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IB DP Chemistry – R3.1.13 pH curves of Acid-Base combinations- Study Notes

IB DP Chemistry - R3.1.13 pH curves of Acid-Base combinations- Study Notes - New Syllabus - 2026, 2027 & 2028

IB DP Chemistry – R3.1.13 pH curves of Acid-Base combinations – Study Notes – New Syllabus

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Reactivity 3.1.13 — pH Curves of Different Combinations of Monoprotic Acids and Bases

Reactivity 3.1.13 — pH Curves of Different Combinations of Monoprotic Acids and Bases 

What Are pH Curves?

  • pH curves (also called titration curves) show how the pH of a solution changes as a titrant (acid or base) is added gradually to a fixed volume of analyte.
  • The overall shape of the curve depends on whether the acid and base are strong or weak and whether acid is added to base or vice versa.
  • This topic only considers monoprotic acids and bases, which donate or accept only one proton per molecule (e.g., HCl, NaOH, CH3COOH, NH3).

   

Standard Assumptions for All pH Curves:

  • Titration is carried out at room temperature (25°C).
  • Equal molar concentrations of acid and base are assumed (typically 0.1 mol dm−3).
  • The volume of analyte used is usually 25.0 cm3, and titrant is added from a burette dropwise.
  • No side reactions occur; neutralization is the dominant reaction.

Factors Determining the Shape of a pH Curve:

  • Strength of the acid: Strong acids dissociate completely, giving very low initial pH. Weak acids partially dissociate, leading to a higher initial pH.
  • Strength of the base: Strong bases dissociate completely, yielding very high final pH. Weak bases accept protons only partially, resulting in a lower final pH.
  • Buffer regions: These appear only when weak acids or weak bases are involved, due to the presence of conjugate acid–base pairs.
  • Equivalence point: Occurs when moles of acid equal moles of base. Its pH depends on the strengths of the acid and base.
  • Gradient near equivalence: A sharp rise (steep vertical section) occurs for titrations involving at least one strong species.

Key Points:

  • Each pH curve reflects the combination of strengths of the acid and base involved.
  • Strong–strong combinations give steep and symmetrical curves centered at pH 7.
  • Weak–strong and strong–weak combinations produce asymmetrical curves with buffer regions and shifted equivalence points.
  • Weak–weak titrations result in shallow curves with less distinct equivalence points, often requiring a pH meter for accuracy.

Interpretation of pH Curves for Strong and Weak Monoprotic Acids and Bases

Essential Concepts in Interpreting pH Curves:

  • Initial pH (y-intercept): Reflects the strength and concentration of the acid or base at the start of titration.
  • Buffer Region: Appears when a weak acid and its conjugate base (or weak base and its conjugate acid) coexist. This region resists pH change and appears as a flattened section of the curve.
  • Half-equivalence Point: Occurs where half the acid or base has reacted. At this point:
    • For weak acid: \( \text{pH} = \text{p}K_a \)
    • For weak base: \( \text{pOH} = \text{p}K_b \)
  • Equivalence Point: Where moles of acid = moles of base. The pH depends on the nature of the ions present in the solution.
  • Post-equivalence: Determined by the excess strong acid or base; pH rises or falls steeply after equivalence.

Interpretation of pH Curves Based on Acid–Base Combinations:

1. Strong Acid + Strong Base (e.g., HCl + NaOH)

  • Initial pH: Very low (~1) due to full dissociation of the strong acid.
  • Buffer Region: Absent — no weak conjugate pairs present.
  • Equivalence Point: pH = 7 — solution contains only neutral ions (e.g., Na+ and Cl–).
  • Curve Shape: Very steep near equivalence (rapid pH rise from ~3 to ~11 in ~1 cm3 volume).
  • Indicator: Phenolphthalein or methyl orange — both are suitable due to sharp pH jump from acidic to basic.

