DAT General Chemistry: Calculating pH at Each Stage of a Titration

DAT High-Yield Concept

Four Distinct Titration Regions: pH calculations change based on position along the titration curve: Initial (weak acid ICE), Buffer region (Henderson-Hasselbalch), Equivalence point (salt hydrolysis), and Post-equivalence (excess strong titrant).
Equivalence Point Basic pH: Titration of a weak acid with a strong base yields a basic equivalence point ( $\text{pH} > 7$ ) because the neutralized conjugate base undergoes hydrolysis ( $\ce{A^- + H2O <=> HA + OH^-}$ ).
Dilution Rule at Equivalence: Salt concentration must be calculated using total solution volume ( $V_\text{total} = V_\text{analyte} + V_\text{titrant}$ ).
5% Small-x Criterion: Simplification of equilibrium expressions is valid when initial concentration divided by $K_\text{a}$ or $K_\text{b}$ is $\ge 400$ .

Selecting the Correct Strategy Across Titration Regions

Solving titration problems on the DAT requires identifying which chemical species predominate at each stage. Neutralization between strong titrants and weak analytes proceeds to completion, altering the primary species present in solution.

Four regions of a titration, four different calculations — A titration curve is not one calculation but four, and the volume of titrant tells you which: a weak base alone needs an ICE table with Kb, the whole stretch before the equivalence point is a buffer and needs only Henderson-Hasselbalch on the mole ratio, the equivalence point leaves only ammonium and needs a fresh ICE table with Ka equal to Kw over Kb, and past it the excess strong acid is simply leftover moles over total volume. That third region is why a weak base titrated with a strong acid always lands below pH 7, at 5.28 here: the flask is left holding a weak acid. Reaching for an ICE table in the buffer region, or for Henderson-Hasselbalch past the equivalence point, is the usual way to lose these points.

Titration Region	Major Species	Governing Equilibrium / Equation	Key Calculation Rule
Initial ( $V_\text{titrant} = \pu{0 mL}$ )	Weak acid $\ce{HA}$	$K_\text{a} = \frac{[\ce{H3O^+}][\ce{A^-}]}{[\ce{HA}]}$	$[\ce{H3O^+}] = \sqrt{K_\text{a}[\ce{HA}]_0}$ if $\frac{[\ce{HA}]_0}{K_\text{a}} \ge 400$
Buffer Region ( $0 < V < V_\text{eq}$ )	$\ce{HA}$ and conjugate base $\ce{A^-}$	$\text{pH} = \text{p}K_\text{a} + \log\left(\frac{n_{\ce{A^-}}}{n_{\ce{HA}}}\right)$	Use molar ratio directly inside logarithm
Equivalence Point ( $V = V_\text{eq}$ )	Conjugate salt $\ce{A^-}$	$K_\text{b} = \frac{K_\text{w}}{K_\text{a}} = \frac{[\ce{OH^-}][\ce{HA}]}{[\ce{A^-}]}$	$[\ce{OH^-}] = \sqrt{K_\text{b}[\ce{A^-}]}$ using total volume
Post-Equivalence ( $V > V_\text{eq}$ )	Excess $\ce{OH^-}$ and $\ce{A^-}$	$[\ce{OH^-}] = \frac{\text{moles excess }\ce{OH^-}}{V_\text{total}}$	Ignore weak hydrolysis of $\ce{A^-}$

Stage-by-Stage pH Calculations for a Weak Acid–Strong Base Titration

Consider the titration of $\pu{50.0 mL}$ of $\pu{0.100 M}$ acetic acid ( $\ce{HC2H3O2}$ , $K_\text{a} = \pu{1.8 \times 10^{-5}}$ , $\text{p}K_\text{a} = 4.74$ ) with $\pu{0.100 M}$ sodium hydroxide ( $\ce{NaOH}$ ).

Core Principle: Strong base reacts completely with weak acid:

\ce{HC2H3O2 + OH^- -> C2H3O2^- + H2O}

. Stoichiometry determines remaining moles, which dictates the appropriate calculation method.

Region 1: Initial Point (0 mL Titrant Added)

Before titrant is added, solution pH depends solely on weak acid equilibrium:

\ce{HC2H3O2 + H2O <=> C2H3O2^- + H3O^+}

Check the 5% approximation criterion: $\frac{[\ce{HC2H3O2}]_0}{K_\text{a}} = \frac{0.100}{1.8 \times 10^{-5}} = 5556 \ge 400$ . Thus, small- $x$ applies:

[\ce{H3O^+}] = \sqrt{K_\text{a}[\ce{HC2H3O2}]_0} = \sqrt{(1.8 \times 10^{-5})(0.100)} = \pu{1.34 \times 10^{-3} M}

\text{pH} = -\log(1.34 \times 10^{-3}) = 2.87

Region 2: Buffer Region (e.g., 20.0 mL NaOH Added)

