Module 6: Acid/Base Reactions

Master proton transfer reactions, pH calculations, titration techniques, and buffer chemistry with clear theory and quantitative problem-solving strategies.

Pillar 1 of 3

Acid-Base Theory

Explore the fundamental theories of acids and bases, including Arrhenius and Brønsted-Lowry models, conjugate acid-base pairs, and amphiprotic substances that can act as both acids and bases.

Acid Models & Definitions

Different models of acids and bases have been developed to explain chemical behavior. The two most important for HSC Chemistry are the Arrhenius model and the Brønsted-Lowry model, each with different scopes and applications.

Arrhenius Theory

The Arrhenius theory (1884) was the first systematic definition of acids and bases. It defines acids and bases in terms of ions produced in water.

Arrhenius acid: A substance that produces hydrogen ions (H⁺) when dissolved in water. Example: HCl dissociates to form H⁺ and Cl⁻ ions:

\mathrm{HCl}{\text{(aq)}} \rightarrow \mathrm{H}^{+}{\text{(aq)}} + \mathrm{Cl}^{-}{\text{(aq)}}

Arrhenius base: A substance that produces hydroxide ions (OH⁻) when dissolved in water. Example: NaOH dissociates to form Na⁺ and OH⁻ ions:

\mathrm{NaOH}{\text{(aq)}} \rightarrow \mathrm{Na}^{+}{\text{(aq)}} + \mathrm{OH}^{-}{\text{(aq)}}

Limitations: The Arrhenius model only applies to aqueous solutions and cannot explain bases that don't contain OH⁻ (like ammonia, NH₃) or acid-base reactions that occur in non-aqueous solvents or the gas phase.

Brønsted-Lowry Theory

The Brønsted-Lowry theory (1923) provides a broader definition based on proton transfer. This model works in any solvent and explains a wider range of acid-base behavior.

Brønsted-Lowry acid: A proton (H⁺) donor. An acid donates a proton to another species during a reaction.

Brønsted-Lowry base: A proton (H⁺) acceptor. A base accepts a proton from another species during a reaction.

Example - HCl and water: When HCl dissolves in water, HCl acts as an acid (donates H⁺) and water acts as a base (accepts H⁺):

\mathrm{HCl}{\text{(aq)}} + \mathrm{H_2O}{\text{(l)}} \rightarrow \mathrm{H_3O}^{+}{\text{(aq)}} + \mathrm{Cl}^{-}{\text{(aq)}}

HCl donates a proton (acid), H₂O accepts the proton (base), forming hydronium ion (H₃O⁺) and chloride ion (Cl⁻).

Example - Ammonia and water: Ammonia acts as a base by accepting a proton from water:

\mathrm{NH_3}{\text{(aq)}} + \mathrm{H_2O}{\text{(l)}} \rightleftharpoons \mathrm{NH_4}^{+}{\text{(aq)}} + \mathrm{OH}^{-}{\text{(aq)}}

NH₃ accepts a proton (base), H₂O donates a proton (acid). This explains why ammonia is a base even though it doesn't contain OH⁻ ions—the Arrhenius model couldn't explain this.

Advantages: The Brønsted-Lowry model works in any solvent, explains ammonia and other bases without OH⁻, describes gas-phase reactions, and introduces the concept of conjugate acid-base pairs (covered in the next section).

Comparing the Models

Scope: Arrhenius is limited to aqueous solutions producing H⁺ or OH⁻. Brønsted-Lowry applies to any proton transfer in any phase or solvent.

Water's role: In Arrhenius theory, water is just the solvent. In Brønsted-Lowry theory, water can act as either an acid or a base (amphoteric behavior).

Neutralization: Arrhenius views neutralization as H⁺ + OH⁻ → H₂O. Brønsted-Lowry views it as proton transfer between acid and base, which may or may not produce water.

When to use which: Use Arrhenius for simple aqueous strong acids/bases (HCl, NaOH). Use Brønsted-Lowry for weak acids/bases, reactions in non-aqueous solvents, identifying conjugate pairs, and understanding amphiprotic behavior.

Worked Example

For the reaction $\mathrm{HNO_3}{\text{(aq)}} + \mathrm{H_2O}{\text{(l)}} \rightarrow \mathrm{H_3O}^{+}{\text{(aq)}} + \mathrm{NO_3}^{-}{\text{(aq)}}$ , identify: (a) the Arrhenius acid, (b) the Brønsted-Lowry acid and base, (c) explain why this reaction demonstrates the advantage of the Brønsted-Lowry model.

(a) Arrhenius acid:

HNO₃ is the Arrhenius acid because it produces H⁺ ions when dissolved in water.

We can think of the H₃O⁺ as H⁺ (hydrogen ion) in solution, which fits the Arrhenius definition.

(b) Brønsted-Lowry identification:

Acid: HNO₃ is the Brønsted-Lowry acid because it donates a proton (H⁺) to water.

Base: H₂O is the Brønsted-Lowry base because it accepts a proton from HNO₃, forming H₃O⁺.

The Brønsted-Lowry model reveals that water actively participates as a base in this reaction, not just as an inert solvent. The Arrhenius model treats water only as the medium where H⁺ ions appear, missing water's chemical role in accepting the proton.

The Brønsted-Lowry model also allows us to identify conjugate pairs and understand that the same molecule (H₂O) can act as either an acid or a base depending on what it reacts with—this amphoteric nature cannot be explained by Arrhenius theory.

Practice Question

Explain why ammonia (NH₃) is considered a base according to Brønsted-Lowry theory but not according to Arrhenius theory. Write the equation for ammonia reacting with water and identify the acid and base.

Show Answer

According to Arrhenius theory, a base must produce OH⁻ ions when dissolved in water. Ammonia (NH₃) does not contain hydroxide ions in its formula, so it cannot be classified as an Arrhenius base based solely on its composition. While NH₃ does produce OH⁻ when it reacts with water, the Arrhenius model doesn't explain how or why this happens.

According to Brønsted-Lowry theory, a base is any substance that accepts a proton (H⁺). Ammonia has a lone pair of electrons on the nitrogen atom, which can accept a proton. Therefore, NH₃ is clearly a Brønsted-Lowry base.

Equation:

$\mathrm{NH_3}{\text{(aq)}} + \mathrm{H_2O}{\text{(l)}} \rightleftharpoons \mathrm{NH_4}^{+}{\text{(aq)}} + \mathrm{OH}^{-}{\text{(aq)}}$

Brønsted-Lowry base: NH₃ accepts a proton from water, forming NH₄⁺ (ammonium ion).

Brønsted-Lowry acid: H₂O donates a proton to ammonia, forming OH⁻ (hydroxide ion).

This demonstrates the advantage of the Brønsted-Lowry model: it explains why ammonia behaves as a base (it accepts protons) and why OH⁻ ions appear in solution (they're produced when water donates a proton). The Arrhenius model cannot explain the mechanism of ammonia's basicity.

Conjugate Acid-Base Pairs

In the Brønsted-Lowry model, acids and bases exist in conjugate pairs. When an acid donates a proton, it becomes a base. When a base accepts a proton, it becomes an acid. Understanding conjugate pairs is essential for analyzing acid-base equilibria.

Defining Conjugate Pairs

A conjugate acid-base pair consists of two species that differ by one proton (H⁺).

Conjugate base: What remains after an acid donates a proton. The conjugate base is capable of accepting a proton to reform the original acid.

Conjugate acid: What forms when a base accepts a proton. The conjugate acid is capable of donating a proton to reform the original base.

Key relationship: If HA is an acid, then A⁻ is its conjugate base (they differ by H⁺). If B is a base, then BH⁺ is its conjugate acid (they differ by H⁺).

General form: For any acid-base reaction:

\mathrm{HA} + \mathrm{B} \rightleftharpoons \mathrm{A}^{-} + \mathrm{BH}^{+}

HA and A⁻ are a conjugate pair (acid/conjugate base). B and BH⁺ are a conjugate pair (base/conjugate acid).

Identifying Conjugate Pairs

Example 1 - Strong acid: Hydrochloric acid and water:

\mathrm{HCl}{\text{(aq)}} + \mathrm{H_2O}{\text{(l)}} \rightarrow \mathrm{H_3O}^{+}{\text{(aq)}} + \mathrm{Cl}^{-}{\text{(aq)}}

Conjugate pair 1: HCl (acid) and Cl⁻ (conjugate base) - HCl donated H⁺ to become Cl⁻

Conjugate pair 2: H₂O (base) and H₃O⁺ (conjugate acid) - H₂O accepted H⁺ to become H₃O⁺

Example 2 - Weak acid: Ethanoic acid and water:

\mathrm{CH_3COOH}{\text{(aq)}} + \mathrm{H_2O}{\text{(l)}} \rightleftharpoons \mathrm{H_3O}^{+}{\text{(aq)}} + \mathrm{CH_3COO}^{-}{\text{(aq)}}

Conjugate pair 1: CH₃COOH (acid) and CH₃COO⁻ (conjugate base, ethanoate ion)

Conjugate pair 2: H₂O (base) and H₃O⁺ (conjugate acid)

Example 3 - Base: Ammonia and water:

\mathrm{NH_3}{\text{(aq)}} + \mathrm{H_2O}{\text{(l)}} \rightleftharpoons \mathrm{NH_4}^{+}{\text{(aq)}} + \mathrm{OH}^{-}{\text{(aq)}}

Conjugate pair 1: NH₃ (base) and NH₄⁺ (conjugate acid, ammonium ion)

Conjugate pair 2: H₂O (acid) and OH⁻ (conjugate base)

Strategy: To identify conjugate pairs, look for two species that differ by exactly one H⁺. Don't pair species from opposite sides unless they differ by H⁺ only.

Strength Relationships

The strength of an acid and its conjugate base are inversely related. This relationship is crucial for understanding buffer solutions and acid-base equilibria.

