Problem setup for triprotic acids and bases calculations
A triprotic acid \(\mathrm{H_3A}\) dissociates in three steps:
\(\mathrm{H_3A \rightleftharpoons H^+ + H_2A^-}\) (with \(K_{a1}\)), \(\mathrm{H_2A^- \rightleftharpoons H^+ + HA^{2-}}\) (with \(K_{a2}\)), \(\mathrm{HA^{2-} \rightleftharpoons H^+ + A^{3-}}\) (with \(K_{a3}\)).
Given: \(K_{a1}=7.1\times 10^{-3}\), \(K_{a2}=6.3\times 10^{-8}\), \(K_{a3}=4.2\times 10^{-13}\).
Tasks: (1) Initial pH of \(0.0500\,\mathrm{M}\) \(\mathrm{H_3A}\). (2) pH at the first equivalence point when \(25.0\,\mathrm{mL}\) of acid is titrated with \(0.0500\,\mathrm{M}\) NaOH.
| Constant | Value | \(pK_a\) | What it typically controls |
|---|---|---|---|
| \(K_{a1}\) | \(7.1\times 10^{-3}\) | \(pK_{a1}=-\log_{10}(K_{a1})\approx 2.149\) | Initial pH (first dissociation dominates) |
| \(K_{a2}\) | \(6.3\times 10^{-8}\) | \(pK_{a2}\approx 7.201\) | Second buffer region / amphiprotic behavior of \(\mathrm{H_2A^-}\) |
| \(K_{a3}\) | \(4.2\times 10^{-13}\) | \(pK_{a3}\approx 12.377\) | Third buffer region at high pH |
1) Initial pH of \(0.0500\,\mathrm{M}\) \(\mathrm{H_3A}\)
In triprotic acids and bases calculations, the initial pH is usually controlled by the first dissociation because \(K_{a1}\gg K_{a2}\gg K_{a3}\). Let \(x=[\mathrm{H^+}]\) produced mainly from the first step.
For \(\mathrm{H_3A \rightleftharpoons H^+ + H_2A^-}\) with initial concentration \(C=0.0500\,\mathrm{M}\):
\([\mathrm{H_3A}]\approx C-x,\quad [\mathrm{H^+}]\approx x,\quad [\mathrm{H_2A^-}]\approx x.\)
Apply the equilibrium expression:
\[ K_{a1}=\frac{[\mathrm{H^+}]\cdot[\mathrm{H_2A^-}]}{[\mathrm{H_3A}]} \approx \frac{x\cdot x}{C-x} =\frac{x^2}{C-x}. \]
Rearranging gives a quadratic:
\[ x^2+K_{a1}\cdot x-K_{a1}\cdot C=0. \]
Solve for the physically meaningful (positive) root:
\[ x=\frac{-K_{a1}+\sqrt{K_{a1}^2+4\cdot K_{a1}\cdot C}}{2} =\frac{-7.1\times 10^{-3}+\sqrt{(7.1\times 10^{-3})^2+4\cdot(7.1\times 10^{-3})\cdot(0.0500)}}{2}. \]
\[ x\approx 1.56\times 10^{-2}\,\mathrm{M} \quad\Longrightarrow\quad \mathrm{pH}=-\log_{10}(x)\approx 1.81. \]
Consistency check: the second dissociation contribution is negligible because \(\frac{K_{a2}}{[\mathrm{H^+}]}\approx \frac{6.3\times 10^{-8}}{1.56\times 10^{-2}}\approx 4\times 10^{-6}\), so \(\mathrm{H_2A^-}\) barely dissociates further at this pH.
2) pH at the first equivalence point (triprotic acid titration)
At the first equivalence point, one mole of \(\mathrm{OH^-}\) has been added per mole of \(\mathrm{H_3A}\), converting essentially all \(\mathrm{H_3A}\) into \(\mathrm{H_2A^-}\):
\[ \mathrm{H_3A + OH^- \rightarrow H_2A^- + H_2O}. \]
Moles of acid initially:
\[ n(\mathrm{H_3A})=0.0500\,\mathrm{mol\cdot L^{-1}}\cdot 0.0250\,\mathrm{L}=1.25\times 10^{-3}\,\mathrm{mol}. \]
Using \(0.0500\,\mathrm{M}\) NaOH, the volume needed for the first equivalence point is:
\[ V_{\mathrm{NaOH}}=\frac{n}{C_{\mathrm{NaOH}}} =\frac{1.25\times 10^{-3}\,\mathrm{mol}}{0.0500\,\mathrm{mol\cdot L^{-1}}} =0.0250\,\mathrm{L}=25.0\,\mathrm{mL}. \]
Total volume at equivalence: \(V_\text{tot}=25.0\,\mathrm{mL}+25.0\,\mathrm{mL}=50.0\,\mathrm{mL}\). Formal concentration of \(\mathrm{H_2A^-}\) is therefore
\[ C^\ast=\frac{1.25\times 10^{-3}\,\mathrm{mol}}{0.0500\,\mathrm{L}}=0.0250\,\mathrm{M}. \]
Key idea: \(\mathrm{H_2A^-}\) is amphiprotic (it can donate a proton via \(K_{a2}\) and accept a proton via \(K_b=\frac{K_w}{K_{a1}}\)). For an amphiprotic species between \(K_{a1}\) and \(K_{a2}\), the pH is well-approximated by \(\,[\mathrm{H^+}]\approx \sqrt{K_{a1}\cdot K_{a2}}\), equivalently \(\mathrm{pH}\approx \frac{pK_{a1}+pK_{a2}}{2}\).
Compute \([\mathrm{H^+}]\) and pH:
\[ [\mathrm{H^+}]\approx \sqrt{K_{a1}\cdot K_{a2}} =\sqrt{(7.1\times 10^{-3})\cdot(6.3\times 10^{-8})} \approx 2.11\times 10^{-5}\,\mathrm{M}. \]
\[ \mathrm{pH}\approx -\log_{10}(2.11\times 10^{-5})\approx 4.67. \]
Visualization: species distribution for a triprotic acid (fraction \(\alpha\) vs pH)
Summary of results
- Initial pH of \(0.0500\,\mathrm{M}\) \(\mathrm{H_3A}\): \(\mathrm{pH}\approx 1.81\) (first dissociation dominates).
- pH at the first equivalence point: \(\mathrm{pH}\approx \tfrac{pK_{a1}+pK_{a2}}{2}\approx 4.67\) (amphiprotic \(\mathrm{H_2A^-}\)).