Initial moles of acid: $\pu{0.0500 L} \times \pu{0.100 M} = \pu{0.00500 mol}$ . Moles of $\ce{OH^-}$ added: $\pu{0.0200 L} \times \pu{0.100 M} = \pu{0.00200 mol}$ . Neutralization produces a conjugate buffer system:

Remaining $\ce{HC2H3O2}$ : $\pu{0.00500 mol} - \pu{0.00200 mol} = \pu{0.00300 mol}$
Formed $\ce{C2H3O2^-}$ : $\pu{0.00200 mol}$

Applying the Henderson-Hasselbalch equation (note: at $\pu{25.0 mL}$ , the half-equivalence point gives $\text{pH} = \text{p}K_\text{a}$ , covered in depth in its own lesson):

\text{pH} = \text{p}K_\text{a} + \log\left(\frac{n_{\ce{C2H3O2^-}}}{n_{\ce{HC2H3O2}}}\right) = 4.74 + \log\left(\frac{0.00200}{0.00300}\right) = 4.74 - 0.18 = 4.56

DAT Exam Trap: Do not waste time converting moles back to molarity in the logarithm; total volume cancels out.

Region 3: Equivalence Point (50.0 mL NaOH Added)

At equivalence, $\pu{0.00500 mol}$ $\ce{OH^-}$ fully converts $\ce{HC2H3O2}$ to $\ce{C2H3O2^-}$ . Total volume is $\pu{50.0 mL} + \pu{50.0 mL} = \pu{100.0 mL} = \pu{0.1000 L}$ .

[\ce{C2H3O2^-}] = \frac{\pu{0.00500 mol}}{\pu{0.1000 L}} = \pu{0.0500 M}

The conjugate base undergoes salt hydrolysis:

\ce{C2H3O2^- + H2O <=> HC2H3O2 + OH^-}

K_\text{b} = \frac{K_\text{w}}{K_\text{a}} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = \pu{5.56 \times 10^{-10}}

Check rule: $\frac{[\ce{C2H3O2^-}]}{K_\text{b}} = \frac{0.0500}{5.56 \times 10^{-10}} = 9.0 \times 10^7 \ge 400$ :

[\ce{OH^-}] = \sqrt{K_\text{b}[\ce{C2H3O2^-}]} = \sqrt{(5.56 \times 10^{-10})(0.0500)} = \pu{5.27 \times 10^{-6} M}

\text{pOH} = -\log(5.27 \times 10^{-6}) = 5.28 \implies \text{pH} = 14.00 - 5.28 = 8.72

DAT Exam Trap: Forgetting to calculate total volume (dilution) is the most common mistake. Always recalculate salt concentration before evaluating $K_\text{b}$ .

Region 4: Post-Equivalence Region (e.g., 60.0 mL NaOH Added)

Addition of $\pu{60.0 mL}$ $\ce{NaOH}$ ( $\pu{0.00600 mol}$ $\ce{OH^-}$ ) yields excess base: $\pu{0.00600 mol} - \pu{0.00500 mol} = \pu{0.00100 mol}$ $\ce{OH^-}$ . Total volume is $\pu{110.0 mL} = \pu{0.1100 L}$ .

[\ce{OH^-}] = \frac{\pu{0.00100 mol}}{\pu{0.1100 L}} = \pu{9.09 \times 10^{-3} M}

\text{pOH} = -\log(9.09 \times 10^{-3}) = 2.04 \implies \text{pH} = 14.00 - 2.04 = 11.96

Mnemonic — I-B-E-P for Titration Regions:
• Initial: ICE table for weak acid.
• Buffer: Buffer equation (Henderson-Hasselbalch).
• Equivalence: Equilibrium hydrolysis of conjugate salt.
• Post-equivalence: POH from excess strong titrant.

DAT Critical Thinking & Application Questions

When titrating $\pu{25.0 mL}$ of $\pu{0.200 M}$ weak acid $\ce{HA}$ ( $K_\text{a} = \pu{1.0 \times 10^{-5}}$ ) with $\pu{0.100 M}$ $\ce{NaOH}$ , why is the equivalence-point volume $\pu{50.0 mL}$ , and what is the salt concentration?
Why does a weak acid titrated with a strong base give a basic equivalence point ( $\text{pH} > 7$ ), whereas a strong acid titrated with a strong base yields $\text{pH} = 7$ ?
In the post-equivalence region, why can we neglect hydroxide ions produced from conjugate salt hydrolysis when calculating pH?

References

Sources returned by Google Search grounding during lesson verification.

Weak Acid-Strong Base Titrations | 17.3b General Chemistry — youtube.com
How to Calculate pH at Equivalence Point - Chemistry Guru — chemistryguru.com.sg

Calculating pH at Each Stage of a Titration

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