Equilibrium · Lesson 12 of 14
Salt Hydrolysis and Buffer Solutions
“Predict the pH of salt solutions and design mixtures that resist small pH disturbances.”
• Identify which salt ions can react with water and predict solution character. • Use conjugate constants to estimate hydrolysis and salt-solution pH. • Explain buffer action and derive the acid and base buffer relationships. • Choose a useful conjugate-pair ratio and account for buffer changes after neutralization.
Dissolving a salt is not the same as hydrolysing its ions
A soluble salt may dissociate completely, yet its ions can react further with water. This proton-transfer reaction is called hydrolysis. The resulting pH depends on the acid–base properties of the ions, not simply on the fact that the material is a salt. Sodium chloride is approximately neutral in the simple model, whereas sodium acetate is basic and ammonium chloride is acidic.
Ions such as Na+ and Cl− are extremely weak acid–base partners in water and are treated as spectators here. Acetate, the conjugate base of a weak acid, accepts a proton from water to form acetic acid and hydroxide. Ammonium, the conjugate acid of a weak base, donates a proton to water to form ammonia and hydronium. This is hydrolysis expressed in modern proton-transfer notation.
| Parent acid and base | Typical salt | Important hydrolysis | Expected character |
|---|---|---|---|
| Strong acid + strong base | NaCl | Negligible in the simple model | Neutral |
| Weak acid + strong base | CH3COONa | Anion accepts a proton | Basic |
| Strong acid + weak base | NH4Cl | Cation donates a proton | Acidic |
| Weak acid + weak base | CH3COONH4 | Both ions participate | Depends on their strengths |
Problem
Estimate pH and degree of hydrolysis of 0.040 M NaNO2 at 298 K when .
- 1.Nitrite is a base: . Its constant is .
- 2.For small hydrolysis, .
- 3.The pOH is about 6.026 and pH about 7.974. The hydrolysed fraction is , about 0.00236%. Water gives a small correction; this is the ordinary hydrolysis approximation.
When both salt ions hydrolyse, compare the original weak acid’s Ka with the original weak base’s Kb. Under the usual dilute weak-acid/weak-base salt model, their competition gives the relationship below. The approximate pH is largely independent of salt concentration within the range where that model holds. At extreme dilution, water becomes important.
Problem
For ammonium acetate at 298 K, use acetic-acid pKa = 4.76 and ammonia pKb = 4.75. Estimate its pH.
- 1.Both acetate and ammonium react with water, so use the weak-acid–weak-base salt relationship.
- 2.Substitute .
- 3.The model gives pH approximately 7.005, very close to neutral. This fine numerical difference should not be interpreted as a highly precise experimental prediction from rounded constants.
A buffer contains partners that consume added acid and base
A buffer is a solution that resists a large pH change when small amounts of acid or base are added. A useful acidic buffer contains a weak acid and a substantial amount of its conjugate base, often supplied by a salt. Added acid is consumed by the conjugate base; added hydroxide is consumed by the weak acid. Both partners must remain after the addition.
Acetic acid and sodium acetate form a familiar pair. Ammonia and ammonium chloride form a basic buffer: added acid consumes NH3, while added base consumes NH4+. Biological fluids and many formulations depend on controlled pH. Buffering is finite, however; a sufficiently large addition can exhaust one component and produce a large change.
Derive the buffer ratio from the acid equilibrium
Start with the weak-acid expression and rearrange it to isolate hydronium: its concentration is Ka times acid concentration divided by conjugate-base concentration. Taking minus logarithms converts the product into a sum and the concentration quotient into a logarithmic ratio. This is the Henderson–Hasselbalch equation in the dilute approximation.
At equal conjugate-base and acid concentrations, the logarithm is zero and pH is approximately pKa. A tenfold larger base-to-acid ratio raises pH by one. Choose an acid with pKa near the desired pH so both partners are present in useful quantities. Dilution that changes both concentrations by the same factor leaves their ratio nearly unchanged, but reduces the amount of acid or base the buffer can absorb per unit volume. At sufficiently extreme dilution the approximation fails.
Problem
Using acetic-acid pKa = 4.76, choose the acetate-to-acetic-acid concentration ratio for pH 5.06.
- 1.Rearrange the buffer equation: .
- 2.Take the antilogarithm: .
- 3.Use approximately twice as much acetate as acetic acid in molar terms, for example analytical concentrations near 0.100 M and 0.050 M after mixing. The ratio determines the approximate pH; the total concentrations also affect capacity.
Keep the basic-buffer ratio in the correct direction
For a base buffer, begin with Kb and isolate hydroxide. This gives pOH as pKb plus the logarithm of conjugate acid divided by base. Converting pOH to pH reverses the sign of that logarithm. Therefore the pH form uses base divided by conjugate acid, not the same ratio as the pOH form.
Problem
At 298 K, a buffer contains 0.100 M ammonia and 0.200 M ammonium. Use ammonia pKb = 4.75.
- 1.The pOH form gives .
- 2.Thus pH is .
- 3.As a cross-check, ammonium pKa is 9.25, so the pH form gives . Both methods agree.
Problem
A 1.00 L buffer initially contains 0.100 mol acetate and 0.100 mol acetic acid. Add 0.0100 mol HCl with negligible volume change. Estimate the new pH using pKa = 4.76.
- 1.The added strong acid reacts with acetate: 0.0100 mol A− is consumed and 0.0100 mol HA is formed.
- 2.Remaining amounts are 0.0900 mol A− and 0.1100 mol HA. Because both occupy the same final volume, their mole ratio equals their concentration ratio.
- 3.The new pH is . The change from 4.76 is small because both buffer partners remain.
To investigate salt hydrolysis, measure comparable salt solutions with pH paper and infer which ions are acidic or basic. To investigate buffering, prepare an acetate mixture of known ratio and compare its response to a small acid or base addition with the response of an equal volume of water. Record initial and final pH and the added amounts. Indicator changes or meter readings must be interpreted with their measurement resolution; the purpose is to compare response, not merely label a solution a buffer.
A salt alone is not automatically a buffer, and a buffer does not keep pH perfectly fixed for unlimited additions. Account for neutralization before using a buffer ratio. For a basic buffer, pOH uses BH+/B while pH uses B/BH+.
Quiz
A sodium acetate solution is basic mainly because
A mixture with equal weak acid and conjugate base concentrations has approximate pH
For a basic buffer, which ratio belongs in pH = pKa(BH+) + log(ratio)?
Ordinary dilution of both buffer components by the same factor generally
Before applying a buffer formula after adding strong acid, calculate
Ammonium chloride is acidic because
Practice Problems
- Predict whether NaCl, KBr, NaCN, NH4NO3, NaNO2 and KF solutions are acidic, neutral or basic in the usual model.
- Calculate approximate pH of 0.10 M sodium acetate using Ka(acetic acid) = at 298 K.
- Find the acetate-to-acid ratio needed for pH 4.46 if pKa = 4.76.
- Explain how an ammonia–ammonium buffer consumes a little added strong acid and a little added hydroxide.
- A buffer has 20 mmol HA and 30 mmol A−. Find its new ratio after addition of 5 mmol HCl, provided both components remain.
- Explain why a measured salt-solution pH can be near neutral even when both ions hydrolyse.
Key Takeaways
Salt ions can hydrolyse even when the salt dissociates completely. Buffers contain a weak species and its conjugate partner, each able to consume a different disturbance. Derive and check the ratio direction, use post-neutralization amounts, and remember that buffer capacity is finite.