Titration of a Weak Acid with a Strong Base — Theory
In a weak-acid/strong-base titration, the curve features four regimes:
(1) the initial weak acid, (2) a buffer region until the weak acid is mostly neutralized,
(3) the equivalence point where the conjugate base hydrolyzes, and
(4) the post-equivalence region dominated by excess strong base.
This calculator uses exact quadratics when needed, Henderson–Hasselbalch in the buffer,
hydrolysis at equivalence, and strong-base excess afterwards. Unless you change assumptions,
it treats solutions at 25 °C with Kw=1.0×10−14 and
activities ≈ concentrations.
HA(aq) + OH⁻(aq) → A⁻(aq) + H₂O(l)
Key relationships
- Equivalence volume (mL):
Ve = (Ca·Va / Cb) × 1000
- Half-equivalence:
Vb = Ve/2, and
at this point pH = pKa for a monoprotic weak acid.
- Buffer identity (mole form of H–H, dilution-invariant):
\( \mathrm{pH}=\mathrm{p}K_a+\log\!\big(\tfrac{n_{\mathrm{A^-}}}{n_{\mathrm{HA}}}\big) \).
- Conjugate-base basicity:
Kb = Kw / Ka.
How pH is computed in each region
-
Initial weak acid (before any base):
Weak acid dissociation \( \mathrm{HA}\rightleftharpoons \mathrm{H^+}+\mathrm{A^-} \) with
\(K_a=\frac{x^2}{C_a-x}\). The exact solution for
\(x=[\mathrm{H^+}]\) is
\[
[\mathrm{H^+}] = \frac{-K_a + \sqrt{K_a^2 + 4K_a C_a}}{2}\,,
\quad \mathrm{pH}=-\log_{10}[\mathrm{H^+}]\,.
\]
(Common approximation when valid: \( \mathrm{pH}\approx \tfrac12(\mathrm{p}K_a - \log C_a) \).)
-
Buffer region (0 < \(V_b\) < \(V_e\)):
Stoichiometry first (OH⁻ fully consumes HA): \( n_{\mathrm{HA}}=n_{\mathrm{HA0}}-n_{\mathrm{OH^-}} \),
\( n_{\mathrm{A^-}}=n_{\mathrm{OH^-}} \).
Then apply Henderson–Hasselbalch with mole ratio:
\[
\mathrm{pH}=\mathrm{p}K_a+\log\!\left(\frac{n_{\mathrm{A^-}}}{n_{\mathrm{HA}}}\right).
\]
Using moles makes the result independent of the total dilution at that instant.
-
Half-equivalence (\(V_b=V_e/2\)):
\(n_{\mathrm{A^-}}=n_{\mathrm{HA}}\Rightarrow \mathrm{pH}=\mathrm{p}K_a\).
The curve is relatively flat near this point, which is often used to determine \( \mathrm{p}K_a \) experimentally.
-
Equivalence point (\(V_b=V_e\), base added equals initial moles of HA):
Only the conjugate base remains in solution at a salt concentration
\( C_{\text{salt}}=\frac{n_{\mathrm{A^-}}}{V_{\text{tot}}} \).
It hydrolyzes: \( \mathrm{A^-}+ \mathrm{H_2O}\rightleftharpoons \mathrm{HA}+\mathrm{OH^-} \) with
\( K_b=\tfrac{K_w}{K_a} \) and
\[
K_b=\frac{x^2}{C_{\text{salt}}-x}\ \Rightarrow\
[\mathrm{OH^-}]=\frac{-K_b+\sqrt{K_b^2+4K_b C_{\text{salt}}}}{2},\quad
\mathrm{pH}=14-\mathrm{pOH}.
\]
For weak acids, equivalence pH is > 7.
-
After equivalence (\(V_b>V_e\)):
Excess strong base dominates:
\[
[\mathrm{OH^-}]=\frac{n_{\mathrm{OH^-}}-n_{\mathrm{HA0}}}{V_{\text{tot}}},\quad
\mathrm{pH}=14-\mathrm{pOH}.
\]
ICE tables used by the calculator
(A) Initial weak acid dissociation (if needed)
| Species | I | C | E |
|---|
| HA | Ca | −x | Ca − x |
| H⁺ | 0 | +x | x |
| A⁻ | 0 | +x | x |
(B) Equivalence (conjugate base hydrolysis)
| Species | I | C | E |
|---|
| A⁻ | Csalt | −x | Csalt − x |
| HA | 0 | +x | x |
| OH⁻ | 0 | +x | x |
Interpreting the titration curve
- Start pH is higher than a strong acid of the same
Ca due to weak dissociation.
- Buffer plateau between 0 and
Ve, with a gentle slope and maximum buffer capacity near pH≈pKa.
- Equivalence pH > 7 (basic), governed by conjugate-base hydrolysis.
- Post-eq rise asymptotically approaches very basic pH as excess OH⁻ grows.
- The tool highlights halfway, equivalence, and ±1 mL around equivalence for accuracy.
Indicator choice
Since equivalence pH is basic, choose an indicator that changes color above 7 (e.g., phenolphthalein,
transition range roughly 8.2–10.0). Aim for an indicator whose range brackets the steep jump region of your specific curve.
What you enter vs. what is computed
- Inputs:
Ca, Va, Cb, and either Ka or pKa; optional custom base-addition volumes (mL).
- Computed automatically:
Ve, Ve/2, pH at 0 mL, halfway, equivalence, ±1 mL about equivalence, each custom point, and the full smoothed curve.
Algorithmic summary (what the calculator does)
- Convert all volumes to liters for mole accounting; compute \(n_{\mathrm{HA0}}\), \(n_{\mathrm{OH^-}}\), and \(V_{\text{tot}}\).
- Select regime by comparing \(n_{\mathrm{OH^-}}\) to \(n_{\mathrm{HA0}}\):
initial (no base), buffer (<), equivalence (=), or base excess (>).
- Use the appropriate model:
exact quadratic for initial HA, Henderson–Hasselbalch in buffer (mole ratio),
quadratic hydrolysis at equivalence with \(K_b=K_w/K_a\), and strong-base excess after equivalence.
- Generate a dense set of volumes with extra sampling near equivalence; compute and plot pH; highlight key points.
Assumptions & notes
- Monoprotic weak acid; strong, monoprotic base; 25 °C (
Kw=1.0×10−14).
- Activities ≈ concentrations; ionic strength and activity-coefficient effects are neglected.
- All inputs must be positive; volumes in mL, concentrations in M.
- For very small
Ka or extremely dilute solutions, approximations may deviate; the calculator falls back to exact formulas whenever necessary.
Common pitfalls
- Using Henderson–Hasselbalch before establishing the mole balance (always perform the stoichiometric reaction with OH⁻ first).
- Forgetting that at equivalence the solution is not neutral: the conjugate base hydrolyzes, so pH > 7.
- Mixing volume units (mL vs L). The tool handles unit conversion internally; make sure you input mL for additions.
