea: whether a solute dissociates fully or partially is a genuine equilibrium property, not a labeling convenience, and it is quantified by $\alpha$ and $K_a$ .

The core insight

A strong electrolyte's dissociation equilibrium lies so far toward ions that it is treated as complete, whereas a weak electrolyte's dissociation is a true, measurable equilibrium governed by a modest $K_a$ or $K_b$ , so the fraction ionized changes noticeably with concentration.

The physical picture

When an ionic solid such as NaCl dissolves, the lattice energy holding Na⁺ and Cl⁻ together is already overcome by hydration; separated ions are the thermodynamically stable form at essentially any dilution reachable in the lab. There is no meaningful "molecular NaCl" species floating in solution. The same is true for strong acids like HCl and strong bases like NaOH: the proton-transfer or dissociation equilibrium constant is enormous ( $K_a \gg 1$ ), so the reverse reaction is negligible. A weak electrolyte, by contrast, contains a covalent bond (O–H in acetic acid, N–H in ammonium formation) whose breaking is only mildly favorable. The equilibrium

\text{HA(aq)} \rightleftharpoons \text{H}^+\text{(aq)} + \text{A}^-\text{(aq)}

sits at a finite $K_a$ , typically $10^{-3}$ to $10^{-10}$ for common weak acids. Most HA molecules remain intact at any moment; only a small fraction ionizes. Because ionization consumes solute and produces two particles from one, dilution shifts the equilibrium toward more dissociation (Le Chatelier), a signature behavior absent in strong electrolytes. The clearest experimental fingerprint is molar conductivity, $\Lambda_m$ , measured as conductivity divided by concentration. This connects directly to Electrolyte Solutions and Ionic Conduction, where $\Lambda_m$ was defined and measured. Here we ask how $\Lambda_m$ depends on concentration itself, and that dependence is qualitatively different for the two classes.

Figure

Molar conductivity versus square-root concentration: strong electrolyte falls linearly, weak electrolyte plunges steeply

Figure

Dissociation equilibrium for a weak acid at two concentrations: dilution increases the ionized fraction

The core relations

\alpha = \frac{n_{\text{dissociated}}}{n_{\text{total dissolved}}}

where: $\alpha$ is the degree of dissociation, a dimensionless number between 0 and 1.

\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- , \qquad K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}

where: $K_a$ is the acid dissociation constant, evaluated at equilibrium concentrations.

K_a = \frac{\alpha^2 c}{1-\alpha}

where: $c$ is the nominal (analytical) concentration of the weak acid before dissociation.

\Lambda_m = \frac{\kappa}{c}

where: $\kappa$ is the measured conductivity and $c$ is the electrolyte concentration.

\alpha \approx \frac{\Lambda_m}{\Lambda_m^{\circ}} \quad \text{(weak electrolytes only)}

where: $\Lambda_m^{\circ}$ is the molar conductivity at infinite dilution, obtained via Kohlrausch's law for weak electrolytes.

Derivation

Derivation AFrom dissociation equilibrium to the concentration-dependence law

Goal. Starting from the mass-action equilibrium for a weak monoprotic acid, derive an explicit relation between the degree of dissociation $\alpha$ , the concentration $c$ , and $K_a$ , and show why strong electrolytes cannot be treated this way.

Step 1. Set up the equilibrium table. Let $c$ be the total analytical concentration of HA added to the solution. Define $\alpha$ as the fraction that ionizes at equilibrium. Then at equilibrium:

[\text{HA}] = c(1-\alpha), \qquad [\text{H}^+] = [\text{A}^-] = c\alpha

This assumes water's autoionization contributes negligible H⁺, valid whenever $c\alpha \gg 10^{-7}\ \text{mol dm}^{-3}$ .

Step 2. Insert into the mass-action expression. Substituting into $K_a = [\text{H}^+][\text{A}^-]/[\text{HA}]$ :

K_a = \frac{(c\alpha)(c\alpha)}{c(1-\alpha)} = \frac{c\alpha^2}{1-\alpha}

This is the general relation linking $\alpha$ , $c$ , and $K_a$ for any weak monoprotic electrolyte, valid at any dilution where the approximation in Step 1 holds.

Step 3. Examine the dilute limit. When $\alpha \ll 1$ (concentrated weak electrolyte), $1-\alpha \approx 1$ , so:

K_a \approx c\alpha^2 \quad\Rightarrow\quad \alpha \approx \sqrt{\frac{K_a}{c}}

This shows $\alpha$ increases as $c$ decreases, scaling as $c^{-1/2}$ — the seed of the Ostwald dilution law, to be developed in full next. Note immediately that $\alpha \to 1$ as $c \to 0$ : any weak electrolyte becomes fully dissociated in the limit of infinite dilution.

