Difficulty: Intermediate to Advanced | Prerequisites: acid-base equilibria, Ka/Kb, pH calculations, ICE tables.
Buffers and titrations are the capstone of the acid-base unit and among the most calculation-heavy topics on the CHM 11200 final. A buffer is a solution that resists pH changes when small amounts of acid or base are added. A titration is the controlled addition of one reactant to another to reach an equivalence point. Both rely on weak acid/base equilibrium, so if you are shaky on Ka, Kb, and ICE tables, revisit those notes first. This material ties together everything from the previous two sets of notes.
A buffer requires a weak acid and its conjugate base (or a weak base and its conjugate acid) in appreciable amounts. The Henderson-Hasselbalch equation (pH = pKa + log([base]/[acid])) is the workhorse for buffer pH calculations. Titration curves show how pH changes as titrant is added; the equivalence point is where moles of acid equal moles of base, and the half-equivalence point is where pH = pKa.
Buffer
A solution that resists changes in pH upon the addition of small amounts of strong acid or strong base. Made from a weak acid and its conjugate base, or a weak base and its conjugate acid.
In simple terms: a pH shock absorber. Add a bit of acid or base and the pH barely moves.
Henderson-Hasselbalch equation
pH = pKa + log([A⁻] / [HA]), where [A⁻] is the conjugate base concentration and [HA] is the weak acid concentration.
Think of it as: the one equation that handles nearly every buffer pH problem.
Buffer capacity
The amount of strong acid or base a buffer can absorb before the pH changes significantly. Higher concentrations of the buffer components give greater capacity.
In simple terms: a more concentrated buffer can soak up more added acid or base before it gives out.
Buffer range
The pH range over which a buffer works effectively, typically pKa ± 1.
Think of it as: a buffer is most useful when the desired pH is within one unit of its acid's pKa.
Equivalence point
The point in a titration where the moles of titrant added exactly equal the moles of analyte present. For a strong acid/strong base titration, pH = 7. For a weak acid titrated with a strong base, pH > 7.
Half-equivalence point
The point in a titration where exactly half the analyte has been neutralised. At this point, [HA] = [A⁻], so pH = pKa.
In simple terms: halfway to the equivalence point, the pH tells you the pKa of the acid.
Titration curve
A plot of pH (y-axis) vs. volume of titrant added (x-axis). The shape reveals whether a strong or weak acid/base is being titrated.
Percent ionisation
([H₃O⁺] / [HA]initial) × 100. In molecular diagrams, count how many HA molecules have split into A⁻ and H₃O⁺ out of the total.
A buffer requires two components present in meaningful amounts:
A weak acid (HA) and its conjugate base (A⁻), or
A weak base (B) and its conjugate acid (BH⁺).
The pair must be conjugates of each other. A strong acid/strong base pair cannot form a buffer. A salt of a strong acid paired with another strong acid salt also fails.
Examples of valid buffer pairs:
NH₃ and NH₄Br (weak base + its conjugate acid as an ammonium salt). Valid.
HCN and NaCN (weak acid + its conjugate base as a sodium salt). Valid.
CH₃COOH and CH₃COONa (acetic acid + sodium acetate). Valid.
Examples that do not work:
NaBr and HCN: NaBr is a salt of a strong acid and strong base (spectator ions). Not a conjugate pair.
KOH and CH₃COOH: KOH is a strong base. It would simply neutralise the weak acid rather than forming a buffer.
HClO₄ and LiClO₄: HClO₄ is a strong acid. No buffer.
pH = pKa + log([A⁻] / [HA])
Calculating buffer pH:
For a buffer of 0.45 M NH₃ and 0.25 M NH₄Cl (Ka of NH₄⁺ = 5.6 × 10⁻¹⁰):
pKa = −log(5.6 × 10⁻¹⁰) = 9.25
Here NH₃ is the base (A⁻ equivalent) and NH₄⁺ is the acid (HA).
pH = 9.25 + log(0.45 / 0.25) = 9.25 + log(1.80) = 9.25 + 0.26 = 9.51
Finding the ratio for a target pH:
For a buffer at pH 5.25 using acetic acid/sodium acetate (Ka = 1.8 × 10⁻⁵, pKa = 4.74):
5.25 = 4.74 + log([CH₃COO⁻] / [CH₃COOH])
log(ratio) = 0.51
ratio = 10⁰·⁵¹ = 3.24
You need about 3.24 mol of sodium acetate for every 1 mol of acetic acid.
