Equilibrium · Lesson 11 of 14
Polyprotic Acids, Acid Strength, and Common Ions
“Explain stepwise proton loss, structural acid-strength trends, and suppression by an ion already present.”
• Write separate ionization steps and constants for polyprotic acids. • Explain why successive proton losses usually become less favourable. • Use bond strength and bond polarity to interpret simple acid-strength trends. • Calculate common-ion suppression after accounting for any neutralization first.
Several protons are lost in separate equilibria
An acid with more than one ionizable proton is polyprotic. Each proton transfer has its own equilibrium constant. A diprotic acid H2A first becomes HA− and then A2−. It is usually harder to remove a positive proton from a species that is already negatively charged, so the second constant is generally smaller than the first. A triprotic acid has three steps rather than one event releasing three protons equally readily.
| Acid | Ka1 | Ka2 | Ka3 where applicable |
|---|---|---|---|
| Oxalic acid | — | ||
| Ascorbic acid | — | ||
| Sulfurous acid | — | ||
| Sulfuric acid | Very large | — | |
| Carbonic acid | — | ||
| Citric acid | |||
| Phosphoric acid |
When successive constants are very different, the first step often supplies most of the hydronium. This does not make every polyprotic acid a strong acid, and it does not mean the later steps vanish. Sulfuric acid illustrates the distinction: its first proton is strongly released in water, but its second must be treated with its own finite constant. Species distributions and common ions can change which contribution matters.
Problem
For a dilute 0.10 M H2S model, use and . Estimate HS− and S2− concentrations without added acid.
- 1.The first step dominates hydronium production. Estimate .
- 2.For the second step, .
- 3.Since the hydronium and HS− concentrations are approximately equal, . The second step contributes negligible additional hydronium, supporting the model.
Bond strength and polarity explain different trends
A proton is transferred more readily when the H–A bond can be broken favourably and the resulting species are stabilised. For simple hydrides down a group, increasing atom size weakens the H–A bond and often dominates the trend. Thus aqueous hydrogen halide acidity increases from HF to HCl, HBr and HI despite fluorine’s greater electronegativity. Hydrogen sulfide is also more acidic than water in this comparison.
Across a period for the simple hydrides, increasing electronegativity of A makes the H–A bond more polar and stabilises negative charge on the conjugate base. The comparison CH4, NH3, H2O and HF therefore shows increasing acidity in that order. These are patterns for related compounds under comparable conditions, not a universal rule that bond polarity alone predicts every acid’s strength.
A common ion suppresses further ionization
Acetic acid produces hydronium and acetate. Adding acetate from a soluble salt supplies an ion already present in the dissociation equilibrium. The resulting quotient initially increases, and net protonation restores the equilibrium ratio. Adding a strong acid similarly suppresses further acetic-acid ionization by supplying hydronium. The degree of ionization changes; Ka at fixed temperature does not.
A change in equilibrium caused by adding a substance that supplies an ion already participating in that equilibrium. For a weak acid or weak base, an appropriate common ion usually suppresses additional ionization.
Problem
A solution is 0.050 M in acetic acid and 0.050 M in sodium acetate. Use to estimate hydronium concentration and pH.
- 1.Let x be the extra acid ionization: , and .
- 2.Write . Since x is tiny compared with 0.050, the ratio of those two terms is approximately one.
- 3.Thus and pH approximately 4.745. The additional ionization is about , far smaller than for the acid alone.
Problem
Estimate HS− and S2− in the 0.10 M H2S solution above when it also contains 0.10 M HCl, using the dilute concentration model.
- 1.The strong acid dominates: . The first step gives .
- 2.Use the second equilibrium: .
- 3.The result is . Added acid suppresses both ionization steps. At 0.10 M, real activity corrections may matter; these are the stated concentration-model results.
Neutralization comes before the equilibrium calculation
When strong acid is mixed with ammonia, it first transfers protons to ammonia almost completely. Use moles or millimoles to determine how much unreacted NH3 and newly formed NH4+ remain. Only then divide by the total volume and apply the weak-base equilibrium. Starting with a square-root formula before the stoichiometric reaction would miss the common ion that the mixing itself creates.
Problem
Find the approximate pH before and after 50.0 mL of 0.100 M NH3 is mixed with 25.0 mL of 0.100 M HCl at 298 K. Use .
- 1.Before mixing, , so pH approximately 11.12.
- 2.There are 5.00 mmol NH3 and 2.50 mmol HCl. Proton transfer consumes 2.50 mmol NH3 and creates 2.50 mmol NH4+.
- 3.In 75.0 mL, both remaining NH3 and formed NH4+ have concentration about 0.0333 M. Thus because the conjugate-acid/base ratio is approximately one.
- 4.The hydroxide concentration is about , giving pOH 4.752 and pH 9.248. The mixture remains basic, but much less so than the initial ammonia.
Adding a common ion does not change Ka or Kb at fixed temperature. For polyprotic species, write the actual ionization step and preserve charges; do not release every proton as though all steps had the same strength.
Quiz
Why is a second proton generally harder to remove from a diprotic acid?
Which order represents increasing hydrogen halide acidity in water?
Adding sodium acetate to acetic acid at fixed temperature generally
Before applying equilibrium to a mixture of strong acid and excess ammonia, first calculate
In a polyprotic-acid solution, different ionization steps share
Across CH4, NH3, H2O and HF, increasing acidity is associated with
Practice Problems
- Write the three sequential proton-loss equilibria of phosphoric acid with correct charges.
- Explain why H2CO3 and HCO3− have different acid constants.
- Compare the roles of H–A bond strength and polarity in down-group and across-period acidity trends.
- Estimate the degree of ionization of 0.050 M acetic acid with Ka = when the solution also contains 0.10 M HCl.
- For 0.020 M dimethylamine with Kb = , compare ionization in pure water and with 0.10 M NaOH using a checked model.
- Mix 40.0 mL of 0.10 M NH3 with 10.0 mL of 0.10 M HCl. Find the amounts of NH3 and NH4+ before calculating the equilibrium pH.
Key Takeaways
Polyprotic acids lose protons in separate equilibria with separate constants. Structural comparisons require the appropriate balance of bond strength, polarity and conjugate-base stability. Common ions suppress additional ionization; when mixing reactive solutions, complete the stoichiometric accounting first.