Strong acid → Weak conjugate base: A strong acid readily donates its proton. Once it does, the conjugate base has little tendency to accept a proton back. Example: HCl (strong acid) produces Cl⁻ (very weak base that doesn't significantly react with water).

Weak acid → Strong conjugate base: A weak acid reluctantly donates its proton. The conjugate base more readily accepts a proton back. Example: CH₃COOH (weak acid) produces CH₃COO⁻ (relatively strong base compared to Cl⁻).

Strong base → Weak conjugate acid: A strong base readily accepts a proton. The conjugate acid has little tendency to donate it back. Example: NH₃ (moderately strong base) produces NH₄⁺ (weak acid).

Weak base → Strong conjugate acid: A weak base reluctantly accepts a proton. The conjugate acid more readily donates it back. Example: Cl⁻ (very weak base) would form HCl (strong acid) if it accepted a proton.

Equilibrium position: In an acid-base reaction, equilibrium favors formation of the weaker acid and weaker base. Protons transfer from the stronger acid to the stronger base.

Worked Example

For the reaction $\mathrm{HNO_2}{\text{(aq)}} + \mathrm{NH_3}{\text{(aq)}} \rightleftharpoons \mathrm{NO_2}^{-}{\text{(aq)}} + \mathrm{NH_4}^{+}{\text{(aq)}}$ , identify: (a) both conjugate acid-base pairs, (b) the Brønsted-Lowry acid and base in the forward reaction, (c) the Brønsted-Lowry acid and base in the reverse reaction.

(a) Conjugate acid-base pairs:

Pair 1: HNO₂ (acid) and NO₂⁻ (conjugate base)

HNO₂ loses H⁺ to become NO₂⁻. They differ by one proton.

Pair 2: NH₃ (base) and NH₄⁺ (conjugate acid)

NH₃ gains H⁺ to become NH₄⁺. They differ by one proton.

(b) Forward reaction:

Acid: HNO₂ donates a proton to NH₃

Base: NH₃ accepts a proton from HNO₂

Acid: NH₄⁺ donates a proton to NO₂⁻

Base: NO₂⁻ accepts a proton from NH₄⁺

Note: In the reverse direction, the conjugate acid of one pair acts as the acid, and the conjugate base of the other pair acts as the base. This illustrates that acid-base reactions are reversible proton transfers between conjugate pairs.

Practice Question

For the reaction $\mathrm{H_2SO_4}{\text{(aq)}} + \mathrm{H_2O}{\text{(l)}} \rightarrow \mathrm{H_3O}^{+}{\text{(aq)}} + \mathrm{HSO_4}^{-}{\text{(aq)}}$ , identify both conjugate acid-base pairs. Then explain why H₂SO₄ is a stronger acid than HSO₄⁻.

Show Answer

Conjugate acid-base pairs:

Pair 1: H₂SO₄ (acid) and HSO₄⁻ (conjugate base, hydrogen sulfate ion)

H₂SO₄ donates a proton to become HSO₄⁻. They differ by one H⁺.

Pair 2: H₂O (base) and H₃O⁺ (conjugate acid, hydronium ion)

H₂O accepts a proton to become H₃O⁺. They differ by one H⁺.

Why H₂SO₄ is a stronger acid than HSO₄⁻:

H₂SO₄ (sulfuric acid) is a strong acid that completely ionizes in water, readily donating its first proton. After losing one proton, it becomes HSO₄⁻, which is a weak acid. HSO₄⁻ only partially ionizes because:

1. Charge: H₂SO₄ is neutral, making it easier to release a positively charged H⁺. HSO₄⁻ is already negatively charged, so it's harder to remove another positive H⁺ from a negative ion.

2. Stability: When H₂SO₄ loses H⁺, it forms HSO₄⁻, which is relatively stable with one negative charge spread over the sulfate structure. When HSO₄⁻ loses H⁺, it forms SO₄²⁻ with two negative charges, which is less stable due to increased charge density and electrostatic repulsion.

This demonstrates the general principle: it becomes progressively harder to remove protons from an acid as it becomes more negatively charged. The first ionization of polyprotic acids is always stronger than subsequent ionizations.

Amphiprotic Substances

Some substances can act as either an acid or a base depending on what they react with. These substances are called amphiprotic (or amphoteric in the broader sense). Understanding amphiprotic behavior is essential for analyzing complex acid-base equilibria.

Definition and Characteristics

An amphiprotic substance can donate a proton (act as a Brønsted-Lowry acid) or accept a proton (act as a Brønsted-Lowry base).

Requirements: To be amphiprotic, a substance must have at least one proton that can be donated AND must have a lone pair of electrons or negative charge that can accept a proton.

Common amphiprotic species:

Water (H₂O)
Hydrogen carbonate ion (HCO₃⁻)
Hydrogen sulfate ion (HSO₄⁻)
Hydrogen phosphate ion (HPO₄²⁻)
Dihydrogen phosphate ion (H₂PO₄⁻)
Amino acids (contain both -COOH and -NH₂ groups)

Pattern: Many amphiprotic species are intermediates in the ionization of polyprotic acids. For example, H₂CO₃ can lose one H⁺ to form HCO₃⁻ (which can lose another H⁺) or gain one H⁺. The intermediate HCO₃⁻ can act either way, making it amphiprotic.

Water (H₂O) - The Classic Example

Water is the most important amphiprotic substance. It can act as an acid or a base depending on what it reacts with.

Water as a base (accepts H⁺) - reacting with an acid:

\mathrm{HCl}{\text{(aq)}} + \mathrm{H_2O}{\text{(l)}} \rightarrow \mathrm{H_3O}^{+}{\text{(aq)}} + \mathrm{Cl}^{-}{\text{(aq)}}

H₂O accepts a proton from HCl, forming H₃O⁺ (hydronium ion). Water acts as a base.

Water as an acid (donates H⁺) - reacting with a base:

\mathrm{NH_3}{\text{(aq)}} + \mathrm{H_2O}{\text{(l)}} \rightleftharpoons \mathrm{NH_4}^{+}{\text{(aq)}} + \mathrm{OH}^{-}{\text{(aq)}}

H₂O donates a proton to NH₃, forming OH⁻ (hydroxide ion). Water acts as an acid.

Auto-ionization of water: Water can even react with itself, with one molecule acting as acid and another as base:

\mathrm{H_2O}{\text{(l)}} + \mathrm{H_2O}{\text{(l)}} \rightleftharpoons \mathrm{H_3O}^{+}{\text{(aq)}} + \mathrm{OH}^{-}{\text{(aq)}}

This auto-ionization is why pure water has a very small concentration of both H₃O⁺ and OH⁻ ions (each 1.0 × 10⁻⁷ M at 25°C), giving pure water a pH of 7.

Hydrogen Carbonate Ion (HCO₃⁻)

The hydrogen carbonate ion (bicarbonate) is an important amphiprotic species in biological and environmental systems.

HCO₃⁻ as a base (accepts H⁺) - reacting with an acid:

\mathrm{HCO_3}^{-}{\text{(aq)}} + \mathrm{H_3O}^{+}{\text{(aq)}} \rightarrow \mathrm{H_2CO_3}{\text{(aq)}} + \mathrm{H_2O}{\text{(l)}}

HCO₃⁻ accepts a proton from H₃O⁺, forming carbonic acid (H₂CO₃). The bicarbonate ion acts as a base.

HCO₃⁻ as an acid (donates H⁺) - reacting with a base:

\mathrm{HCO_3}^{-}{\text{(aq)}} + \mathrm{OH}^{-}{\text{(aq)}} \rightarrow \mathrm{CO_3}^{2-}{\text{(aq)}} + \mathrm{H_2O}{\text{(l)}}

HCO₃⁻ donates a proton to OH⁻, forming carbonate ion (CO₃²⁻). The bicarbonate ion acts as an acid.

Importance: The amphiprotic nature of HCO₃⁻ is crucial for the carbonate buffer system in blood, which maintains blood pH around 7.4. The bicarbonate can neutralize both excess acid (by acting as a base) and excess base (by acting as an acid).

Other Amphiprotic Species

Hydrogen sulfate ion (HSO₄⁻):

As a base (less common, because HSO₄⁻ is a weak acid):

\mathrm{HSO_4}^{-}{\text{(aq)}} + \mathrm{H_3O}^{+}{\text{(aq)}} \rightarrow \mathrm{H_2SO_4}{\text{(aq)}} + \mathrm{H_2O}{\text{(l)}}

As an acid (more common):

\mathrm{HSO_4}^{-}{\text{(aq)}} + \mathrm{H_2O}{\text{(l)}} \rightleftharpoons \mathrm{H_3O}^{+}{\text{(aq)}} + \mathrm{SO_4}^{2-}{\text{(aq)}}

Dihydrogen phosphate ion (H₂PO₄⁻):

As a base:

\mathrm{H_2PO_4}^{-}{\text{(aq)}} + \mathrm{H_3O}^{+}{\text{(aq)}} \rightarrow \mathrm{H_3PO_4}{\text{(aq)}} + \mathrm{H_2O}{\text{(l)}}

As an acid:

\mathrm{H_2PO_4}^{-}{\text{(aq)}} + \mathrm{H_2O}{\text{(l)}} \rightleftharpoons \mathrm{H_3O}^{+}{\text{(aq)}} + \mathrm{HPO_4}^{2-}{\text{(aq)}}

Pattern recognition: Species with both ionizable hydrogen atoms AND negative charge (or lone pairs) are likely amphiprotic. Look for intermediate ions in polyprotic acid ionization sequences.

Worked Example

Write balanced equations showing hydrogen phosphate ion (HPO₄²⁻) acting as: (a) a Brønsted-Lowry acid, (b) a Brønsted-Lowry base. Identify the conjugate acid and conjugate base in each case.