Finding grams of salt to add:
For an 85 mL solution of 0.75 M HF, target pH 3.45, Ka = 7.1 × 10⁻⁴ (pKa = 3.15):
3.45 = 3.15 + log([F⁻] / [HF])
log(ratio) = 0.30, ratio = 10⁰·³⁰ = 2.00
Moles of HF = 0.085 × 0.75 = 0.06375 mol
Moles of F⁻ needed = 2.00 × 0.06375 = 0.1275 mol
Grams of NaF = 0.1275 × 41.99 g/mol ≈ 5.35 g
Select a buffer whose pKa is closest to the target pH.
For a target pH of 2.00, compare:
H₂SO₃/Na₂SO₃: Ka = 1.3 × 10⁻², pKa = 1.89. Closest to 2.00.
CH₃COOH/CH₃COONa: Ka = 1.8 × 10⁻⁵, pKa = 4.74. Too far.
HCN/NaCN: Ka = 4.9 × 10⁻¹⁰, pKa = 9.31. Far too high.
HF/NaF: Ka = 7.1 × 10⁻⁴, pKa = 3.15. Not as close.
Best choice: H₂SO₃/Na₂SO₃.
A buffer is effective within pKa ± 1.
For benzoic acid (Ka = 6.5 × 10⁻⁵): pKa = −log(6.5 × 10⁻⁵) = 4.19. Ideal pH range: 3.19 to 5.19.
Increasing buffer capacity: add more of the component that is present in smaller amount (to bring the ratio closer to 1:1), or increase the total concentrations of both components. Adding water dilutes the buffer and decreases capacity. Adding a small amount of strong acid or base uses up capacity rather than increasing it.
For diagrams showing HA (undissociated acid) and A⁻ (conjugate base) particles:
The solution with the most total particles (HA + A⁻) has the highest buffer capacity.
A solution with only A⁻ and no HA (or vice versa) cannot act as a buffer, because it lacks one of the two required components.
A solution where [HA] = [A⁻] (equal counts) has pH = pKa.
Effect of adding strong acid to a buffer: the strong acid reacts with the conjugate base (A⁻), converting it to HA. This shifts the ratio slightly, and the pH decreases by a small amount. It does not stay exactly the same, but the change is much smaller than it would be in an unbuffered solution.
When given molecular diagrams of weak acids (HX, HY, HZ), count:
Large paired particles = HA (undissociated acid)
Small single particles = H₃O⁺
Large single particles = A⁻
The acid with the most H₃O⁺ and A⁻ relative to HA is the strongest (highest Ka). The acid with the least ionisation has the smallest Ka.
Percent ionisation = (number of H₃O⁺ / total acid molecules originally present) × 100.
If HY has 12 total original molecules and 3 are ionised: percent ionisation = (3/12) × 100 = 25%.
Strong acid + strong base titration (e.g. NaOH added to HNO₃):
Equivalence point pH = 7.00 exactly.
At the equivalence point, only water and a neutral salt remain.
Weak acid + strong base titration (e.g. NaOH added to a weak acid):
Equivalence point pH > 7 (basic), because the conjugate base of the weak acid remains in solution and acts as a base.
The buffer region is the gradually sloping section before the equivalence point.
The half-equivalence point is where half the acid has been neutralised: [HA] = [A⁻], so pH = pKa.
From a titration curve, the equivalence point is the steepest part of the curve (the inflection point in the sharp rise). The half-equivalence point is at half the volume of titrant needed to reach the equivalence point.
Reading a titration curve (weak acid titrated with NaOH, equivalence point at about 13 mL):
Equivalence point pH ≈ 8.5 to 9 (above 7, confirming a weak acid).
Half-equivalence point at about 6.5 mL: pH at this point ≈ pKa of the acid (read from the y-axis, often around 4 to 5).
Buffer region: from 0 mL to near the equivalence point (the relatively flat section).
After the equivalence point, pH rises steeply because excess NaOH accumulates.
Finding the concentration of a titrant:
At the equivalence point, moles of acid = moles of base.
Moles acid = Macid × Vacid. Moles base = Mbase × Vbase.
Example: 20 mL of 0.50 M HCl titrated with 16 mL NaOH to reach the equivalence point.
