Concept
A strong acid or strong base dissociates stoichiometrically in water. If the
solute releases \(n_{\mathrm H}\) protons (acid) or \(n_{\mathrm{OH}}\) hydroxides (base) per
formula unit and its formal molarity is \(C_0\) (mol·L⁻¹), the formal amount of the
acidic or basic species added to water is
\[
C_{\text{formal}} =
\begin{cases}
n_{\mathrm H}\cdot C_0 & \text{(acid)}\\[4pt]
n_{\mathrm{OH}}\cdot C_0 & \text{(base)}
\end{cases}
\]
For very dilute solutions, the autoionization of water must also be included via
\(K_{\mathrm w} = \big[\text{H}_3\text{O}^{+}\big]\cdot \big[\text{OH}^{-}\big]\).
Including water autoionization
Let \(x\) be the contribution from water. Mass balance for the two cases:
\[
\textbf{Acid:}\quad
\big[\text{H}_3\text{O}^{+}\big] = C_{\text{formal}} + x,\qquad
\big[\text{OH}^{-}\big] = x
\]
\[
\textbf{Base:}\quad
\big[\text{OH}^{-}\big] = C_{\text{formal}} + x,\qquad
\big[\text{H}_3\text{O}^{+}\big] = x
\]
In both cases the same quadratic arises from \(K_{\mathrm w}=(C_{\text{formal}}+x)\cdot x\):
\[
x^2 + C_{\text{formal}}\,x - K_{\mathrm w} = 0
\quad\Longrightarrow\quad
x = \frac{-\,C_{\text{formal}} + \sqrt{C_{\text{formal}}^{2} + 4\cdot K_{\mathrm w}}}{2}
\]
Therefore,
\[
\begin{aligned}
\text{Acid:}\quad
\big[\text{H}_3\text{O}^{+}\big] &= C_{\text{formal}} + x, &
\big[\text{OH}^{-}\big] &= x \\[6pt]
\text{Base:}\quad
\big[\text{H}_3\text{O}^{+}\big] &= x, &
\big[\text{OH}^{-}\big] &= C_{\text{formal}} + x
\end{aligned}
\]
pH, pOH, and \(pK_{\mathrm w}\)
\[
\mathrm{pH} = -\log_{10}\!\big(\big[\text{H}_3\text{O}^{+}\big]\big),\qquad
\mathrm{pOH} = -\log_{10}\!\big(\big[\text{OH}^{-}\big]\big),\qquad
\mathrm{pH} + \mathrm{pOH} = pK_{\mathrm w} = -\log_{10}(K_{\mathrm w})
\]
At \(25\,^{\circ}\text{C}\), \(K_{\mathrm w}\approx 1.0\times 10^{-14}\) so \(pK_{\mathrm w}\approx 14.00\).
Away from \(25\,^{\circ}\text{C}\) you must use the appropriate \(K_{\mathrm w}(T)\).
Useful approximation (when water is negligible)
If \(C_{\text{formal}}\gg \sqrt{K_{\mathrm w}}\) (e.g., \(C_{\text{formal}}\gtrsim 100\times 10^{-7}\,\mathrm{mol\cdot L^{-1}}\) at \(25^{\circ}\text{C}\)),
then \(x\) is tiny and
\[
\big[\text{H}_3\text{O}^{+}\big]\approx
\begin{cases}
C_{\text{formal}} & \text{(acid)}\\[4pt]
\dfrac{K_{\mathrm w}}{C_{\text{formal}}} & \text{(base)}
\end{cases}
\qquad
\big[\text{OH}^{-}\big]\approx
\begin{cases}
\dfrac{K_{\mathrm w}}{C_{\text{formal}}} & \text{(acid)}\\[10pt]
C_{\text{formal}} & \text{(base)}
\end{cases}
\]
The calculator automatically switches to the full quadratic so that extremely dilute strong
solutions are handled correctly.
Notes & pitfalls
-
Polyprotic/ polyvalent species. For sulfuric acid, the first proton is effectively strong
in water; the second is not. Set \(n_{\mathrm H}\) equal to the number of strongly
released protons you wish to count. For bases such as calcium hydroxide, use \(n_{\mathrm{OH}}=2\).
-
Units and symbols.
Concentrations are in \(\mathrm{mol\cdot L^{-1}}\).
Activities are approximated by molarities (ideal solution assumption).
-
Temperature. \(K_{\mathrm w}\) is temperature–dependent; using the correct value is crucial
when computing \(\mathrm{pH}\) and \(\mathrm{pOH}\) away from \(25^{\circ}\text{C}\).
Symbol key
\(C_0\): formal molarity of the solute; \(\,n_{\mathrm H}, n_{\mathrm{OH}}\): number of released
\(\text{H}^{+}\) or \(\text{OH}^{-}\) per formula unit; \(\,C_{\text{formal}} = n\cdot C_0\);
\(K_{\mathrm w}\): ionic product of water; \(x\): contribution from water autoionization.