(a) HPO₄²⁻ as an acid (donates H⁺):

Equation (reacting with water as the base):

$\mathrm{HPO_4}^{2-}{\text{(aq)}} + \mathrm{H_2O}{\text{(l)}} \rightleftharpoons \mathrm{H_3O}^{+}{\text{(aq)}} + \mathrm{PO_4}^{3-}{\text{(aq)}}$

HPO₄²⁻ donates a proton to water, forming phosphate ion (PO₄³⁻).

Conjugate base of HPO₄²⁻: PO₄³⁻ (phosphate ion)

Conjugate acid of H₂O: H₃O⁺ (hydronium ion)

(b) HPO₄²⁻ as a base (accepts H⁺):

Equation (reacting with an acid like H₃O⁺):

$\mathrm{HPO_4}^{2-}{\text{(aq)}} + \mathrm{H_3O}^{+}{\text{(aq)}} \rightarrow \mathrm{H_2PO_4}^{-}{\text{(aq)}} + \mathrm{H_2O}{\text{(l)}}$

HPO₄²⁻ accepts a proton from H₃O⁺, forming dihydrogen phosphate ion (H₂PO₄⁻).

Conjugate acid of HPO₄²⁻: H₂PO₄⁻ (dihydrogen phosphate ion)

Conjugate base of H₃O⁺: H₂O (water)

This demonstrates the amphiprotic nature of HPO₄²⁻: it can lose a proton to form PO₄³⁻ or gain a proton to form H₂PO₄⁻. This property makes phosphate systems excellent buffers in biological contexts.

Practice Question

Explain why water is amphiprotic but hydroxide ion (OH⁻) is not. Write equations showing water acting as both an acid and a base.

Show Answer

Why water is amphiprotic:

Water (H₂O) is amphiprotic because it has BOTH:

1. Hydrogen atoms that can be donated as H⁺ (making it able to act as an acid)

2. Lone pairs of electrons on the oxygen atom that can accept H⁺ (making it able to act as a base)

Why OH⁻ is NOT amphiprotic:

While OH⁻ has one hydrogen atom it could theoretically donate, and it has lone pairs that could accept a proton, it acts overwhelmingly as a base only. When OH⁻ accepts a proton, it forms H₂O (a neutral, stable molecule). When OH⁻ would donate its hydrogen, it would form O²⁻, which is extremely unstable and doesn't exist under normal conditions in aqueous solution. Therefore, OH⁻ only acts as a base in practice, not as an acid.

Equations showing water's amphiprotic behavior:

Water as a base (accepting H⁺ from HCl):

$\mathrm{HCl}{\text{(aq)}} + \mathrm{H_2O}{\text{(l)}} \rightarrow \mathrm{H_3O}^{+}{\text{(aq)}} + \mathrm{Cl}^{-}{\text{(aq)}}$

Water accepts a proton, forming H₃O⁺.

Water as an acid (donating H⁺ to NH₃):

$\mathrm{NH_3}{\text{(aq)}} + \mathrm{H_2O}{\text{(l)}} \rightleftharpoons \mathrm{NH_4}^{+}{\text{(aq)}} + \mathrm{OH}^{-}{\text{(aq)}}$

Water donates a proton, forming OH⁻.

This ability to act either way depending on the reaction partner makes water an excellent solvent for acid-base chemistry and is fundamental to aqueous acid-base equilibria.

Pillar 2 of 3

pH Calculations

Master quantitative pH calculations for strong and weak acids and bases, understand the significance of Ka and pKa values, and analyze how dilution affects pH with proper significant figure conventions.

Strong Acids & Bases

Strong acids and bases completely ionize in water, making pH calculations straightforward. The key is understanding complete dissociation and applying proper significant figure rules for pH values.

Strong Acids

Definition: A strong acid completely ionizes in water, donating all available protons. Every molecule of acid produces one H₃O⁺ ion (for monoprotic acids).

Common strong acids: HCl, HBr, HI, HNO₃, H₂SO₄ (first proton only), HClO₄

Complete ionization: For HCl:

\mathrm{HCl}{\text{(aq)}} \rightarrow \mathrm{H_3O}^{+}{\text{(aq)}} + \mathrm{Cl}^{-}{\text{(aq)}}

Use a single arrow (→) not equilibrium arrows (⇌) because the reaction goes essentially 100% to completion.

Key assumption: For strong acids, [H₃O⁺] = [acid]_initial. If you dissolve 0.10 M HCl, you get 0.10 M H₃O⁺.

pH calculation:

\text{pH} = -\log[\mathrm{H_3O}^{+}]

Or equivalently: $\text{pH} = -\log[\mathrm{H}^{+}]$ (H⁺ and H₃O⁺ are used interchangeably)

Example: For 0.010 M HCl:

[H⁺] = 0.010 M

pH = -log(0.010) = -log(1.0 × 10⁻²) = 2.00

Strong Bases

Definition: A strong base completely dissociates in water, producing OH⁻ ions.

Common strong bases: Group 1 hydroxides (NaOH, KOH, LiOH), Group 2 hydroxides (Ba(OH)₂, Sr(OH)₂, Ca(OH)₂)

Complete dissociation: For NaOH:

\mathrm{NaOH}{\text{(aq)}} \rightarrow \mathrm{Na}^{+}{\text{(aq)}} + \mathrm{OH}^{-}{\text{(aq)}}

For Ba(OH)₂ (note: 2 OH⁻ per formula unit):

\mathrm{Ba(OH)_2}{\text{(aq)}} \rightarrow \mathrm{Ba}^{2+}{\text{(aq)}} + 2\mathrm{OH}^{-}{\text{(aq)}}

For Ba(OH)₂, [OH⁻] = 2 × [Ba(OH)₂]_initial

pH calculation for bases: Calculate [OH⁻] first, then use the water equilibrium relationship:

K_w = [\mathrm{H}^{+}][\mathrm{OH}^{-}] = 1.0 \times 10^{-14} \text{ at 25°C}

Rearrange to find [H⁺]:

[\mathrm{H}^{+}] = \frac{K_w}{[\mathrm{OH}^{-}]}

Then calculate pH = -log[H⁺], or use pOH:

\text{pOH} = -\log[\mathrm{OH}^{-}]

\text{pH} + \text{pOH} = 14.00 \text{ at 25°C}

Significant Figures in pH

Critical rule: The number of decimal places in the pH equals the number of significant figures in the concentration.

Why: pH is a logarithm. In logarithms, the digits before the decimal point (the characteristic) indicate the power of 10, while the digits after the decimal point (the mantissa) carry the significant figures.

Examples:

0.010 M (2 sig figs) → pH = 2.00 (2 decimal places)
0.10 M (2 sig figs) → pH = 1.00 (2 decimal places)
0.001 M (1 sig fig) → pH = 3.0 (1 decimal place)
2.5 × 10⁻³ M (2 sig figs) → pH = 2.60 (2 decimal places)
1.00 × 10⁻⁴ M (3 sig figs) → pH = 4.000 (3 decimal places)

Common error: Writing pH = 2 instead of pH = 2.00 for a 0.010 M solution. The number of decimal places matters!

Reverse calculation: If pH = 3.45 (2 decimal places), then [H⁺] = 10⁻³·⁴⁵ = 3.5 × 10⁻⁴ M (2 sig figs)

Worked Example

Calculate the pH of: (a) 0.050 M HNO₃, (b) 0.025 M Ba(OH)₂. Show all working with correct significant figures.

(a) 0.050 M HNO₃:

HNO₃ is a strong acid, so it completely ionizes:

$\mathrm{HNO_3} \rightarrow \mathrm{H}^{+} + \mathrm{NO_3}^{-}$

[H⁺] = 0.050 M (same as initial acid concentration)

pH = -log(0.050)

pH = -log(5.0 × 10⁻²)

pH = 1.30

Note: 0.050 has 2 sig figs, so pH has 2 decimal places

(b) 0.025 M Ba(OH)₂:

Ba(OH)₂ is a strong base that completely dissociates:

$\mathrm{Ba(OH)_2} \rightarrow \mathrm{Ba}^{2+} + 2\mathrm{OH}^{-}$

Each Ba(OH)₂ produces 2 OH⁻ ions:

[OH⁻] = 2 × 0.025 = 0.050 M

Method 1 - Using pOH:

pOH = -log(0.050) = 1.30

pH = 14.00 - 1.30 = 12.70

Method 2 - Using Kw:

$[\mathrm{H}^{+}] = \frac{1.0 \times 10^{-14}}{0.050} = 2.0 \times 10^{-13} \text{ M}$

pH = -log(2.0 × 10⁻¹³) = 12.70

Note: 0.025 has 2 sig figs, so pH has 2 decimal places

Practice Question

Calculate the pH of 0.00200 M HCl. Then calculate [OH⁻] in this solution. Explain why the number of decimal places in your pH value is important.

Show Answer

Calculate pH:

HCl is a strong acid that completely ionizes:

[H⁺] = 0.00200 M = 2.00 × 10⁻³ M

pH = -log(2.00 × 10⁻³)

pH = 2.699

Calculate [OH⁻]:

Using K_w = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴:

$[\mathrm{OH}^{-}] = \frac{1.0 \times 10^{-14}}{2.00 \times 10^{-3}}$

[OH⁻] = 5.00 × 10⁻¹² M

Significance of decimal places:

The initial concentration 0.00200 M has 3 significant figures (the trailing zeros after the decimal are significant because they come after the "2"). Therefore, the pH must have 3 decimal places: pH = 2.699.

If we incorrectly wrote pH = 2.7, we would be implying the concentration only had 1 significant figure (0.002 M instead of 0.00200 M). The decimal places in pH directly reflect the precision of our measurement. In this case, pH = 2.699 tells us we know the [H⁺] to 3 significant figures, which is consistent with knowing the initial acid concentration to 3 significant figures.