 

2. Weak Acid + Strong Base (e.g., CH3COOH + NaOH)

  • Initial pH: Moderate (~3–4) due to partial dissociation of the weak acid.
  • Buffer Region: Clearly visible — from ~25% to ~75% neutralization.
  • Half-equivalence Point: \( \text{pH} = \text{p}K_a \) — ideal for determining \( K_a \) experimentally.
  • Equivalence Point: pH > 7 (~8.5) — conjugate base hydrolyzes to produce OH–.
  • Curve Shape: Gentle rise through buffer region, moderate vertical section at equivalence.
  • Indicator: Phenolphthalein — changes color around pH 8.3–10, matching the equivalence range.

 

 

 

3. Strong Acid + Weak Base (e.g., HCl + NH3)

  • Initial pH: Very low (~1) due to strong acid in excess.
  • Buffer Region: Appears as NH3 neutralizes acid to form NH4+.
  • Half-equivalence Point: \( \text{pOH} = \text{p}K_b \); use to determine base strength experimentally.
  • Equivalence Point: pH < 7 (~5.5) — conjugate acid (NH4+) hydrolyzes to release H+.
  • Curve Shape: Gradual slope through buffer, smaller pH jump at equivalence.
  • Indicator: Methyl orange — color change occurs around pH 3.1–4.4, ideal for detecting equivalence.

 

4. Weak Acid + Weak Base (e.g., CH3COOH + NH3)

  • Initial pH: Moderate (~4) — weak acid partially dissociates.
  • Buffer Regions: Present on both sides of equivalence — conjugate pairs coexist before and after the midpoint.
  • Half-equivalence Point: Occurs for both acid and base, depending on direction of titration.
  • Equivalence Point: Depends on relative strength:
    • If \( K_a = K_b \): pH = 7
    • If \( K_a > K_b \): pH < 7
    • If \( K_b > K_a \): pH > 7
  • Curve Shape: Very shallow — no sharp vertical rise; changes are gradual and smooth.
  • Indicator: No reliable indicator — use a pH meter instead.

Comparison Table: pH Curve Interpretation

CombinationInitial pHBuffer RegionpH = pKa / pOH = pKbEquivalence pHBest Indicator
Strong Acid + Strong Base
(e.g., HCl + NaOH)		~1NoneNot applicable7Methyl orange / Phenolphthalein
Weak Acid + Strong Base
(e.g., CH3COOH + NaOH)		~3–4Yes (before equivalence)pH = pKa>7 (~8.5)Phenolphthalein
Strong Acid + Weak Base
(e.g., HCl + NH3)		~1Yes (before equivalence)pOH = pKb<7 (~5.5)Methyl orange
Weak Acid + Weak Base
(e.g., CH3COOH + NH3)		~3–4Yes (both sides)Direction dependentVariable (5–9)None (use pH meter)

Example

25.0 cm3 of 0.100 mol dm–3 ethanoic acid (CH3COOH, a weak monoprotic acid, \( K_a = 1.8 \times 10^{–5} \)) is titrated with 0.100 mol dm–3 aqueous ammonia (NH3, a weak base, \( K_b = 1.8 \times 10^{–5} \)). Predict the general shape of the pH curve and identify:

  • Initial pH
  • Presence and nature of buffer regions
  • Volume at equivalence point
  • pH at equivalence point
  • Appropriate indicator
▶️Answer/Explanation

Step 1: Initial pH (Before any base added)

Use \( K_a \) to find pH of weak acid:
\( K_a = \dfrac{[H^+]^2}{[HA]} \Rightarrow [H^+] = \sqrt{K_a \cdot [HA]} = \sqrt{1.8 \times 10^{–5} \cdot 0.100} = 1.34 \times 10^{–3} \)
\( \text{pH} = -\log(1.34 \times 10^{–3}) \approx 2.87 \)

Step 2: Buffer Region Before Equivalence

As NH3 is added, CH3COOH reacts to form CH3COO–, forming a buffer solution. The reaction: CH3COOH + NH3 ⇌ CH3COO– + NH4+
Buffer region appears from ~5 cm3 to ~20 cm3 of NH3 added.

