Acid-Base Equilibria: pKa, Strong vs Weak Acids, and Ammonium Chloride Case Study
Key concepts
- Strong acid definition: a strong acid completely dissociates in water, donating its proton to water. For HCl, this means the acid exists essentially as H3O+ and Cl− in solution, with negligible undissociated HCl remaining.
- Protonation state vs pH: the protonation state of an acid/base in solution depends on pH relative to the acid’s pKa. If pH is low (acidic), the conjugate base form is suppressed; if pH is high (basic), the protonated form is suppressed.
- pH scale and extremes: negative pH values are theoretically possible for very strong acids or highly concentrated solutions, but in many classroom contexts, pH is treated within the 0–14 range for practical aqueous solutions. If pH equals the acid’s pKa, the protonated and deprotonated forms are present in equal amounts.
pH = pKa and acid-base distribution
- For a weak acid HA with conjugate base A− in water, the ratio is governed by the Henderson–Hasselbalch relationship in the implied form:
- rac{[A^-]}{[HA]} = 10^{\mathrm{pH} - \mathrm{p}K_a(HA)}.
- Therefore:
- At pH = pKa, [HA] = [A−] (equal protonated and deprotonated forms).
- If pH < pKa, the protonated form HA dominates.
- If pH > pKa, the deprotonated form A− dominates.
- Example: HF with pKa ≈ 3.2. If pH = 3.2, [HF] ≈ [F−]. If you lower the pH to 1, the equilibrium shifts toward HF (more protonated) because the solution becomes more acidic.
pKa values (selected references from the discussion)
- Hydronium acid: \mathrm{p}Ka(\mathrm{H3O^+}) \approx -1.74.
- Water as an acid (to form OH−): \mathrm{p}Ka(\mathrm{H2O}) \approx 15.7.
- Hydrofluoric acid: \mathrm{p}K_a(\mathrm{HF}) \approx 3.17\text{–}3.2.
- Hydrochloric acid: \mathrm{p}K_a(\mathrm{HCl}) \approx -7.
- Hydrobromic acid: \mathrm{p}K_a(\mathrm{HBr})\approx -9\text{ (approx.)}
- Hydroiodic acid: \mathrm{p}K_a(\mathrm{HI})\approx -10.
- Ammonium ion: \mathrm{p}Ka(\mathrm{NH4^+}) \approx 9.25.
- Ammonia (as a base has related value): Kb(\mathrm{NH3}) \approx 1.8\times 10^{-5}.
Relationships between pKa and equilibrium constants for acid-base reactions
- Consider an acid-base exchange: HA1 + B− ⇌ A1− + HB2, where HA1 and HB2 are acids with conjugate bases A1− and B2−.
- The equilibrium constant for this reaction can be estimated from the acids’ strengths as:
- K = 10^{\mathrm{p}Ka(\text{HB2}) - \mathrm{p}Ka(\text{HA1})}
- This follows from the idea that the reaction transfers a proton from HA1 (acid on the left) to HB2 (acid on the right).
- If you know the two pKa values, you can estimate K this way. A short derivation is available in the course materials (D2L).
Numerical example using the NH4+/H2O system
- Reaction: \mathrm{NH4^+} + \mathrm{H2O} \rightleftharpoons \mathrm{NH3} + \mathrm{H3O^+}
- Here, the reactant acid is NH4+ (pKa ≈ 9.25) and the product acid is H3O+ (pKa ≈ -1.74).
- Therefore, the equilibrium constant is:
- K = 10^{\mathrm{p}Ka(\mathrm{H3O^+}) - \mathrm{p}Ka(\mathrm{NH4^+})} = 10^{(-1.74) - (9.25)} \approx 10^{-10.99} \approx 10^{-11}.
- Interpretation: This equilibrium lies overwhelmingly to the left (ammonium is a much stronger acid than water’s conjugate acid here), so only a tiny fraction of NH4+ reacts with water to form NH3 and H3O+; dissolution of ammonium chloride in water mainly yields NH4+ and Cl−.
A practical exercise: identifying acids and bases and judging which reactions are favorable
- Given an acid-base pair, identify potential acids and bases, and decide which proton transfer will dominate by comparing pKa values.
- Example considerations from the transcript:
- HCl is a strong acid; Cl− is a weak base.
- NH4+ could act as an acid, and NH3 could act as a base.
- Cl− cannot act as an acid (no proton to donate) but can act as a base under some scenarios.
- NH4+ cannot act as a base in this context because it is already protonated (it would need to lose a proton to act as a base, which is not favorable here).
- For the ammonium chloride/water system, three potential processes were considered:
1) NH4+ + H2O ⇌ NH3 + H3O+ with K \approx 10^{(-1.74) - 9.25} \approx 10^{-11}.
2) NH3 + HCl ⇌ NH4+ + Cl− with K \approx 10^{9.25 - (-7)} = 10^{16.25}, a very large value, so the reaction proceeds strongly to the right (NH4+ formation) when NH3 and HCl are present.
3) NH4+ + Cl− ⇌ NH3 + HCl with K \approx 10^{(-7) - 9.25} = 10^{-16.25}, a very small value, so this reaction is strongly suppressed. - Ranking by favorability (largest K to smallest): 2 >> 1 >> 3.
- Practical conclusion: among these, the most favorable reaction is NH3 + HCl → NH4+ + Cl−; NH4+ + H2O ⇌ NH3 + H3O+ is the least favorable of the three, and NH4+ + Cl− ⇌ NH3 + HCl is negligible under typical conditions.
Real-world implications and notes
- In a solution of ammonium chloride in water, the predominant species are NH4+ and Cl−; only a tiny fraction of NH4+ will deprotonate to NH3 by reaction with water, due to the large pKa gap.
- Spectator ions: Na+ or K+ often accompany such salts in solution, but they do not participate directly in the acid-base equilibria discussed here.
- The pKa table is not exhaustive for every acid encountered in a course, but it is sufficiently populated to estimate relative acid strengths and to predict the direction of proton transfers in many common cases.
Quick summary recap
- Strong acids dissociate completely; weak acids exist in equilibrium with their conjugate bases.
- The position of an acid-base equilibrium can be estimated using pKa values: for HA1 + B− ⇌ A1− + HB2, K ≈ 10^{\mathrm{p}Ka(\mathrm{HB2}) - \mathrm{p}Ka(\mathrm{HA1})}.
- When pH = pKa, you have equal amounts of the acid and its conjugate base.
- In systems like ammonium chloride in water, the most favorable proton-transfer process is NH3 + HCl → NH4+ + Cl−; direct proton transfer from NH4+ to water is very unfavorable (K ≈ 10^{-11}).
Additional context (experimental mindset)
- For any given acid-base question in this course, identify all species that can act as acids or bases, compare their pKa values, and predict which proton transfers are favored.
- If you want to see a derivation of the K = 10^{pKa(product) − pKa(reactant)} relationship, refer to the D2L materials or ask during office hours.