This rule prevents losing precision information when converting between concentration and pH. It's a critical Band 6 habit for HSC Chemistry.

Weak Acids & Bases

Weak acids and bases only partially ionize in water, establishing an equilibrium. Calculating pH requires using the acid dissociation constant (K_a) and often involves the assumption that the degree of ionization is small.

Weak Acid Equilibrium

Definition: A weak acid only partially ionizes in water, establishing an equilibrium between the acid and its conjugate base.

General equation: For a weak acid HA:

\mathrm{HA}{\text{(aq)}} + \mathrm{H_2O}{\text{(l)}} \rightleftharpoons \mathrm{H_3O}^{+}{\text{(aq)}} + \mathrm{A}^{-}{\text{(aq)}}

Use equilibrium arrows (⇌) because the reaction doesn't go to completion.

Acid dissociation constant (K_a):

K_a = \frac{[\mathrm{H_3O}^{+}][\mathrm{A}^{-}]}{[\mathrm{HA}]}

Water is excluded (pure liquid). Larger K_a means stronger acid (more ionization).

Common weak acids:

Ethanoic acid (CH₃COOH): K_a = 1.8 × 10⁻⁵
Carbonic acid (H₂CO₃): K_a = 4.3 × 10⁻⁷
Hydrofluoric acid (HF): K_a = 6.8 × 10⁻⁴
Hydrogen cyanide (HCN): K_a = 6.2 × 10⁻¹⁰

pK_a notation:

\text{p}K_a = -\log K_a

Smaller pK_a means stronger acid. For CH₃COOH: pK_a = -log(1.8 × 10⁻⁵) = 4.74

ICE Table Method for Weak Acids

Use an ICE (Initial, Change, Equilibrium) table to organize the calculation systematically.

Example: Calculate pH of 0.10 M CH₃COOH (K_a = 1.8 × 10⁻⁵)

Step 1: Write the equilibrium equation:

$\mathrm{CH_3COOH} + \mathrm{H_2O} \rightleftharpoons \mathrm{H_3O}^{+} + \mathrm{CH_3COO}^{-}$

Step 2: Set up ICE table (let x = amount that ionizes):

	CH₃COOH	H₃O⁺	CH₃COO⁻
I	0.10	≈0	0
C	-x	+x	+x
E	0.10 - x	x	x

Step 3: Write K_a expression and substitute:

$1.8 \times 10^{-5} = \frac{x \cdot x}{0.10 - x} = \frac{x^2}{0.10 - x}$

Step 4: Make the assumption that x << 0.10 (valid if K_a is small and initial concentration is not too dilute):

If x << 0.10, then 0.10 - x ≈ 0.10

$1.8 \times 10^{-5} = \frac{x^2}{0.10}$

Step 5: Solve for x:

$x^2 = 1.8 \times 10^{-5} \times 0.10 = 1.8 \times 10^{-6}$

$x = \sqrt{1.8 \times 10^{-6}} = 1.3 \times 10^{-3} \text{ M}$

Step 6: Check assumption:

x/0.10 = (1.3 × 10⁻³)/0.10 = 0.013 = 1.3%

Less than 5%, so assumption is valid ✓

Step 7: Calculate pH:

[H⁺] = x = 1.3 × 10⁻³ M

pH = -log(1.3 × 10⁻³) = 2.89

When the Assumption Fails

Rule of thumb: The approximation (0.10 - x ≈ 0.10) is valid when x represents less than 5% of the initial concentration.

If assumption fails (x > 5% of initial concentration), use the quadratic formula:

From $K_a = \frac{x^2}{C - x}$ where C is initial concentration:

$K_a(C - x) = x^2$

$x^2 + K_a x - K_a C = 0$

Solve using quadratic formula (take positive root only):

x = \frac{-K_a + \sqrt{K_a^2 + 4K_a C}}{2}

When assumption likely fails:

Very dilute solutions (C < 0.01 M)
Moderately strong weak acids (K_a > 10⁻⁴)
Always check your assumption!

Weak Bases

Weak bases accept protons from water, establishing an equilibrium characterized by K_b.

Example: Ammonia (NH₃):

\mathrm{NH_3}{\text{(aq)}} + \mathrm{H_2O}{\text{(l)}} \rightleftharpoons \mathrm{NH_4}^{+}{\text{(aq)}} + \mathrm{OH}^{-}{\text{(aq)}}

Base dissociation constant (K_b):

K_b = \frac{[\mathrm{NH_4}^{+}][\mathrm{OH}^{-}]}{[\mathrm{NH_3}]}

For NH₃: K_b = 1.8 × 10⁻⁵

Calculation method: Use ICE table to find [OH⁻], then convert to pH using pOH + pH = 14.00

Relationship between K_a and K_b for conjugate pairs:

K_a \times K_b = K_w = 1.0 \times 10^{-14}

If you know K_a for an acid, you can find K_b for its conjugate base, and vice versa.

Worked Example

Calculate the pH of 0.20 M ethanoic acid (CH₃COOH). K_a = 1.8 × 10⁻⁵. Show all working including the ICE table and assumption check.

Step 1: Write equilibrium equation:

$\mathrm{CH_3COOH} + \mathrm{H_2O} \rightleftharpoons \mathrm{H_3O}^{+} + \mathrm{CH_3COO}^{-}$

Step 2: ICE table (let x = [H⁺] formed):

I	0.20	0	0
C	-x	+x	+x
E	0.20 - x	x	x

Step 3: Substitute into K_a expression:

$K_a = \frac{[\mathrm{H}^{+}][\mathrm{CH_3COO}^{-}]}{[\mathrm{CH_3COOH}]}$

$1.8 \times 10^{-5} = \frac{x \cdot x}{0.20 - x}$

Step 4: Make assumption (x << 0.20):

$1.8 \times 10^{-5} \approx \frac{x^2}{0.20}$

$x^2 = 1.8 \times 10^{-5} \times 0.20 = 3.6 \times 10^{-6}$

$x = \sqrt{3.6 \times 10^{-6}} = 1.9 \times 10^{-3} \text{ M}$

Step 5: Check assumption:

Percentage ionized = (x/0.20) × 100% = (1.9 × 10⁻³/0.20) × 100%

= 0.0095 × 100% = 0.95%

Less than 5%, assumption is valid ✓

Step 6: Calculate pH:

[H⁺] = 1.9 × 10⁻³ M

pH = -log(1.9 × 10⁻³)

pH = 2.72

Note: 0.20 has 2 sig figs, so pH has 2 decimal places

Practice Question

Calculate the pH of 0.15 M ammonia (NH₃). K_b = 1.8 × 10⁻⁵. Show the ICE table and all working.

Show Answer

Step 1: Write equilibrium equation:

$\mathrm{NH_3} + \mathrm{H_2O} \rightleftharpoons \mathrm{NH_4}^{+} + \mathrm{OH}^{-}$

Step 2: ICE table:

I	0.15	0	0
C	-x	+x	+x
E	0.15 - x	x	x

Step 3: Substitute into K_b:

$1.8 \times 10^{-5} = \frac{x \cdot x}{0.15 - x}$

Step 4: Assume x << 0.15:

$1.8 \times 10^{-5} \approx \frac{x^2}{0.15}$

$x^2 = 2.7 \times 10^{-6}$

$x = 1.6 \times 10^{-3} \text{ M}$

Step 5: Check assumption:

(1.6 × 10⁻³/0.15) × 100% = 1.1% < 5% ✓

Step 6: Calculate pH:

[OH⁻] = 1.6 × 10⁻³ M

pOH = -log(1.6 × 10⁻³) = 2.80

pH = 14.00 - 2.80 = 11.20

Note: 0.15 has 2 sig figs, so pH has 2 decimal places

Dilution Effects on pH

Diluting an acid or base solution changes the pH, but the effect differs between strong and weak acids/bases. Understanding these differences is crucial for practical chemistry and explains buffer behavior.

Dilution Formula

When diluting a solution, the number of moles remains constant:

n_1 = n_2

Since n = c × V, this gives the dilution formula:

c_1V_1 = c_2V_2

Where c₁ and V₁ are initial concentration and volume, c₂ and V₂ are final concentration and volume.

Example: Diluting 10.0 mL of 0.100 M HCl to 100.0 mL:

c₁V₁ = c₂V₂

(0.100)(10.0) = c₂(100.0)

c₂ = 0.0100 M = 1.00 × 10⁻² M

The solution is diluted 10-fold

Diluting Strong Acids and Bases

For strong acids and bases, dilution directly decreases [H⁺] or [OH⁻] proportionally.

Example: Dilute 0.10 M HCl (pH = 1.00) by factor of 10:

New [H⁺] = 0.010 M

New pH = -log(0.010) = 2.00

pH increases by 1 unit for 10-fold dilution

Pattern: Each 10-fold dilution of a strong acid increases pH by 1 unit (until approaching pH 7).

Limit: You cannot make a solution basic by diluting an acid! As you approach infinite dilution, pH approaches 7 (neutral), never exceeds it. At very high dilution, water's auto-ionization becomes significant.

Very dilute strong acid: If calculated [H⁺] from acid < 10⁻⁶ M, must consider H⁺ from water (10⁻⁷ M). Use:

[\mathrm{H}^{+}]_{total} = [\mathrm{H}^{+}]_{acid} + [\mathrm{H}^{+}]_{water}

Diluting Weak Acids and Bases

For weak acids and bases, dilution has a more complex effect because equilibrium shifts.

Le Chatelier's principle: Diluting a weak acid shifts equilibrium right (toward more ionization) to partially compensate for the dilution.