Step 3: Half-Equivalence Point

Occurs when 12.5 cm3 NH3 has been added (half the acid neutralized). At this point: \( [\text{CH}_3\text{COOH}] = [\text{CH}_3\text{COO}^-] \), so \( \text{pH} = \text{p}K_a = 4.74 \)

Step 4: Equivalence Point Volume

Moles of CH3COOH = \( 0.025 \times 0.100 = 2.50 \times 10^{–3} \)
Since [NH3] = 0.100 mol dm–3, volume required = \( \dfrac{2.50 \times 10^{–3}}{0.100} = 25.0 \, \text{cm}^3 \)

Step 5: pH at Equivalence Point

At this point, solution contains CH3COO– and NH4+. Both undergo hydrolysis.
Since \( K_a = K_b \), the hydrolysis of conjugate acid and base contributes equally.
Therefore: \( \text{pH} \approx 7 \) (though slightly offset depending on temperature and ionic strength).

Step 6: Curve Shape and Indicator

The pH curve is shallow at the start, has a gentle buffer region, and no sharp vertical rise at equivalence.
Because of the lack of steepness, no indicator is ideal. A pH meter should be used instead.

Example

A student titrates 20.0 mL of 0.060 mol dm−3 ammonia (NH₃), a weak base, with 0.060 mol dm−3 perchloric acid (HClO₄), a strong acid. Given: \( K_b \, \text{(NH}_3) = 1.8 \times 10^{-5} \)

  1. Identify whether the acid and base are strong or weak.
  2. Determine if the acid:base mole ratio is 1:1 or otherwise.
  3. Calculate the initial pH before titration begins.
  4. Calculate the volume of HClO₄ required to reach the equivalence point.
  5. Identify the major species at the equivalence point and whether the pH is greater than, less than, or equal to 7.
  6. On the pH curve:
    1. Label the axes
    2. Sketch the general shape of the curve
    3. Identify and label the buffering region
    4. Mark the half-equivalence point and give its pH
    5. Mark the equivalence point with an X and approximate its pH
▶️Answer/Explanation
  1. Strength of acid and base:
    Ammonia (NH₃) is a weak base and perchloric acid (HClO₄) is a strong acid.
  2. Mole ratio:
    The reaction between NH₃ and HClO₄ proceeds in a 1:1 mole ratio:
    \( \text{NH}_3 + \text{HClO}_4 \rightarrow \text{NH}_4^+ + \text{ClO}_4^- \)
  3. Initial pH of NH₃ before titration:
    Use \( K_b = \frac{x^2}{[NH_3]} \Rightarrow 1.8 \times 10^{-5} = \frac{x^2}{0.060} \)
    Solving: \( x = 1.0 \times 10^{-3} \Rightarrow [OH^-] = 1.0 \times 10^{-3} \)
    \( \text{pOH} = -\log(1.0 \times 10^{-3}) = 3 \Rightarrow \text{pH} = 14 – 3 = \mathbf{11.0} \)
  4. Volume of acid to reach equivalence:
    Moles NH₃ = \( 0.060 \times 0.020 = 1.2 \times 10^{-3} \, \text{mol} \)
    Required moles of HClO₄ = same
    Volume = \( \frac{1.2 \times 10^{-3}}{0.060} = 0.020 \, \text{L} = \mathbf{20.0 \, mL} \)
  5. Species at equivalence point and pH:
    At equivalence, NH₃ is fully converted into \( \text{NH}_4^+ \), a weak acid.
    The solution contains \( \text{NH}_4^+ \) and \( \text{ClO}_4^- \) (a neutral conjugate base).
    The pH is therefore less than 7.
  6. Graph and pH curve interpretation:

    1. x-axis = Volume of base added (mL), y-axis = pH
    2. Curve starts, gently slopes down, shows a buffer region, then drops sharply at 20 mL and levels around pH ≈ 2
    3. Buffering occurs from start to ~10 mL (half-equivalence point)
    4. At half-equivalence, pH = pKa of NH₄⁺
      \( K_b = 1.8 \times 10^{-5} \Rightarrow K_a = \frac{10^{-14}}{K_b} = 5.6 \times 10^{-10} \)
      \( \text{pKa} = -\log(5.6 \times 10^{-10}) ≈ \mathbf{9.25} \)
    5. At 20.0 mL (equivalence point), pH ≈ 5.25 (acidic)
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