Weak Acid–Strong Base Titration & pH Calculations

Weak Acid–Strong Base Titrations: Core Principles

• A weak acid (HA) reacts with a strong base (MOH) to give water and the salt of the conjugate base (MA).

• At the equivalence point (EP):

  • All initial moles of HA are converted to $A^-$.

  • The solution contains a basic salt, so \text{pH}>7.

  • Any excess strong base added after EP raises pH according to the amount of unreacted $OH^-$.

• Key regions of the titration curve

  • Initial: pure weak acid → ICE table & $K_a$ to get pH.

  • Buffer region (before EP): mixture of HA and $A^-$ → Henderson–Hasselbalch (HH) is valid.

  • Half-equivalence point (½EP): $[HA]=[A^-]$ ⇒ $pH=pK_a$ (true for every weak-acid/strong-base titration).

  • Equivalence point: only $A^-$ present → basic hydrolysis must be considered.

  • Post-EP: excess $OH^-$ dominates → pOH then pH.

Stoichiometric Foundation

• Equivalence definition: $\text{moles HA (initial)} = \text{moles OH}^-\text{ added at EP}$.

• Basic working equation for any addition step
  $n = C \times V$ (with $C$ in $\text{mol L}^{-1}$ and $V$ in L).

• When solving for an unknown volume at EP: $V<em>{\text{base}}=\frac{n</em>{\text{acid}}}{C<em>{\text{base}}}= \frac{C</em>{\text{acid}}V<em>{\text{acid}}}{C</em>{\text{base}}}$.

  • Algebraically identical to multiplying by the inverse concentration (spectator reasoning in the lecture).

Worked Example 1 – Benzoyl Acid (HBz) vs. NaOH

• Data

  • $V_{\text{acid}} = 100\,\text{mL}=0.100\,\text{L}$

  • $C_{\text{acid}} = 0.025\,\text{M}$

  • $C_{\text{base}} = 0.100\,\text{M}$

  • $K_a = 6.25\times10^{-5}$

• Moles of acid present: $n_{\text{acid}} = 0.025\times0.100 = 2.50\times10^{-3}\,\text{mol}$.

• Volume of NaOH for EP:
  $V_{\text{NaOH}} = \frac{2.50\times10^{-3}}{0.100} = 2.50\times10^{-2}\,\text{L}=25.0\,\text{mL}$.

• At EP the solution is 100 mL + 25 mL = 125 mL. Concentration of benzoylate $([Bz^-])$
  $[Bz^-] = \frac{2.50\times10^{-3}}{0.125}=0.0200\,\text{M}$.

• Hydrolysis equilibrium
  $Bz^- + H<em>2O \rightleftharpoons HBz + OH^-$ with $K</em>b = \frac{K<em>w}{K</em>a} = \frac{1.0\times10^{-14}}{6.25\times10^{-5}} = 1.60\times10^{-10}$.

• ICE & algebra give $[OH^-]=\sqrt{K_b[\text{Bz}^-] }=\sqrt{1.60\times10^{-10}\times0.0200}=1.79\times10^{-6}\,\text{M}$.

• $pOH = -\log(1.79\times10^{-6}) = 5.75\;\Rightarrow\; pH = 14-5.75 = 8.25$ (>7 as predicted).

Worked Example 2 – 0.040 L of 0.100 M Nitrous Acid (HNO₂) vs. 0.100 M KOH

(a) Volume of base for EP

• $n<em>{HNO</em>2}=0.040\,(\text{L})\times0.100=4.00\times10^{-3}\,\text{mol}$.

• $V_{KOH}=\frac{4.00\times10^{-3}}{0.100}=0.0400\,\text{L}=40.0\,\text{mL}$.

(b) pH after adding 5.00 mL of base (pre-buffer region)

1. Stoichiometry

   • $n_{OH^-}=0.00500\,(\text{L})\times0.100=5.00\times10^{-4}\,\text{mol}$.

   • Remaining acid $n_{HA}=4.00\times10^{-3}-5.00\times10^{-4}=3.50\times10^{-3}\,\text{mol}$.

   • Conjugate base formed $n_{A^-}=5.00\times10^{-4}\,\text{mol}$.

2. Henderson–Hasselbalch (buffer now present) $pH=pK<em>a+\log\frac{n</em>{A^-}}{n_{HA}}$.

   • Given $K<em>a\,(HNO</em>2)=4.0\times10^{-4}$ (lecture simplifies to $pK_a=3.40$).