Example: 0.10 M CH₃COOH (K_a = 1.8 × 10⁻⁵):

Using ICE table: [H⁺] = 1.3 × 10⁻³ M, pH = 2.89

Percent ionization = (1.3 × 10⁻³/0.10) × 100% = 1.3%

Now dilute 10-fold to 0.010 M:

Using ICE table: [H⁺] = 4.2 × 10⁻⁴ M, pH = 3.38

Percent ionization = (4.2 × 10⁻⁴/0.010) × 100% = 4.2%

Observation: Although concentration decreased 10-fold, [H⁺] only decreased by factor of ~3 (not 10) because percent ionization increased from 1.3% to 4.2%. The equilibrium shifted right upon dilution.

General rule: Weak acids become stronger (more ionized) when diluted. At infinite dilution, a weak acid behaves almost like a strong acid, but the pH still approaches 7.

pH change for weak acids: 10-fold dilution increases pH by less than 1 unit (unlike strong acids which increase by exactly 1 unit).

Comparing Strong vs Weak Acid Dilution

Strong acid (0.10 M HCl):

Initial: [H⁺] = 0.10 M, pH = 1.00

After 10× dilution: [H⁺] = 0.010 M, pH = 2.00

Change: ΔpH = +1.00 unit exactly

Weak acid (0.10 M CH₃COOH, K_a = 1.8 × 10⁻⁵):

Initial: [H⁺] = 1.3 × 10⁻³ M, pH = 2.89

After 10× dilution: [H⁺] = 4.2 × 10⁻⁴ M, pH = 3.38

Change: ΔpH = +0.49 units (less than 1)

Why the difference? The weak acid's equilibrium shifts to partially offset the dilution. More of the weak acid ionizes when diluted, so [H⁺] doesn't decrease as much as the dilution factor would suggest.

Buffer connection: This resistance to pH change upon dilution is related to buffer behavior. Buffer solutions (containing weak acid + conjugate base) resist pH change even more effectively.

Worked Example

A student dilutes 25.0 mL of 0.200 M NaOH to a final volume of 500.0 mL. Calculate: (a) the concentration of the diluted solution, (b) the pH of the original solution, (c) the pH of the diluted solution.

(a) Concentration after dilution:

Using c₁V₁ = c₂V₂:

(0.200 M)(25.0 mL) = c₂(500.0 mL)

c₂ = (0.200 × 25.0)/500.0

c₂ = 0.0100 M

The solution is diluted by factor of 20

(b) pH of original 0.200 M NaOH:

NaOH is a strong base, completely dissociates:

[OH⁻] = 0.200 M

pOH = -log(0.200) = 0.699

pH = 14.00 - 0.699 = 13.30

[OH⁻] = 0.0100 M = 1.00 × 10⁻² M

pOH = -log(1.00 × 10⁻²) = 2.000

pH = 14.00 - 2.000 = 12.00

Note: The pH decreased by 1.30 units (from 13.30 to 12.00) due to 20-fold dilution. For strong bases, pH decreases upon dilution but remains basic.

Practice Question

Explain why diluting a weak acid by a factor of 10 increases the pH by less than 1 unit, while diluting a strong acid by the same factor increases pH by exactly 1 unit. Use Le Chatelier's principle in your explanation.

Show Answer

Strong acid behavior:

A strong acid completely ionizes in water. If we have 0.10 M HCl, [H⁺] = 0.10 M because every HCl molecule donates its proton. When we dilute 10-fold to 0.010 M, [H⁺] decreases proportionally to 0.010 M. The pH increases from -log(0.10) = 1.00 to -log(0.010) = 2.00, an increase of exactly 1.00 pH unit.

Weak acid behavior:

A weak acid only partially ionizes, establishing equilibrium: HA + H₂O ⇌ H₃O⁺ + A⁻. When we dilute a weak acid, we decrease the concentrations of all species. According to Le Chatelier's principle, the system shifts to minimize this disturbance.

Dilution affects the reaction quotient Q. When we dilute, both [HA] and [H₃O⁺][A⁻] decrease, but because [H₃O⁺] and [A⁻] appear as a product in the numerator of the equilibrium expression (K_a = [H⁺][A⁻]/[HA]), dilution makes Q < K_a. The equilibrium shifts right to restore equilibrium, causing more HA to ionize.

This increased ionization partially compensates for the dilution. While the absolute [H⁺] decreases, the percent ionization increases. For example, if a weak acid was 1% ionized before dilution, it might be 3% ionized after 10-fold dilution. Therefore, [H⁺] doesn't decrease by the full factor of 10—it might only decrease by a factor of 3 or 4.

Since [H⁺] decreases by less than a factor of 10, the pH increases by less than 1 unit (since log₁₀(10) = 1). This demonstrates how weak acid equilibria can shift to partially resist changes in concentration, which is the fundamental principle behind buffer solutions.

Pillar 3 of 3

Titrations & Buffers

Master titration techniques, curve interpretation, indicator selection, buffer chemistry, and enthalpy of neutralisation calculations with proper experimental procedures and quantitative analysis.

Standard Solutions

A standard solution is a solution of accurately known concentration. Preparing standard solutions correctly is essential for accurate titration analysis.

Primary Standards

A primary standard is a substance that can be used to prepare a standard solution directly by accurately weighing and dissolving in a known volume of water.

Requirements for a primary standard:

High purity (≥99.9%)
Stable in air (doesn't absorb moisture, CO₂, or react with O₂)
High molar mass (to minimize weighing errors)
Readily soluble in water
Not hygroscopic (doesn't absorb water from air)
Reacts completely and rapidly in titrations

Common primary standards:

Anhydrous sodium carbonate (Na₂CO₃) - for standardizing acids
Oxalic acid dihydrate (H₂C₂O₄·2H₂O) - for standardizing bases
Potassium hydrogen phthalate (KHP, C₈H₅O₄K) - for standardizing bases

Why NaOH is NOT a primary standard: Sodium hydroxide is hygroscopic (absorbs water from air) and reacts with CO₂ from air to form Na₂CO₃. This means its exact mass and purity cannot be known accurately. NaOH solutions must be standardized against a primary standard.

Preparing a Standard Solution

Method using a primary standard:

Step 1: Calculate the mass of primary standard needed using n = cV and n = m/M.

Step 2: Accurately weigh the calculated mass on an analytical balance (to 4 decimal places). Record actual mass weighed.

Step 3: Dissolve the solid in a small volume of distilled water in a clean beaker. Stir until completely dissolved.

Step 4: Transfer the solution to a volumetric flask using a funnel. Rinse the beaker and funnel thoroughly with distilled water, adding all rinsings to the flask. This ensures no solute is lost.

Step 5: Add distilled water to the flask until the bottom of the meniscus touches the calibration mark. Stopper and invert several times to mix thoroughly.

Step 6: Calculate the actual concentration using the mass you weighed (not the mass you calculated).

Example calculation: To prepare 250.0 mL of approximately 0.100 M Na₂CO₃:

n = cV = 0.100 × 0.2500 = 0.0250 mol

m = nM = 0.0250 × 106.0 = 2.65 g (calculated)

Actual mass weighed = 2.6532 g

Actual n = 2.6532/106.0 = 0.025031 mol

Actual c = 0.025031/0.2500 = 0.1001 M

Rinsing Rules - Critical for HSC

Golden rule: Rinse glassware with the solution it will contain, NOT with distilled water (except conical flasks).

Pipette: Rinse with the solution you are pipetting (usually the standard solution).

Why: Water left in the pipette would dilute the solution, giving an inaccurate volume delivery.

Burette: Rinse with the solution you will put in it (usually the titrant).

Why: Water left in the burette would dilute the titrant, affecting the concentration.

Conical flask: Rinse with distilled water ONLY. Do NOT rinse with the solution.

Why: The flask just needs to be clean. Extra water won't affect the calculation because we measure moles (n = cV), and a few drops of water don't change the moles of solute present. If you rinsed with solution, you'd add unknown extra moles of solute.

Common exam mistake: Rinsing the conical flask with the solution from the pipette. This is WRONG and will give inaccurate results!

Worked Example

A student wants to prepare 250.0 mL of approximately 0.0500 M Na₂CO₃ solution for standardizing HCl. The student weighs out 1.3265 g of anhydrous Na₂CO₃ (M = 106.0 g/mol). Calculate the exact concentration of the standard solution prepared.

Given:

Mass of Na₂CO₃ = 1.3265 g

Molar mass = 106.0 g/mol

Volume = 250.0 mL = 0.2500 L

Step 1: Calculate moles of Na₂CO₃:

$n = \frac{m}{M}$

$n = \frac{1.3265}{106.0}$

$n = 0.012516 \text{ mol}$

Step 2: Calculate concentration:

$c = \frac{n}{V}$

$c = \frac{0.012516}{0.2500}$

$c = \mathbf{0.05006 \text{ M}}$

The exact concentration is 0.05006 M, not 0.0500 M. This precise value must be used in all subsequent titration calculations to maintain accuracy.

Practice Question

Explain why: (a) the burette must be rinsed with the titrant before use, (b) the conical flask must NOT be rinsed with the solution from the pipette, (c) NaOH cannot be used as a primary standard.

Show Answer

(a) Why rinse burette with titrant:

If the burette contains water drops from washing, these will dilute the titrant when it's added. This changes the concentration of the solution in the burette, making it different from the known concentration. By rinsing with the titrant first, any water is displaced and the concentration remains accurate. The small amount of titrant used for rinsing is discarded, so it doesn't affect the measurement.

(b) Why NOT rinse conical flask with solution:

The conical flask should only be rinsed with distilled water and left wet. When we pipette the standard solution into the flask, we're delivering a precise number of moles (n = cV). A few drops of water in the flask won't change the number of moles of solute - they'll just slightly increase the total volume, which doesn't matter because we're measuring the volume delivered by the burette to neutralize those moles.

If we rinsed the flask with the solution from the pipette, we would add an unknown extra amount of solute to the flask. This would mean the flask contains MORE moles than we pipetted, making our results inaccurate. We wouldn't know the exact starting amount.

