Ionisation and dilution law

Strong vs weak electrolytes. Strong electrolytes ($\text{HCl}$, $\text{NaOH}$, $\text{NaCl}$) ionise almost completely at any reasonable concentration — there is no real equilibrium to speak of, just a one-way split. Weak electrolytes ($\text{CH}_3\text{COOH}$, $\text{NH}_4\text{OH}$) ionise only partially: unionised molecules and ions coexist at a true dynamic equilibrium, and the balance point shifts with conditions.

Degree of ionisation. The degree of ionisation $\alpha$ is the fraction of the total moles of electrolyte taken that have actually ionised at equilibrium:

where: $\alpha$ is a pure number between $0$ and $1$ (often quoted as a percentage); $\alpha = 0$ means no ionisation at all, $\alpha = 1$ means complete ionisation.

Setting up the ICE table. For a weak monobasic acid $HA$ ionising as $HA \rightleftharpoons H^+ + A^-$, start with initial concentration $c$ (mol L$^{-1}$) of $HA$ and zero for the ions. At equilibrium, a fraction $\alpha$ has ionised, so:

where: the change row of the ICE table is $-c\alpha$ for $HA$ and $+c\alpha$ for each ion, so the three equilibrium concentrations here are what feed directly into the equilibrium-constant expression below.

Substituting the ICE-table concentrations into $K_a = \dfrac{[H^+][A^-]}{[HA]}$ gives Ostwald's dilution law, derived in full below. When $\alpha$ is small (weak electrolytes at ordinary concentrations), the approximation $1 - \alpha \approx 1$ lets $\alpha$ be pulled out cleanly as $\alpha \approx \sqrt{K_a/c}$. Since $K_a$ is fixed at a given temperature, this shows $\alpha \propto 1/\sqrt{c}$: diluting a weak electrolyte increases its degree of ionisation, and $\alpha \to 1$ as $c \to 0$. Le Chatelier's principle gives the same conclusion qualitatively — diluting the solution lowers all concentrations, and the equilibrium shifts toward the side with more particles ($H^+ + A^-$, two particles) to partially oppose that drop, i.e. toward more ionisation. The same law applies to a weak base $BOH \rightleftharpoons B^+ + OH^-$, with $K_b \approx c\alpha^2$.

Comparing $\alpha$ at two concentrations of acetic acid. Acetic acid has $K_a = 1.8\times10^{-5}$ at 298 K. Find $\alpha$ at $c_1 = 0.10\ M$ and at $c_2 = 0.0010\ M$ (a 100-fold dilution), and check the relationship between them.

Step 1. At $c_1 = 0.10\ M$, apply $\alpha \approx \sqrt{K_a/c}$: $\alpha_1 \approx \sqrt{(1.8\times10^{-5})/0.10} = \sqrt{1.8\times10^{-4}} \approx 0.0134$, i.e. about $1.34\%$ ionised.

Step 2. At $c_2 = 0.0010\ M$, the same formula gives $\alpha_2 \approx \sqrt{(1.8\times10^{-5})/0.0010} = \sqrt{1.8\times10^{-2}} \approx 0.134$, i.e. about $13.4\%$ ionised.

Step 3. Compare: diluting 100-fold increased $\alpha$ tenfold ($0.134/0.0134 = 10.0$), exactly matching the prediction $\alpha \propto 1/\sqrt{c}$, since $\sqrt{100} = 10$.

Step 4. Check the approximation at each point. At $c_1$, $\alpha_1 = 0.0134 \ll 1$, so $1-\alpha \approx 0.987 \approx 1$ is safe. At $c_2$, $\alpha_2 = 0.134$ is over $5\%$, so the approximation is getting shaky — a more careful answer would solve the exact quadratic $K_a = \dfrac{c\alpha^2}{1-\alpha}$ instead.

Final answer: $\alpha_1 \approx 0.0134$ (1.34%) at $0.10\ M$, and $\alpha_2 \approx 0.134$ (13.4%) at $0.0010\ M$ — a tenfold increase in ionisation on hundredfold dilution.

• Applying Ostwald's dilution law to strong electrolytes — the law describes the partial, equilibrium-governed ionisation of weak electrolytes only; strong electrolytes are already almost fully ionised regardless of concentration, so there is no meaningful equilibrium for the law to describe.
• Using $\alpha \approx \sqrt{K_a/c}$ without checking that $\alpha \ll 1$ — at low concentrations or large $K_a$, $\alpha$ can grow too large for $1-\alpha \approx 1$ to hold, and the exact quadratic form is needed instead; skipping this check is the single most common source of a wrong final answer.
• Confusing $K_a$ with $\alpha$ — $K_a$ is constant at a given temperature and depends only on the identity of the acid; $\alpha$ changes with concentration, increasing as the solution is diluted, even though $K_a$ itself never moves.
• Assuming ion concentration rises with dilution — $\alpha$ rises, but the absolute ion concentration $c\alpha$ typically still falls overall as $c$ drops sharply; a bigger fraction of a much smaller total can still be a smaller number.
• Forgetting that $\alpha$ depends on temperature only through $K_a$ — at a fixed temperature, changing the volume of solvent (dilution) is the only lever that moves $\alpha$ for a given electrolyte; changing $K_a$ itself requires changing the temperature.

Ask the tutor: Want to see how this degree of ionisation feeds directly into a pH calculation for a weak acid, or how the common-ion effect works against dilution's push on $\alpha$?