Percent Ionization of a Weak Acid

% ion=[H+]C×100%\%\,\text{ion} = \frac{[\mathrm{H^+}]}{C} \times 100\%

Worked example: [H+] 1.34e-3 in 0.100 M acid → 1.34% ionized — press Try an example to run it live, then adjust anything.

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Percent Ionization of a Weak Acid explained

C[H+]% ion

Ka tells you how a weak acid behaves in principle; percent ionization tells you what actually happened in the beaker. In 0.100 M acetic acid the equilibrium [H⁺] is 1.34 × 10⁻³ M, so only 1.34% of the molecules have given up a proton and 98.7% are still intact — a vivid picture of what "weak" means. It is also the number that justifies the x-is-small approximation, which is generally trusted below about 5% ionization.

The counterintuitive part is that percent ionization rises as you dilute. Because [H⁺] goes as the square root of concentration while C goes linearly, halving the concentration multiplies the fraction ionized by roughly √2. Dilute that acetic acid to 0.0010 M and it is about 13% ionized even though the solution is far less acidic in absolute terms. Ostwald's dilution law captures the same effect, and it is why a weak acid approaches complete dissociation in the limit of infinite dilution.

Percent Ionization of a Weak Acid formula

% ion=[H+]C×100%\%\,\text{ion} = \frac{[\mathrm{H^+}]}{C} \times 100\%
Where
  • % ion\%\,\text{ion}= Percent ionization (%)
  • [H+][\mathrm{H^+}]= Hydrogen ion concentration at equilibrium (M)
  • CC= Formal acid concentration (M)

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