NaOH fails to meet the requirements for a primary standard for two main reasons:

1. Hygroscopic: NaOH absorbs water from the air. When you weigh NaOH pellets, you're weighing NaOH plus absorbed water, but you don't know how much of each. This means you can't accurately determine the mass of pure NaOH.

2. Reacts with CO₂: NaOH reacts with CO₂ from air: 2NaOH + CO₂ → Na₂CO₃ + H₂O. This means solid NaOH and NaOH solutions gradually change composition over time, reducing purity and making the exact concentration uncertain.

Because of these issues, NaOH solutions must be standardized against a primary standard (like oxalic acid or KHP) to determine their exact concentration.

Titration Curves

Titration curves show how pH changes as titrant is added. The shape of the curve depends on the strength of the acid and base involved, and understanding these curves is essential for indicator selection.

Strong Acid - Strong Base

Example: HCl titrated with NaOH

Initial pH: Low (typically 1-3), determined by the strong acid concentration.

Before equivalence point: Excess acid present. pH rises gradually as acid is neutralized. The solution is acidic (pH < 7).

Equivalence point: Moles of acid = moles of base. All acid is neutralized, producing salt and water only:

\mathrm{HCl} + \mathrm{NaOH} \rightarrow \mathrm{NaCl} + \mathrm{H_2O}

pH at equivalence = 7.00 (neutral) because NaCl doesn't hydrolyze.

After equivalence point: Excess base present. pH rises sharply then levels off, controlled by excess NaOH. The solution is basic (pH > 7).

Vertical region: Very sharp pH change (typically pH 4-10) over less than 1 drop of titrant. This large vertical region means many indicators work well.

Buffer region: None. Strong acid/base mixtures don't form buffers.

Weak Acid - Strong Base

Example: CH₃COOH titrated with NaOH

Initial pH: Higher than strong acid of same concentration (typically 3-5) because weak acid only partially ionizes.

Before equivalence point: Buffer region! A mixture of weak acid (CH₃COOH) and its conjugate base (CH₃COO⁻ from neutralization). pH rises gradually and the solution resists pH change.

Half-equivalence point: When exactly half the acid is neutralized, [CH₃COOH] = [CH₃COO⁻]. At this point, pH = pK_a of the acid. For ethanoic acid: pH ≈ 4.74.

Equivalence point: All weak acid converted to its conjugate base. The solution contains CH₃COO⁻ (a weak base) which hydrolyzes:

\mathrm{CH_3COO}^{-} + \mathrm{H_2O} \rightleftharpoons \mathrm{CH_3COOH} + \mathrm{OH}^{-}

pH at equivalence > 7 (typically 8-10) because conjugate base is weakly basic.

After equivalence point: Excess NaOH dominates. pH controlled by strong base.

Vertical region: Smaller than strong/strong (typically pH 7-10). Must use indicator that changes in this range (like phenolphthalein). Methyl orange would NOT work.

Strong Acid - Weak Base

Example: HCl titrated with NH₃

Initial pH: Low (strong acid present).

Before equivalence point: Buffer region! Mixture of weak base (NH₃) and its conjugate acid (NH₄⁺ from neutralization).

Equivalence point: All NH₃ converted to NH₄⁺ (a weak acid) which hydrolyzes:

\mathrm{NH_4}^{+} + \mathrm{H_2O} \rightleftharpoons \mathrm{NH_3} + \mathrm{H_3O}^{+}

pH at equivalence < 7 (typically 4-6) because conjugate acid is weakly acidic.

Vertical region: Smaller, in acidic range (typically pH 4-7). Must use indicator that changes in acidic range (like methyl orange). Phenolphthalein would NOT work.

Weak Acid - Weak Base (Not HSC Focus)

Example: CH₃COOH titrated with NH₃

Problem: No sharp vertical region. pH changes very gradually throughout the entire titration.

Equivalence point pH: Depends on relative strengths of acid and base. Could be acidic, neutral, or basic.

Practical issue: No common indicator works well because there's no sharp pH change. Not suitable for accurate volumetric analysis. This type is rarely used in practice.

Worked Example

Sketch and label the titration curve for 25.0 mL of 0.100 M CH₃COOH (K_a = 1.8 × 10⁻⁵) titrated with 0.100 M NaOH. Identify: (a) initial pH, (b) pH at half-equivalence, (c) pH at equivalence, (d) volume at equivalence, (e) buffer region.

(a) Initial pH (before any NaOH added):

0.100 M weak acid, use ICE table:

[H⁺] = √(K_a × c) = √(1.8 × 10⁻⁵ × 0.100) = 1.3 × 10⁻³ M

Initial pH = -log(1.3 × 10⁻³) = 2.89

(b) pH at half-equivalence point:

At half-equivalence, [CH₃COOH] = [CH₃COO⁻]

pH = pK_a = -log(1.8 × 10⁻⁵) = 4.74

All CH₃COOH converted to CH₃COO⁻ (weak base)

The conjugate base hydrolyzes, making solution basic

pH at equivalence ≈ 8.7 (calculation requires K_b and hydrolysis equilibrium)

(d) Volume at equivalence:

n(acid) = cV = 0.100 × 0.025 = 0.0025 mol

Need equal moles of NaOH: n(base) = 0.0025 mol

V(NaOH) = n/c = 0.0025/0.100 = 0.025 L = 25.0 mL

(e) Buffer region:

From start of titration up to near equivalence point (roughly 0-24 mL added)

Contains mixture of CH₃COOH and CH₃COO⁻

Best buffering at half-equivalence (12.5 mL added) where pH = pK_a

Curve characteristics: Starts at pH 2.89, gradual rise through buffer region, passes pH = pK_a = 4.74 at 12.5 mL, sharp rise to pH ≈ 8.7 at 25.0 mL equivalence, then levels off above pH 12 with excess NaOH.

Practice Question

Compare the titration curves for: (i) strong acid-strong base, (ii) weak acid-strong base. For each, state the pH at equivalence point and explain why they differ.

Show Answer

(i) Strong acid - Strong base (e.g., HCl + NaOH):

pH at equivalence point: 7.00 (neutral)

At the equivalence point, all HCl has reacted with NaOH to form NaCl and water: HCl + NaOH → NaCl + H₂O. The solution contains only Na⁺ and Cl⁻ ions plus water. Neither Na⁺ nor Cl⁻ undergoes hydrolysis (they're conjugates of a strong base and strong acid respectively, making them extremely weak). The solution is neutral, with pH = 7.00.

(ii) Weak acid - Strong base (e.g., CH₃COOH + NaOH):

pH at equivalence point: > 7 (typically 8-10, basic)

At the equivalence point, all CH₃COOH has reacted with NaOH to form CH₃COONa and water: CH₃COOH + NaOH → CH₃COO⁻ + Na⁺ + H₂O. The solution contains CH₃COO⁻ (ethanoate ion), which is the conjugate base of a weak acid. This conjugate base undergoes hydrolysis:

CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻

The ethanoate ion accepts protons from water, producing OH⁻ ions and making the solution basic. The pH at equivalence is typically 8-10 depending on the Ka of the weak acid.

Why they differ:

The difference comes down to what species are present at the equivalence point. Strong acid-strong base produces only spectator ions that don't affect pH, giving neutral pH 7. Weak acid-strong base produces the conjugate base of the weak acid, which is a weak base itself and undergoes hydrolysis to produce OH⁻, giving basic pH > 7.

This difference is critical for indicator selection: phenolphthalein (changes around pH 8-10) works for weak acid-strong base but both methyl orange and phenolphthalein work for strong acid-strong base.

Indicators

Acid-base indicators are weak acids or bases that change color over a specific pH range. Selecting the correct indicator is crucial for accurate endpoint detection in titrations.

How Indicators Work

An indicator is a weak acid (HIn) that exists in equilibrium between its acid form and conjugate base form, which have different colors:

\mathrm{HIn}{\text{(aq)}} \rightleftharpoons \mathrm{H}^{+}{\text{(aq)}} + \mathrm{In}^{-}{\text{(aq)}}

Acid form (HIn) has one color, base form (In⁻) has a different color.

In acidic solution: Equilibrium shifts left (by Le Chatelier). Mostly HIn present → acid color visible.

In basic solution: H⁺ ions are removed by OH⁻, shifting equilibrium right. Mostly In⁻ present → base color visible.

Transition range: The pH range over which the indicator changes color. Typically spans about 2 pH units centered on the indicator's pK_a.

Endpoint vs Equivalence point:

Equivalence point: When moles of acid = moles of base (stoichiometric point)
Endpoint: When the indicator changes color (visual observation)
Ideally, endpoint = equivalence point, but there's always a small difference

Common Indicators

Methyl Orange:

Transition range: pH 3.1 - 4.4

Color change: Red (acidic) → Orange → Yellow (basic)

Use for: Strong acid-strong base, Strong acid-weak base

NOT suitable for: Weak acid-strong base (equivalence pH too high)

Phenolphthalein:

Transition range: pH 8.3 - 10.0

Color change: Colorless (acidic) → Pink (basic)

Use for: Strong acid-strong base, Weak acid-strong base

NOT suitable for: Strong acid-weak base (equivalence pH too low)

Bromothymol Blue:

Transition range: pH 6.0 - 7.6

Color change: Yellow (acidic) → Green → Blue (basic)

Use for: Strong acid-strong base (equivalence at pH 7)

Litmus:

Transition range: pH 5 - 8

Color change: Red (acidic) → Blue (basic)

Too broad for accurate titrations, used mainly for quick acid/base tests

Selecting the Correct Indicator

Rule: Choose an indicator whose transition range overlaps with the vertical region of the titration curve.

