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Acids and Bases - pH curves (HL)

Grade 11IBChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The pH curve for a strong acid-strong base titration is characterized by a very large vertical region (the pH jump) around the equivalence point. The equivalence point occurs exactly at pH=7pH = 7 at 298K298 K because the salt formed (e.g., NaClNaCl) does not undergo hydrolysis.

pH curve of a strong acid vs strong base titration showing the equivalence point at pH 7.
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In a weak acid-strong base titration, the curve starts at a higher pH. It features a 'buffer region' where the pH changes slowly despite the addition of base. The half-equivalence point is a critical feature where [HA]=[A−][HA] = [A^-] and pH=pKapH = pK_a. The equivalence point occurs at pH>7pH > 7 due to the hydrolysis of the conjugate base.

pH curve of a weak acid vs strong base titration highlighting the pKa at the half-equivalence point.
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For a weak base-strong acid titration, the curve starts at a high pH (around 1111). The pH drops slowly in the buffer region. The equivalence point occurs at pH<7pH < 7 because the conjugate acid of the weak base (e.g., NH4+NH_4^+) hydrolyzes to produce H+H^+ ions.

pH curve of a weak base vs strong acid titration showing the equivalence point below pH 7.
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The choice of indicator is determined by the pH range of the vertical section of the titration curve. An indicator is suitable if its pKin±1pK_{in} \pm 1 range falls within the steep vertical portion of the curve. For example, Phenolphthalein (range 8.3−10.08.3 - 10.0) is ideal for weak acid-strong base titrations.

📐Formulae

pH=−log⁡10[H+]pH = -\log_{10}[H^+]

pOH=−log⁡10[OH−]pOH = -\log_{10}[OH^-]

pH+pOH=pKw=14.00 at 298KpH + pOH = pK_w = 14.00 \text{ at } 298K

pH=pKa+log⁡10([A−][HA])pH = pK_a + \log_{10}\left(\frac{[A^-]}{[HA]}\right) (Henderson-Hasselbalch Equation)

pKa=−log⁡10KapK_a = -\log_{10}K_a

Kw=Ka×KbK_w = K_a \times K_b

💡Examples

Problem 1:

Calculate the pHpH at the half-equivalence point when 25.0 cm325.0\,cm^3 of 0.100 mol dm−30.100\,mol\,dm^{-3} CH3COOHCH_3COOH is titrated with 0.100 mol dm−30.100\,mol\,dm^{-3} NaOHNaOH. The acid dissociation constant for ethanoic acid is Ka=1.8×10−5K_a = 1.8 \times 10^{-5} at 298K298K.

Solution:

pH=pKa=−log⁡10(1.8×10−5)≈4.74pH = pK_a = -\log_{10}(1.8 \times 10^{-5}) \approx 4.74

Explanation:

At the half-equivalence point, exactly half of the CH3COOHCH_3COOH has reacted to form CH3COO−CH_3COO^-. Thus, [CH3COOH]=[CH3COO−][CH_3COOH] = [CH_3COO^-]. Substituting these equal values into the Henderson-Hasselbalch equation makes the log term log⁡10(1)=0\log_{10}(1) = 0, leaving pH=pKapH = pK_a.

Problem 2:

Explain why the pHpH at the equivalence point for the titration of NH3(aq)NH_3(aq) with HCl(aq)HCl(aq) is less than 77.

Solution:

The salt formed is NH4ClNH_4Cl. The NH4+NH_4^+ ion is the conjugate acid of a weak base and undergoes hydrolysis: NH4+(aq)+H2O(l)⇌NH3(aq)+H3O+(aq)NH_4^+(aq) + H_2O(l) \rightleftharpoons NH_3(aq) + H_3O^+(aq).

Explanation:

Since the hydrolysis of the ammonium ion produces hydronium ions (H3O+H_3O^+), the concentration of [H+][H^+] increases, resulting in an acidic pHpH at the stoichiometric equivalence point.

Problem 3:

Identify the type of titration represented by the provided pH curve where the initial pHpH is 11.1311.13, the half-equivalence point occurs at pH=9.25pH = 9.25 after adding 10.0cm310.0 cm^3 of titrant, and the equivalence point is reached at pH=5.28pH = 5.28 with 20.0cm320.0 cm^3 of titrant.

Graph of a weak base being titrated by a strong acid.

Solution:

  1. Initial pH=11.13pH = 11.13 indicates a weak base.
  2. The titrant added is an acid since the pHpH decreases.
  3. The equivalence point pH=5.28pH = 5.28 (less than 77) confirms it is a weak base-strong acid titration.
  4. At 10.0cm310.0 cm^3 (half-equivalence), pOH=pKbpOH = pK_b. Since pH=9.25pH = 9.25, pOH=14−9.25=4.75pOH = 14 - 9.25 = 4.75. Thus, pKb=4.75pK_b = 4.75.

Explanation:

The shape shows a buffer region starting from a basic pH and a steep drop through the acidic range, characteristic of NH3NH_3 titrated with HClHCl.

Problem 4:

A 25.0cm325.0 cm^3 sample of 0.10mol⋅dm−30.10 mol \cdot dm^{-3} methanoic acid (HCOOHHCOOH, pKa=3.75pK_a = 3.75) is titrated with 0.10mol⋅dm−30.10 mol \cdot dm^{-3} KOHKOH. Sketch the curve and calculate the pHpH after 12.5cm312.5 cm^3 and 25.0cm325.0 cm^3 of KOHKOH have been added.

pH curve for methanoic acid titration showing points at pH 3.75 and 8.22.

Solution:

  1. At V=12.5cm3V = 12.5 cm^3: This is the half-equivalence point (25.0/225.0 / 2). At this point, pH=pKa=3.75pH = pK_a = 3.75.
  2. At V=25.0cm3V = 25.0 cm^3: This is the equivalence point. All HCOOHHCOOH is converted to HCOO−HCOO^-. The concentration of HCOO−HCOO^- is 0.05mol⋅dm−30.05 mol \cdot dm^{-3} (due to doubling the volume).
  3. pOH=12(pKb−log⁡[HCOO−])pOH = \frac{1}{2}(pK_b - \log[HCOO^-]). Since pKb=14−3.75=10.25pK_b = 14 - 3.75 = 10.25, pOH=12(10.25−log⁡(0.05))≈5.78pOH = \frac{1}{2}(10.25 - \log(0.05)) \approx 5.78.
  4. pH=14−5.78=8.22pH = 14 - 5.78 = 8.22.

Explanation:

The curve starts at a low (but not very low) pH, rises through a buffer region to the half-equivalence point, and has an equivalence point in the basic region.