Chemistry · Lesson
Weak acids, weak bases and polyprotic systems
Solve a weak-acid or weak-base equilibrium with the required inventory. Check reduced-model assumptions and treat each polyprotic transfer as a separate equilibrium.
Back to the Chemistry mapBy the end, you can
- Combine ionization with charge and analytical material balance.
- Check a small-change approximation against a reduced quadratic model.
- Use step-specific constants for polyprotic species.
Write the species and balances first
A hypothetical neutral monoprotic acid follows HA + H₂O ⇌ H₃O⁺ + A⁻. In a dilute ideal model with water activity near 1, fixed temperature, no other solutes and negligible water autoionization, set [H₃O⁺] = [A⁻] = x and [HA] = C − x. The reduced charge balance and analytical acid total then match.
Use dimensionless y = x/c° and C_n = C/c°, with c° = 1 mol L⁻¹. The reduced equation is K_a° = y²/(C_n − y). The word reduced matters: including water’s hydroxide contribution changes the charge equation slightly. Check that it is small before treating H₃O⁺ and A⁻ as equal.
Test the tempting square-root shortcut
The original hypothetical model chooses C = 0.0100 mol L⁻¹ and K_a° = 1.00 × 10⁻⁵. Treat these as exact chosen numerical parameters within the reduction. Solving y² + K_a°y − K_a°C_n = 0 gives x ≈ 3.11267 × 10⁻⁴ mol L⁻¹ and remaining HA ≈ 0.00968873 mol L⁻¹. Model pH is approximately 3.507.
Assuming x is tiny compared with C gives x ≈ c°√(K_a°C_n) ≈ 3.16228 × 10⁻⁴ mol L⁻¹, about 1.59% higher than the quadratic result. The calculated ionization is about 3.11%; a five-percent inventory check does not mean every quoted pH digit is unaffected by the shortcut.
| Quantity | Reduced quadratic result | Square-root shortcut | Interpretation |
|---|---|---|---|
| Hydronium x / mol L⁻¹ | 3.11267 × 10⁻⁴ | 3.16228 × 10⁻⁴ | Shortcut is about 1.59% higher |
| Remaining HA / mol L⁻¹ | 0.00968873 | Use C − estimated x if checking | HA + A⁻ must retain the total |
| Fraction ionized | About 3.11% | Assumes x ≪ C | A small fraction still needs a precision check |
| Water-model check at 25 °C | OH⁻ ≈ 3.21 × 10⁻¹¹ mol L⁻¹ | Ordinary K_w ≈ 10⁻¹⁴ | Negligible compared with this model hydronium |
Original ideal reduced HA model, fixed stipulated temperature, C=0.0100 mol L⁻¹, c°=1 mol L⁻¹, K_a°=1.00×10⁻⁵ as exact chosen parameters. No other solutes; water activity≈1 and autoionization negligible. Displayed values are rounded model calculations, not measured acid properties.
A weak base needs the hydroxide ledger
For a separate neutral base B + H₂O ⇌ BH⁺ + OH⁻, the corresponding reduced model uses [BH⁺] = [OH⁻] = x, [B] = C − x and K_b° = y²/(C_n − y). If C and K_b° have the same chosen numerical values as the acid example, x is the same numerical concentration, now of hydroxide and BH⁺.
This gives pOH ≈ 3.507. With approximate pK_w = 14.00 at 25 °C, pH ≈ 10.49. Calling x hydronium in this base calculation would reverse the interpretation. The water convention and its reporting precision must be retained when converting pOH to pH.
Each transferred proton has its own relation
A hypothetical diprotic species can have H₂A + H₂O ⇌ HA⁻ + H₃O⁺, followed by HA⁻ + H₂O ⇌ A²⁻ + H₃O⁺. Each step conserves the constituent A and transfers one proton; the charge balances are 0 = −1 + 1 and −1 = −2 + 1. The two step constants describe different equilibria.
If a supplied state has K_a2° = 1.00 × 10⁻⁷ and a_H₃O⁺ = 1.00 × 10⁻³ with water activity near 1, then a_A²⁻/a_HA⁻ = 1.00 × 10⁻⁴. That ratio does not provide the whole composition. Full speciation also needs the first constant, total inventory and charge balance. Do not count every H in a molecular formula as an independently fully released proton.
Worked examples
Solve the reduced acid model
Use exact chosen C_n = 0.0100 and K_a° = 1.00 × 10⁻⁵ with c° = 1 mol L⁻¹ in the declared reduced model. Find hydronium and model pH.
- Solve y² + K_a°y − K_a°C_n = 0 for the nonnegative root.
- y = [−K_a° + √((K_a°)² + 4K_a°C_n)]/2 ≈ 0.0003112673.
- x = c°y ≈ 3.11267 × 10⁻⁴ mol L⁻¹.
- Check y²/(C_n − y) ≈ 1.00 × 10⁻⁵ and HA+A⁻ = C.
- Model pH = −log₁₀y ≈ 3.5068665.
Hydronium ≈ 3.11 × 10⁻⁴ mol L⁻¹ and model pH ≈ 3.507. This is exact-quadratic algebra within a declared approximation, not an exact real-water measurement.
Try it yourself
What analytical balance holds for the reduced neutral HA model?
Show the answer and reasoning
[HA] + [A⁻] = C.
Ionization redistributes the conserved acid constituent.
What omitted contribution must be checked before setting [H₃O⁺] = [A⁻]?
Show the answer and reasoning
Water’s hydroxide contribution in the charge balance.
The reduced equality is adequate only when that contribution is negligible.
In the stated neutral weak-base reduction, does x represent hydronium or hydroxide?
Show the answer and reasoning
Hydroxide, alongside BH⁺.
The base receives a proton from water and produces OH⁻.
Does a diprotic formula imply two complete proton releases in every solution?
Show the answer and reasoning
No.
Each step has its own equilibrium relation and composition-dependent extent.
What to carry forward
- Use analytical and charge balances alongside ionization constants.
- A quadratic can be exact within a deliberately reduced model.
- Assess approximation errors before reporting extra digits.
- Treat polyprotic transfers step by step.
References
Teaching resources and reference material supporting this lesson’s explanations and examples.