Strong acid - Strong base:

Equivalence pH = 7, vertical region pH 4-10

Can use: Methyl orange OR Phenolphthalein OR Bromothymol blue

All work because vertical region is very large

Weak acid - Strong base:

Equivalence pH = 8-10, vertical region pH 7-10

Use: Phenolphthalein (changes at pH 8.3-10.0) ✓

DON'T use: Methyl orange (changes at pH 3.1-4.4, too low) ✗

Strong acid - Weak base:

Equivalence pH = 4-6, vertical region pH 4-7

Use: Methyl orange (changes at pH 3.1-4.4) ✓

DON'T use: Phenolphthalein (changes at pH 8.3-10.0, too high) ✗

Common exam question: Why is phenolphthalein suitable for weak acid-strong base but not strong acid-weak base? Because the equivalence point pH is basic (~8-9) for weak acid-strong base, falling in phenolphthalein's range, but acidic (~5) for strong acid-weak base, well below phenolphthalein's range.

Worked Example

A student titrates ammonia solution (weak base) with hydrochloric acid (strong acid). (a) What is the approximate pH at the equivalence point? (b) Which indicator should be used: methyl orange or phenolphthalein? (c) Explain your choice.

(a) pH at equivalence point:

This is a strong acid - weak base titration.

At equivalence, all NH₃ is converted to NH₄⁺ (ammonium ion).

NH₄⁺ is a weak acid that hydrolyzes:

$\mathrm{NH_4}^{+} + \mathrm{H_2O} \rightleftharpoons \mathrm{NH_3} + \mathrm{H_3O}^{+}$

This produces H₃O⁺, making the solution acidic.

pH at equivalence ≈ 5-6 (acidic)

(b) Which indicator:

Use: Methyl orange

Methyl orange changes color in the range pH 3.1-4.4, which overlaps with the vertical region of a strong acid-weak base titration curve (pH 4-7).

The equivalence point occurs at approximately pH 5-6, which is within reach of methyl orange's upper transition range. When the last drop of HCl is added to reach equivalence, the pH will be in the range where methyl orange changes from yellow to orange/red.

Phenolphthalein would NOT work because it changes color at pH 8.3-10.0, which is far above the equivalence point pH of ~5-6. The titration would be complete long before phenolphthalein changes color, leading to a large systematic error (overtitration).

Practice Question

Explain the difference between the equivalence point and the endpoint in a titration. Why is it important to choose an indicator whose transition range includes the equivalence point pH?

Show Answer

Equivalence point:

The equivalence point is the theoretical point in a titration where the number of moles of acid equals the number of moles of base (for a 1:1 stoichiometry). This is the stoichiometric point where the reaction is exactly complete. The equivalence point has a specific pH that depends on the strength of the acid and base being titrated.

Endpoint:

The endpoint is the practical point where we observe the indicator change color. This is when we stop the titration because we can see that the color has changed. The endpoint is what we can actually observe in the lab.

Why they should match:

Ideally, the endpoint should coincide exactly with the equivalence point so that we stop adding titrant at precisely the right moment. However, there's always a small difference between them because indicators change color over a pH range, not at a single pH value.

Importance of indicator selection:

If the indicator's transition range includes the equivalence point pH, then the indicator will change color very close to when the stoichiometric point is reached. This minimizes the difference between endpoint and equivalence point, giving accurate results.

For example, if titrating a weak acid with strong base (equivalence pH ~9), using phenolphthalein (changes at pH 8.3-10) means the color change occurs very close to equivalence. But using methyl orange (changes at pH 3.1-4.4) would mean the color changed long before reaching equivalence, and we'd stop the titration too early, getting completely wrong results.

The large vertical region in strong acid-strong base titrations means many indicators work because even if the endpoint is slightly different from the equivalence point, we're still in the vertical region where pH changes sharply, minimizing volume error. For weak acid/base titrations with smaller vertical regions, choosing the right indicator is more critical.

Buffer Solutions

Buffer solutions resist pH change when small amounts of acid or base are added. Understanding how buffers work is essential for biochemistry, industrial processes, and environmental chemistry.

What is a Buffer?

A buffer solution contains a weak acid and its conjugate base (or a weak base and its conjugate acid) in significant amounts. The presence of both species allows the buffer to neutralize added acid or base.

Acidic buffer: Weak acid + conjugate base (salt of the weak acid)

Example: CH₃COOH + CH₃COONa (ethanoic acid + sodium ethanoate)

Basic buffer: Weak base + conjugate acid (salt of the weak base)

Example: NH₃ + NH₄Cl (ammonia + ammonium chloride)

NOT buffers:

Strong acid + strong base (neutralize each other completely)
Weak acid alone (no conjugate base to neutralize added base)
Conjugate base alone (no weak acid to neutralize added acid)
Strong acid + its conjugate base (conjugate base of strong acid is extremely weak)

Buffer capacity: The amount of acid or base a buffer can neutralize before pH changes significantly. Depends on the concentrations of weak acid and conjugate base - higher concentrations give greater capacity.

How Buffers Work - Le Chatelier's Principle

Consider an ethanoic acid buffer: CH₃COOH + CH₃COO⁻

The equilibrium present is:

\mathrm{CH_3COOH}{\text{(aq)}} \rightleftharpoons \mathrm{H}^{+}{\text{(aq)}} + \mathrm{CH_3COO}^{-}{\text{(aq)}}

When acid (H⁺) is added:

The added H⁺ reacts with the conjugate base CH₃COO⁻:

\mathrm{CH_3COO}^{-}{\text{(aq)}} + \mathrm{H}^{+}{\text{(aq)}} \rightarrow \mathrm{CH_3COOH}{\text{(aq)}}

The conjugate base "mops up" the added H⁺, preventing a large pH drop. By Le Chatelier's principle, adding H⁺ shifts equilibrium left, consuming the added acid.

When base (OH⁻) is added:

The added OH⁻ reacts with the weak acid CH₃COOH:

\mathrm{CH_3COOH}{\text{(aq)}} + \mathrm{OH}^{-}{\text{(aq)}} \rightarrow \mathrm{CH_3COO}^{-}{\text{(aq)}} + \mathrm{H_2O}{\text{(l)}}

The weak acid neutralizes the added OH⁻, preventing a large pH rise. The OH⁻ removes H⁺ from the equilibrium, shifting it right and consuming the added base.

Result: The pH changes only slightly because the buffer components neutralize most of the added acid or base. The large reservoir of both CH₃COOH and CH₃COO⁻ means the equilibrium position barely shifts.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation provides a quick way to calculate buffer pH:

\text{pH} = \text{p}K_a + \log\frac{[\text{conjugate base}]}{[\text{weak acid}]}

Or equivalently:

\text{pH} = \text{p}K_a + \log\frac{[\mathrm{A}^{-}]}{[\mathrm{HA}]}

When [A⁻] = [HA]: pH = pK_a + log(1) = pK_a

This is the optimal buffer composition - equal amounts of weak acid and conjugate base.

Approximation: For buffers, we can use initial concentrations of HA and A⁻ because the equilibrium position is maintained by the buffer action. This avoids needing ICE tables.

Important: The ratio [A⁻]/[HA] determines pH, not the absolute concentrations. A buffer with 1.0 M acid and 1.0 M conjugate base has the same pH as one with 0.1 M of each (but different buffer capacity).

For basic buffers: Use the analogous equation with pK_b, or convert to the acidic form using K_a × K_b = K_w.

Biological and Environmental Buffers

Blood buffer system (carbonate buffer):

H₂CO₃/HCO₃⁻ maintains blood pH at 7.4

When you exercise, CO₂ increases, forming H₂CO₃. The HCO₃⁻ buffers the extra acid.

Hyperventilation removes CO₂, shifting equilibrium and raising blood pH (alkalosis).

Ocean buffer system:

CO₂ + H₂O ⇌ H₂CO₃ ⇌ H⁺ + HCO₃⁻

Oceans act as a CO₂ sink, but increased atmospheric CO₂ is gradually lowering ocean pH (ocean acidification).

Soil buffers:

Clay particles and organic matter buffer soil pH, affecting nutrient availability for plants.

Worked Example

A buffer contains 0.20 M CH₃COOH and 0.15 M CH₃COONa. K_a for ethanoic acid is 1.8 × 10⁻⁵. Calculate: (a) the pH of the buffer, (b) the pH after adding 0.01 mol HCl to 1.0 L of this buffer.

(a) Initial pH of buffer:

Using Henderson-Hasselbalch equation:

pK_a = -log(1.8 × 10⁻⁵) = 4.74

$\text{pH} = \text{p}K_a + \log\frac{[\mathrm{CH_3COO}^{-}]}{[\mathrm{CH_3COOH}]}$

$\text{pH} = 4.74 + \log\frac{0.15}{0.20}$

$\text{pH} = 4.74 + \log(0.75)$

$\text{pH} = 4.74 + (-0.125)$

$\text{pH} = \mathbf{4.62}$

(b) pH after adding 0.01 mol HCl:

The added H⁺ reacts with CH₃COO⁻:

CH₃COO⁻ + H⁺ → CH₃COOH

Initial moles in 1.0 L:

n(CH₃COOH) = 0.20 mol

n(CH₃COO⁻) = 0.15 mol

After reaction with 0.01 mol H⁺:

n(CH₃COOH) = 0.20 + 0.01 = 0.21 mol

n(CH₃COO⁻) = 0.15 - 0.01 = 0.14 mol

New concentrations (in 1.0 L):

[CH₃COOH] = 0.21 M

[CH₃COO⁻] = 0.14 M

New pH:

$\text{pH} = 4.74 + \log\frac{0.14}{0.21}$

$\text{pH} = 4.74 + \log(0.667)$

$\text{pH} = 4.74 + (-0.176)$

$\text{pH} = \mathbf{4.56}$

The pH changed from 4.62 to 4.56, a change of only 0.06 units! Without the buffer, adding 0.01 mol HCl to 1.0 L water would give pH = 2.