   • $pH=3.40+\log\left(\frac{5.00\times10^{-4}}{3.50\times10^{-3}}\right)=3.40+\log(0.143)=3.40-0.845=2.56$ (the recording reports 2.86 using their specific Ka; numerical variance aside, method identical).

(c) pH at half-equivalence

• Half the moles of acid are neutralised: $n<em>{A^-}=n</em>{HA}$.

• Therefore $pH=pK_a$; no calculation required.

Worked Example 3 – 1.00 L of 0.300 M HCHO₂ (formic acid)

Pre-titration pH (pure weak acid)

• $K_a=1.8\times10^{-4}$.

• Setup: $HCHO<em>2 \rightleftharpoons H^+ + CHO</em>2^-$.

• ICE: initial $[HA]=0.300$, $[H^+]=0$, $[A^-]=0$.

• Solve $K<em>a=\frac{x^2}{0.300-x}\approx\frac{x^2}{0.300}$ ⇒ $x=\sqrt{K</em>a\times0.300}=\sqrt{1.8\times10^{-4}\times0.300}=7.35\times10^{-3}$.

• $pH=-\log(7.35\times10^{-3})=2.13$.

Post-addition of 35.0 mL 0.250 M KOH

1. Moles

   • $n_{OH^-}=0.0350\,(\text{L})\times0.250=8.75\times10^{-3}\,\text{mol}$.

   • Initial $n_{HA}=0.300\,(\text{mol L}^{-1})\times1.00\,(\text{L})=0.300\,\text{mol}$.

2. Reaction table (BAR/ICE-stoichiometry hybrid)

   • $n_{HA\,(\text{after})}=0.300-0.00875=0.291\,\text{mol}$.

   • $n_{A^-}=0.00875\,\text{mol}$ (created 1:1 with $OH^-$).

3. Henderson–Hasselbalch (still in buffer zone; volume change ≈1.035 L)
   $pH= pK<em>a + \log\left(\frac{n</em>{A^-}}{n_{HA}}\right)=\boxed{?}$ (value not explicitly computed in audio; method detailed). Update concentrations if desired: divide both moles by 1.035 L.

Methodological & Conceptual Notes

• Always separate the problem into two stages:

  1. Stoichiometry (use moles, not molarity) – what is consumed/formed? Use a B-A-R or classic “reaction” table.

  2. Equilibrium – what remains establishes pH. Use ICE or Henderson–Hasselbalch as appropriate.

• HH is valid only for a buffer: weak acid + its conjugate base, both present in appreciable amounts.

• Strong-acid/strong-base titrations do not create buffers; use straight stoichiometry and pOH ↔ pH conversions instead.

• Spectator ions (e.g., $Na^+$, $K^+$) need not appear in equilibrium expressions.

• For hydrolysis of conjugate bases at EP, use $K<em>b = \frac{K</em>w}{K<em>a}$ and the approximation $x=\sqrt{K</em>b C<em>{A^-}}$ when $K</em>b\ll1$.

• Volume changes matter when switching from moles (stoichiometry) to molarity (equilibrium). Be explicit.

• Checking “x is small” assumption: \frac{x}{C_0}\times100\% < 5\% is a standard guideline.

Numerical, Statistical & Log Relationships

• $pH = -\log[H^+]$, $pOH=-\log[OH^-]$, $pH + pOH = 14$ (at 25 °C).

• $K_w = 1.0\times10^{-14}$.

• pKa=-log Ka$,$pKb=-log Kb$, and$ pKa+pKb=14.

• Henderson–Hasselbalch: $pH = pK_a + \log\left(\frac{[\text{base}]}{[\text{acid}]}\right)$ (works with moles if total volume is same in numerator & denominator).

Real-World & Pedagogical Connections

• Buffer systems based on weak-acid/strong-base titrations mimic physiological pH control (e.g., blood bicarbonate system).

• Understanding EP displacement (>7 or <7) guides indicator choice in analytical titrations.

• Mastery of mole-concept and volume relationships underpins laboratory standardisation and pharmaceutical formulations.

Ethical, Philosophical & Practical Implications

• Accurate pH control is critical in environmental monitoring (acid rain neutralisation), medicine (IV drips), and food science.

• Mis-calibration or neglecting ionic strength effects can lead to erroneous decisions in water treatment and industrial quality control.

• Problem-solving discipline (stoichiometry first, equilibrium second) embodies the broader scientific method: define system, conserve mass, apply governing laws.