Ionization constant and percent ionization (monoprotic acids/bases)
This note explains the relations used by the calculator to determine the equilibrium constant
\(K_a\) or \(K_b\), the extent of ionization \(x\), pH/pOH, and the percent ionization for a
weak monoprotic acid \(\mathrm{HA}\) or a weak monobasic base \(\mathrm{B}\) in water at
\(25\,^{\circ}\mathrm{C}\) (so \(K_w = 1.0\times 10^{-14}\)).
1) Model equilibria and definitions
- Acid: \(\mathrm{HA(aq) + H_2O(l) \rightleftharpoons A^-(aq) + H_3O^+(aq)}\),
\[
K_a \;=\; \frac{[H_3O^+][A^-]}{[HA]}\,.
\]
- Base: \(\mathrm{B(aq) + H_2O(l) \rightleftharpoons BH^+(aq) + OH^-(aq)}\),
\[
K_b \;=\; \frac{[BH^+][OH^-]}{[B]}\,.
\]
- For conjugate pairs: \(K_a(\mathrm{HA})\,K_b(\mathrm{A^-}) = K_w\).
2) ICE table for a single, dilute weak electrolyte
Let the analytical concentration be \(C_0\) (mol·L\(^{-1}\)), and let \(x\) be the amount ionized at equilibrium.
Neglecting autoionization of water in the mass balance:
Acid:
\[
\begin{array}{l|ccc}
& [\mathrm{HA}] & [\mathrm{A^-}] & [\mathrm{H_3O^+}] \\ \hline
\text{I} & C_0 & 0 & 0 \\
\text{C} & -x & +x & +x \\
\text{E} & C_0-x & x & x
\end{array}
\qquad\Rightarrow\qquad
K_a \;=\; \frac{x^2}{C_0 - x}.
\]
Base:
\[
\begin{array}{l|ccc}
& [\mathrm{B}] & [\mathrm{BH^+}] & [\mathrm{OH^-}] \\ \hline
\text{I} & C_0 & 0 & 0 \\
\text{C} & -x & +x & +x \\
\text{E} & C_0-x & x & x
\end{array}
\qquad\Rightarrow\qquad
K_b \;=\; \frac{x^2}{C_0 - x}.
\]
3) Solving for the unknown
- Given \(C_0\) and \(x\) (e.g. from measured \([H_3O^+]\) or \([OH^-]\)):
\(K_a\) or \(K_b = \dfrac{x^2}{C_0 - x}\).
- Given \(C_0\) and \(K_a\) (or \(K_b\)):
solve the exact quadratic
\[
x^2 + K\,x - K\,C_0 = 0
\quad\Rightarrow\quad
x = \frac{-K + \sqrt{K^2 + 4KC_0}}{2}\ (>0).
\]
(No weak-approximation needed.)
- Given pH/pOH:
\([H_3O^+] = 10^{-\mathrm{pH}}\), \([OH^-] = 10^{-\mathrm{pOH}}\), and
\(\mathrm{pH}+\mathrm{pOH}=14\).
For an acid, set \(x=[H_3O^+]\); for a base, \(x=[OH^-]\); then use the expression above for \(K\).
4) Degree and percent ionization
The degree of ionization is \(\alpha = \dfrac{x}{C_0}\) (fraction of molecules that ionize).
The percent ionization is
\[
\%\,\text{ionization} \;=\; 100\,\alpha \;=\; 100\,\frac{x}{C_0}.
\]
If \(K\) and \(C_0\) are given, one can first find \(x\) from the quadratic, then compute \(\alpha\).
For very weak species (\(x \ll C_0\)) the common approximation gives
\[
x \approx \sqrt{K\,C_0},\qquad \alpha \approx \sqrt{\frac{K}{C_0}}.
\]
5) From percent ionization to \(K\)
If the percent ionization is reported (say \(\% = 100\,\alpha\)), then \(x=\alpha C_0\) and
\[
K \;=\; \frac{x^2}{C_0 - x} \;=\; \frac{\alpha^2\,C_0}{1-\alpha},
\qquad \alpha=\frac{\%\text{ ionization}}{100}.
\]
6) Interpreting pH/pOH and consistency checks
- Acidic solutions: \(\mathrm{pH}<7.0\) and typically \(x=[H_3O^+] \ll C_0\) for weak acids.
- Basic solutions: \(\mathrm{pOH}<7.0\) and typically \(x=[OH^-] \ll C_0\) for weak bases.
- If \(x \ge C_0\), the weak-electrolyte model breaks (the acid/base behaves strong under these conditions).
7) Assumptions and limitations
- Monoprotic/monobasic, dilute solutions; activity coefficients are taken as \(1\).
- \(K_w\) is taken as \(1.0\times 10^{-14}\) at \(25\,^{\circ}\mathrm{C}\); it varies with temperature.
- Water autoionization is neglected in the mass balance but used to relate pH and pOH.
- For polyprotic acids or polybasic bases, coupled equilibria require extended treatment.