Step 4. Contrast with a strong electrolyte. For a strong electrolyte, there is no meaningful equilibrium constant to invert because the reverse (association) reaction essentially does not occur; formally $K_a \to \infty$ . Taking that limit in Step 3's formula:

\alpha = \sqrt{\frac{K_a}{c}} \to 1 \quad \text{for all finite } c \text{ as } K_a \to \infty

So the same mass-action framework, pushed to its extreme, reproduces "complete dissociation regardless of concentration" — confirming that strong and weak electrolytes are two ends of one continuum, distinguished by the size of $K$ , not by different physics.

Step 5. Connect to conductivity. Since each ion pair produced carries the current, $\Lambda_m$ for a weak electrolyte is proportional to the fraction actually dissociated:

\boxed{\alpha \approx \frac{\Lambda_m}{\Lambda_m^{\circ}}, \qquad K_a = \frac{c\,\alpha^2}{1-\alpha} = \frac{c\,(\Lambda_m/\Lambda_m^{\circ})^2}{1 - \Lambda_m/\Lambda_m^{\circ}}}

This boxed pair is the operational bridge: measure $\Lambda_m$ at concentration $c$ , know $\Lambda_m^{\circ}$ from Kohlrausch's law, and obtain both $\alpha$ and $K_a$ from conductivity data alone — no pH electrode required.

Worked examples

Example 1 — Foundational: degree of dissociation of acetic acid from $K_a$ . Acetic acid has $K_a = 1.8 \times 10^{-5}$ at 298 K. Find $\alpha$ and the pH for $c = 0.100\ \text{mol dm}^{-3}$ . Using the dilute-limit approximation first: $\alpha \approx \sqrt{K_a/c}$ .

\alpha \approx \sqrt{\frac{1.8\times10^{-5}}{0.100}} = \sqrt{1.8\times10^{-4}} = 0.0134

Check self-consistency: $1-\alpha = 0.987 \approx 1$ , so the approximation is justified (error under 2%). Solve exactly for comparison: $K_a(1-\alpha) = c\alpha^2$ gives $c\alpha^2 + K_a\alpha - K_a = 0$ .

\alpha = \frac{-K_a + \sqrt{K_a^2 + 4cK_a}}{2c} = \frac{-1.8\times10^{-5} + \sqrt{(1.8\times10^{-5})^2 + 4(0.100)(1.8\times10^{-5})}}{2(0.100)}

\alpha = \frac{-1.8\times10^{-5} + \sqrt{3.24\times10^{-10} + 7.2\times10^{-6}}}{0.200} = \frac{-1.8\times10^{-5} + 2.683\times10^{-3}}{0.200} = 0.01325

The two methods agree to three significant figures ( $\alpha = 0.0132$ – $0.0134$ ). Take $\alpha = 0.0132$ .

[\text{H}^+] = c\alpha = (0.100)(0.0132) = 1.32\times10^{-3}\ \text{mol dm}^{-3}

\text{pH} = -\log_{10}(1.32\times10^{-3}) = 2.88

Sanity check: pure 0.100 M strong acid would give pH = 1.00. Acetic acid's pH of 2.88 is much higher, confirming only a small fraction (1.32%) has ionized — consistent with "weak."

Example 2 — Applied: distinguishing strong from weak using conductivity data. At 298 K, a 0.0100 mol dm⁻³ solution of acetic acid has measured conductivity $\kappa = 1.63\times10^{-4}\ \text{S cm}^{-1}$ . Independent Kohlrausch-law extrapolation for acetic acid gives $\Lambda_m^{\circ} = 390.7\ \text{S cm}^2\ \text{mol}^{-1}$ (built from $\lambda^{\circ}(\text{H}^+) = 349.6$ and $\lambda^{\circ}(\text{CH}_3\text{COO}^-) = 40.9\ \text{S cm}^2\ \text{mol}^{-1}$ , plus small correction; here we use the standard tabulated composite value). A 0.0100 mol dm⁻³ solution of NaCl at the same temperature has $\Lambda_m = 118.5\ \text{S cm}^2\ \text{mol}^{-1}$ , close to its $\Lambda_m^{\circ} = 126.5\ \text{S cm}^2\ \text{mol}^{-1}$ . Compare the two electrolytes quantitatively. Step A: convert $c$ to consistent units. $c = 0.0100\ \text{mol dm}^{-3} = 1.00\times10^{-5}\ \text{mol cm}^{-3}$ . Step B: compute $\Lambda_m$ for acetic acid.

\Lambda_m = \frac{\kappa}{c} = \frac{1.63\times10^{-4}\ \text{S cm}^{-1}}{1.00\times10^{-5}\ \text{mol cm}^{-3}} = 16.3\ \text{S cm}^2\ \text{mol}^{-1}

Step C: compute $\alpha$ for acetic acid.

\alpha = \frac{\Lambda_m}{\Lambda_m^{\circ}} = \frac{16.3}{390.7} = 0.0417

Step D: compute $\alpha$ for NaCl the same way.