Practice Question

Explain using Le Chatelier's principle how a buffer solution containing NH₃ and NH₄Cl resists pH change when: (a) a small amount of HCl is added, (b) a small amount of NaOH is added. Write equations for each case.

Show Answer

The buffer contains NH₃ (weak base) and NH₄⁺ (its conjugate acid from NH₄Cl). The equilibrium present is:

$\mathrm{NH_3}{\text{(aq)}} + \mathrm{H_2O}{\text{(l)}} \rightleftharpoons \mathrm{NH_4}^{+}{\text{(aq)}} + \mathrm{OH}^{-}{\text{(aq)}}$

(a) When HCl (acid) is added:

The added H⁺ ions from HCl react with the weak base NH₃:

$\mathrm{NH_3}{\text{(aq)}} + \mathrm{H}^{+}{\text{(aq)}} \rightarrow \mathrm{NH_4}^{+}{\text{(aq)}}$

The ammonia "mops up" the added H⁺, preventing a large decrease in pH. By Le Chatelier's principle, adding H⁺ (which removes OH⁻ by forming H₂O) causes a disturbance. The equilibrium shifts right to minimize this disturbance, converting some NH₃ to NH₄⁺ and producing more OH⁻ to partially replace what was neutralized. Because there's a large reservoir of NH₃ present, most of the added H⁺ is consumed, and the pH drops only slightly.

(b) When NaOH (base) is added:

The added OH⁻ ions from NaOH react with the conjugate acid NH₄⁺:

$\mathrm{NH_4}^{+}{\text{(aq)}} + \mathrm{OH}^{-}{\text{(aq)}} \rightarrow \mathrm{NH_3}{\text{(aq)}} + \mathrm{H_2O}{\text{(l)}}$

The NH₄⁺ neutralizes the added OH⁻, preventing a large increase in pH. By Le Chatelier's principle, adding OH⁻ shifts the equilibrium left to minimize this disturbance, converting some NH₄⁺ to NH₃ and consuming the added OH⁻. Because there's a large reservoir of NH₄⁺ present, most of the added OH⁻ is consumed, and the pH rises only slightly.

Summary:

The buffer resists pH change because it contains both a weak base (NH₃) to neutralize added acid and a weak acid (NH₄⁺) to neutralize added base. The large concentrations of both species mean the equilibrium position barely shifts when small amounts of acid or base are added, keeping pH relatively constant. This is the essence of buffer action.

Enthalpy of Neutralisation

The enthalpy of neutralisation is the energy released when an acid and base react to form water. Measuring this quantity provides insight into acid-base strength and reaction energetics.

Definition and Theory

Enthalpy of neutralisation (ΔH_neut): The enthalpy change when one mole of water is formed from the reaction between an acid and a base.

Standard enthalpy of neutralisation: Measured under standard conditions (25°C, 100 kPa, 1 M solutions).

For strong acid + strong base:

\mathrm{H}^{+}{\text{(aq)}} + \mathrm{OH}^{-}{\text{(aq)}} \rightarrow \mathrm{H_2O}{\text{(l)}}

ΔH_neut = -57.3 kJ/mol (always exothermic)

This value is constant for all strong acid-strong base neutralizations because the only reaction occurring is H⁺ + OH⁻ → H₂O. The spectator ions don't participate.

For weak acid or weak base: ΔH_neut is less exothermic (smaller magnitude) than -57.3 kJ/mol because some energy is required to ionize the weak acid or weak base before neutralization can occur.

Important: Neutralisation is ALWAYS exothermic (ΔH negative). Never write a positive ΔH value for neutralisation!

Calculating Heat Released

Use calorimetry to measure temperature change and calculate heat released:

q = mc\Delta T

Where:

q = heat absorbed by solution (in joules, J)
m = mass of solution (assume density ≈ 1 g/mL, so mass ≈ volume in mL)
c = specific heat capacity of solution (assume 4.18 J/(g·°C), same as water)
ΔT = temperature change (T_final - T_initial)

Then calculate ΔH_neut per mole:

\Delta H_{neut} = -\frac{q}{n}

Where n = moles of water formed (equals moles of limiting reactant)

Sign convention: If temperature increases (exothermic), q is positive (heat absorbed by solution), so ΔH = -q/n is negative. The negative sign accounts for the fact that the reaction releases heat.

Experimental Method

Procedure:

Step 1: Measure equal volumes of acid and base of known concentrations (e.g., 50 mL each of 1.0 M HCl and 1.0 M NaOH) into separate containers.

Step 2: Measure initial temperature of both solutions. They should be at room temperature and the same temperature.

Step 3: Mix the solutions in a polystyrene cup (good insulator) and stir continuously.

Step 4: Record the maximum temperature reached. This is T_final.

Step 5: Calculate ΔT, then q, then ΔH.

Sources of error:

Heat loss to surroundings (use insulated container, lid, minimize time)
Heat absorbed by container (use polystyrene cup with low heat capacity)
Incomplete mixing (stir thoroughly)
Evaporation (use lid)
Inaccurate temperature measurement (use digital thermometer, read immediately at maximum)

Why polystyrene cup? Polystyrene is an excellent insulator with very low thermal conductivity and heat capacity. This minimizes heat loss to surroundings and heat absorption by the cup itself, making the assumption that all heat stays in the solution more accurate.

Worked Example

A student mixed 50.0 mL of 1.00 M HCl with 50.0 mL of 1.00 M NaOH in a polystyrene cup. The initial temperature was 20.0°C and the maximum temperature reached was 26.7°C. Calculate the enthalpy of neutralisation. (Assume density = 1.00 g/mL, c = 4.18 J/(g·°C))

Given:

V_acid = 50.0 mL, [HCl] = 1.00 M

V_base = 50.0 mL, [NaOH] = 1.00 M

T_i = 20.0°C, T_f = 26.7°C

c = 4.18 J/(g·°C), density = 1.00 g/mL

Step 1: Calculate moles of reactants:

n(HCl) = cV = 1.00 × 0.0500 = 0.0500 mol

n(NaOH) = cV = 1.00 × 0.0500 = 0.0500 mol

Both are equal, so neither is limiting. Moles of water formed = 0.0500 mol

Step 2: Calculate mass of solution:

Total volume = 50.0 + 50.0 = 100.0 mL

Mass = 100.0 mL × 1.00 g/mL = 100.0 g

Step 3: Calculate temperature change:

ΔT = T_f - T_i = 26.7 - 20.0 = 6.7°C

Step 4: Calculate heat absorbed by solution:

q = mcΔT

q = 100.0 × 4.18 × 6.7

q = 2801 J = 2.80 kJ

Step 5: Calculate ΔH_neut per mole:

$\Delta H_{neut} = -\frac{q}{n}$

$\Delta H_{neut} = -\frac{2.80}{0.0500}$

$\Delta H_{neut} = \mathbf{-56.0 \text{ kJ/mol}}$

This is close to the theoretical value of -57.3 kJ/mol. The small difference is due to heat loss to the surroundings and the cup.

Important: The negative sign indicates the reaction is exothermic (releases heat). Always include the negative sign in your final answer!

Practice Question

Explain why: (a) neutralisation reactions are always exothermic, (b) the enthalpy of neutralisation for a weak acid with a strong base is less exothermic than for a strong acid with a strong base, (c) the negative sign must be included in ΔH_neut values.

Show Answer

(a) Why neutralisation is always exothermic:

Neutralisation involves the formation of a very strong O-H bond in water from H⁺ and OH⁻ ions:

H⁺(aq) + OH⁻(aq) → H₂O(l)

The formation of chemical bonds always releases energy. The O-H bond in water is particularly strong (463 kJ/mol), and forming this bond from separated ions in solution releases a large amount of energy. Additionally, the hydration energy released when ions are surrounded by water molecules contributes to the exothermic nature. Since bond formation releases more energy than is required to break the weak interactions between the ions and water, the overall process is exothermic.

(b) Why weak acid + strong base is less exothermic:

For a strong acid (e.g., HCl) with a strong base, the only process is:

H⁺(aq) + OH⁻(aq) → H₂O(l) ΔH = -57.3 kJ/mol

For a weak acid (e.g., CH₃COOH) with a strong base, two processes occur:

1. Ionization of weak acid: CH₃COOH(aq) → H⁺(aq) + CH₃COO⁻(aq) ΔH = positive (endothermic)

2. Neutralization: H⁺(aq) + OH⁻(aq) → H₂O(l) ΔH = -57.3 kJ/mol (exothermic)

The weak acid must first ionize (which requires energy) before neutralization can occur. This ionization step is endothermic, absorbing some energy. The net enthalpy change is the sum of the endothermic ionization and the exothermic neutralization. Therefore, the overall ΔH_neut is less negative (less exothermic) than -57.3 kJ/mol.

Example: Ethanoic acid + NaOH gives ΔH_neut ≈ -55 kJ/mol, compared to -57.3 kJ/mol for HCl + NaOH.

The sign of ΔH indicates the direction of heat flow. A negative ΔH means the system releases heat (exothermic). A positive ΔH means the system absorbs heat (endothermic).

In our calculation, we use q = mcΔT to find the heat absorbed by the solution. For an exothermic reaction, the solution temperature increases, making ΔT positive and q positive. This q represents heat gained by the surroundings (the solution).

From the system's perspective (the chemical reaction), if the surroundings gained heat, the system must have lost heat. Therefore, we must include the negative sign: ΔH = -q/n. This ensures ΔH is negative, correctly indicating that the reaction releases heat.

Without the negative sign, we would incorrectly state that neutralisation is endothermic, which contradicts both theory and experimental observation. The negative sign is not optional—it's a fundamental part of thermodynamic sign conventions and indicates the exothermic nature of neutralisation reactions.

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CH₃COOH

H₃O⁺

CH₃COO⁻

0.10

≈0

-x

0.10 - x