Stepwise ionisation

Why "stepwise" and not "all at once". A polybasic (polyprotic) acid such as $H_2A$ has more than one ionisable hydrogen, but chemically it cannot lose them simultaneously — each proton departs in its own distinct equilibrium, with its own equilibrium constant. Writing a single combined equation such as $H_2A \rightleftharpoons 2H^+ + A^{2-}$ hides this staged structure and is not how the ionisation constants are defined or used. For a diprotic acid, the two steps and their constants are:

where: $K_{a1}$ and $K_{a2}$ are the first and second (stepwise) ionisation constants, and $[\ ]$ denotes equilibrium molar concentration. For a triprotic acid such as $\text{H}_3\text{PO}_4$ a third analogous step follows, with constant $K_{a3}$.

The universal trend. For any polybasic acid, successive constants fall sharply — typically by a factor of $10^4$ to $10^6$ at each step:

The same staged logic, and the same decreasing trend in successive $K_b$ values, applies to polyacidic bases (species that accept more than one proton), though ISC problems most commonly work with polybasic acids.

Why each step is harder. Once the first proton leaves, the remaining species ($HA^-$) is negatively charged, and this negative charge electrostatically attracts the departing proton more strongly than the neutral parent $H_2A$ did — so removing the second proton takes more energy and happens far less readily. Remove that proton too, and the species is even more negative, making the third departure (where relevant) harder still. The pattern is not an arbitrary numerical coincidence; it is charge bookkeeping.

The practical payoff. Because $K_{a1} \gg K_{a2}$, almost all of the $H^+$ in a polybasic acid solution comes from the first ionisation step alone — the second (and further) steps contribute a concentration of $H^+$ that is negligible by comparison. This is what lets pH calculations for polybasic acids be done, to excellent approximation, as if the acid were monoprotic using only $K_{a1}$. It is also why, when comparing acid strength across different acids, chemists compare $K_{a1}$ (or the single $K_a$ of a monobasic acid) — a larger $K_a$ signals a more extensively ionised, stronger acid, and likewise a larger $K_b$ signals a stronger base.

Extending carbonic acid to check the approximation numerically. A $0.10\ M$ solution of $\text{H}_2\text{CO}_3$ has $K_{a1} = 4.3\times10^{-7}$ and $K_{a2} = 4.8\times10^{-11}$. Find $[H^+]$ using only the first ionisation, confirm the second step is negligible, and then use the ratio $K_{a1}/K_{a2}$ to comment on relative step sizes.

Step 1. Write the first-ionisation ICE table: $[\text{H}_2\text{CO}_3]_{eq} \approx 0.10 - x$, $[H^+] = x$, $[\text{HCO}_3^-] = x$, with $x \ll 0.10$ expected since $K_{a1}$ is small.

Step 2. Apply the small-$x$ approximation: $K_{a1} \approx \dfrac{x^2}{0.10} = 4.3\times10^{-7} \implies x^2 = 4.3\times10^{-8}$.

Step 3. Solve: $x = \sqrt{4.3\times10^{-8}} \approx 2.07\times10^{-4}\ M$, so $[H^+] \approx 2.1\times10^{-4}\ M$ from the first step, and $[\text{HCO}_3^-] \approx 2.1\times10^{-4}\ M$.

Step 4. Estimate the second step's extra $H^+$: let $y$ be the additional ionisation from $\text{HCO}_3^-$; $y^2 \approx K_{a2} \times [\text{HCO}_3^-] \approx 4.8\times10^{-11}\times 2.1\times10^{-4} \approx 1.0\times10^{-14}$, so $y \approx 1.0\times10^{-7}\ M$ — over $2000$ times smaller than $x$, confirming the approximation.

Step 5. Compute $K_{a1}/K_{a2} = \dfrac{4.3\times10^{-7}}{4.8\times10^{-11}} \approx 9.0\times10^{3}$: the first ionisation is roughly $9000$ times more extensive than the second, which is exactly why $y \ll x$ above and why the electrostatic-attraction argument (the doubly-negative $\text{CO}_3^{2-}$ that would form holds its neighbouring proton far more tightly than neutral $\text{H}_2\text{CO}_3$ does) is quantitatively confirmed, not just qualitatively plausible. Final answer: $[H^+] \approx 2.1\times10^{-4}\ M$ (pH $\approx 3.68$), essentially unchanged by including the second ionisation step.

Competitive stretch

For a triprotic acid like $\text{H}_3\text{PO}_4$ ($K_{a1} \approx 7.5\times10^{-3}$, $K_{a2} \approx 6.2\times10^{-8}$, $K_{a3} \approx 4.2\times10^{-13}$), the same reasoning applies twice over: $K_{a1}/K_{a2} \approx 1.2\times10^{5}$ and $K_{a2}/K_{a3} \approx 1.5\times10^{5}$, so essentially all the $H^+$ in a phosphoric acid solution again comes from the first step. Note that $K_{a1}$ here is far larger than either carbonic constant — phosphoric acid is simply a stronger acid at its first step, an entirely separate comparison from how sharply its own steps fall relative to each other.

• Adding all steps as if each contributes equally to $[H^+]$. With $K_{a1} \gg K_{a2} \gg K_{a3}$, only the first step is needed for an accurate $[H^+]$; summing all three overstates the acid's strength.
• Writing one combined equation instead of separate stepwise equilibria. $\text{H}_2\text{CO}_3 \rightleftharpoons 2H^+ + \text{CO}_3^{2-}$ in a single step is chemically wrong; ionisation genuinely proceeds one proton at a time, each with its own $K_a$ and its own equilibrium.
• Treating the falling $K_a$ values as an unexplained pattern. The drop from $K_{a1}$ to $K_{a2}$ (and to $K_{a3}$) follows directly from increasing electrostatic attraction between the departing proton and a progressively more negative conjugate base — always be ready to say why, not just quote the trend.
• Assuming $[HA^-]$ from step one is untouched by step two. Strictly, $HA^-$ is produced in step one and partly consumed in step two; the amount consumed is negligible only because $K_{a2} \ll K_{a1}$ — that inequality is the justification for ignoring it, not a free pass to ignore it always.

Ask the tutor: Want to see how the first-ionisation shortcut extends to a full phosphoric-acid pH calculation, or how stepwise ionisation feeds into the next hydrolysis and buffer articles?