\alpha_{\text{NaCl}} = \frac{\Lambda_m}{\Lambda_m^{\circ}} = \frac{118.5}{126.5} = 0.937

Step E: interpret. Acetic acid at this concentration is only 4.17% dissociated — the vast majority exists as neutral molecules. NaCl is nominally "93.7% dissociated" by this formula, but the shortfall from 100% is not due to molecular NaCl forming; it reflects interionic drag (ion-atmosphere relaxation and electrophoretic retardation) slowing the fully dissociated ions down, not a change in the number of free ions. This is precisely why the ratio $\Lambda_m/\Lambda_m^{\circ}$ is only meaningful as a dissociation fraction for weak electrolytes; for strong electrolytes its shortfall from unity has an entirely different physical origin. Step F: back out $K_a$ for a check.

K_a = \frac{c\,\alpha^2}{1-\alpha} = \frac{(0.0100)(0.0417)^2}{1-0.0417} = \frac{1.74\times10^{-5}}{0.958} = 1.81\times10^{-5}

This matches the accepted literature value $K_a = 1.8\times10^{-5}$ to two significant figures — a strong consistency check that the conductivity-based method correctly recovers the equilibrium constant.

Limiting cases & common traps

Situation	What actually happens	Common error
Strong electrolyte, any dilute $c$	$\alpha \approx 1$ always; $\Lambda_m$ decreases mildly with $c$ due to ion-ion interactions	Assuming falling $\Lambda_m$ means "less dissociation"
Weak electrolyte, $c \to 0$	$\alpha \to 1$ ; even weak electrolytes fully dissociate at infinite dilution	Thinking weak electrolytes never fully ionize
Weak electrolyte, $c$ large	$\alpha$ small, scales as $\sqrt{K_a/c}$	Forgetting the square-root, not linear, dependence
Polyprotic weak acids (H2SO4 second step, H3PO4)	Each step has its own $K_a$ ; first step of H2SO4 is effectively strong, later steps weak	Treating the whole molecule as uniformly "strong" or "weak"
Weak base in water (NH3)	Same formalism with $K_b$ , $[\text{OH}^-] = c\alpha$	Using $K_a$ expression directly without switching to $K_b$

The chemist behind it

Svante Arrhenius proposed in 1884 that electrolytes dissociate into ions in solution — a radical idea at the time, since it implied that a neutral salt crystal produces charged species spontaneously in water, with no external electric field required. His theory explained why solutions of electrolytes conduct electricity and why colligative properties (freezing-point depression, osmotic pressure) of electrolyte solutions exceed simple predictions based on solute particle count. Wilhelm Ostwald extended Arrhenius's ideas quantitatively, applying mass-action reasoning to weak-electrolyte dissociation — the same reasoning used in the derivation above — and this became the Ostwald dilution law, the natural next topic.

Origins · 1 minSvante August Arrhenius and how this was discovered

Svante August Arrhenius

1859–1927 · Proposed ionic dissociation theory in 1884, later foundational to electrolyte theory and awarded the 1903 Nobel Prize in Chemistry

Ask the tutor: How does the Ostwald dilution law let us extract $K_a$ from a series of conductivity measurements alone?

Is degree of dissociation the same as percent ionization

Yes — percent ionization is simply $\alpha \times 100\%$ . Both describe the fraction of dissolved solute existing as free ions at equilibrium, at a specified analytical concentration. Because $\alpha$ depends on concentration for weak electrolytes, always quote it alongside the concentration at which it was measured or calculated; an isolated percentage without a concentration is not fully meaningful.

Why does molar conductivity of strong electrolytes still decrease with concentration if dissociation is complete

Even with 100% dissociation, ions in a more concentrated solution experience stronger electrostatic drag from their surrounding "ionic atmosphere" of opposite charge, which slows their drift under an applied field. This is captured by the Debye–Hückel–Onsager theory, giving $\Lambda_m$ a mild $\sqrt{c}$ decrease rather than the steep drop seen for weak electrolytes, whose decrease reflects genuinely fewer ions being present.

Can a substance be a strong electrolyte in one solvent and weak in another

Yes. Dissociation depends on solvent polarity and the solute-solvent interactions, not solely on the solute. HCl is a strong acid in water but behaves as a weak acid in acetic acid solvent, because acetic acid is a much poorer proton acceptor than water. Classification as "strong" or "weak" is always implicitly solvent-specific, with water assumed by default in general chemistry.

How does this connect to the Ostwald dilution law mentioned throughout

The Ostwald dilution law is exactly the relation $K_a = c\alpha^2/(1-\alpha)$ derived above, applied specifically to predict how $\alpha$ (and hence $\Lambda_m$ ) changes as concentration is systematically varied. This article built the equilibrium foundation; the dedicated treatment develops the full dilution series, its graphical linearization, and how it is used to extract $K_a$ purely from conductivity measurements at multiple concentrations.

Are all salts either clearly strong or clearly weak

Mostly, but not universally. Salts of divalent ions such as MgSO4 show measurable "ion association" — a significant fraction exists as neutral ion pairs (Mg2+·SO4^2-) even though no covalent bond forms, due to strong electrostatic attraction between divalent ions. Such species are usually still classified as strong electrolytes for simplicity, since the association is not proton-transfer chemistry, but conductivity data shows real deviation from complete dissociation.