Moles HCl = 0.020 × 0.50 = 0.010 mol
Moles NaOH = 0.010 mol = Mbase × 0.016
Mbase = 0.010 / 0.016 = 0.625 M
Titrating a diprotic acid:
For H₂CO₃ (diprotic) titrated with NaOH, the first equivalence point uses 1 mol NaOH per mol H₂CO₃, and the second uses another mol. To reach the first equivalence point:
50.0 mL of 0.0462 M H₂CO₃ = 0.00231 mol H₂CO₃
Moles NaOH needed = 0.00231 mol (1:1 for first proton)
Volume NaOH = 0.00231 / 0.0750 = 0.0308 L = 30.8 mL
For a weak acid/strong base titration with diagrams showing HA and A⁻ particles:
Before titration: mostly HA with very few A⁻ and H₃O⁺ (weak acid barely ionised).
At the half-equivalence point: equal amounts of HA and A⁻.
At the equivalence point: all HA has been converted to A⁻. Only conjugate base remains (plus water).
Henderson-Hasselbalch: pH = pKa + log([A⁻] / [HA])
Buffer capacity = moles of acid or base added / (|ΔpH| × litres of buffer)
At the half-equivalence point: [HA] = [A⁻], so pH = pKa
At the equivalence point: moles acid = moles base, so Macid × Vacid = Mbase × Vbase
Buffer range: pKa ± 1
pKa = −log(Ka)
Blood is buffered by the carbonic acid/bicarbonate system (H₂CO₃/HCO₃⁻), holding pH near 7.4. Swimming pool chemistry relies on buffer systems to keep pH stable despite the continuous addition of chlorine and the introduction of acids from swimmers. In pharmaceutical manufacturing, buffer solutions maintain the correct pH for drug stability and solubility.
Students often think a buffer keeps pH perfectly constant. It does not. The pH shifts slightly when acid or base is added; the buffer simply minimises the change.
A common mistake is using Ka where pKa belongs in the Henderson-Hasselbalch equation (or forgetting to take the negative log).
Students sometimes select a buffer pair based on matching the Ka value to the target pH numerically, when they should match the pKa to the target pH.
On titration curves, students confuse the equivalence point with pH 7. For weak acid/strong base titrations, the equivalence point pH is above 7; for weak base/strong acid titrations, it is below 7. Only strong/strong titrations land at pH 7.
⚠️ Henderson-Hasselbalch problems appear in several forms: find pH given concentrations, find the ratio for a target pH, find grams of salt to add. Practice all three.
⚠️ Buffer pair identification (which of the following can form a buffer?) is a quick-hit multiple choice question. Eliminate any pair involving a strong acid or strong base, and any pair that is not a conjugate acid-base pair.
⚠️ Titration curve reading is heavily tested: locate the buffer region, equivalence point, and half-equivalence point. Know what the pH at each reveals.
⚠️ For titration stoichiometry, always set up moles acid = moles base at the equivalence point. Watch for diprotic acids that need twice the moles of base.
⚠️ Molecular-level diagrams (particle pictures) of buffers and titrations require you to count HA vs. A⁻ particles and connect the count to buffer capacity, whether a solution can act as a buffer, and whether pH = pKa.
True or false: a buffer can be made from HCl and NaCl. (False; HCl is a strong acid)
Fill in the blank: at the half-equivalence point of a weak acid titration, pH = ___. (pKa)
True or false: the equivalence point pH of a weak acid/strong base titration is exactly 7. (False; it is above 7)
Fill in the blank: the Henderson-Hasselbalch equation is pH = ___ + log([base]/[acid]). (pKa)
True or false: adding water to a buffer increases its buffer capacity. (False; dilution decreases capacity)
Q: Which of the following pairs can form a buffer: (a) NaBr and HCN, (b) KOH and CH₃COOH, (c) HClO₄ and LiClO₄, (d) NH₃ and NH₄Br?
A: (d) NH₃ and NH₄Br. NH₃ is a weak base and NH₄⁺ is its conjugate acid. The other pairs fail because they involve strong acids, strong bases, or non-conjugate combinations.
Q: Calculate the pH of a buffer made from 0.45 M NH₃ and 0.25 M NH₄Cl (Ka of NH₄⁺ = 5.6 × 10⁻¹⁰).
A: pKa = 9.25. pH = 9.25 + log(0.45/0.25) = 9.25 + 0.26 = 9.51.
Q: How would you increase the buffer capacity of the NH₃/NH₄Cl buffer above?
A: Add more ammonium chloride (the component present in smaller amount), bringing the ratio closer to 1:1 and increasing total concentration.
Q: How does the pH change when a small amount of strong acid is added to this buffer?
A: The pH decreases slightly. The strong acid reacts with NH₃, converting some of it to NH₄⁺.
Q: What is the pH at the equivalence point when NaOH is added to HNO₃?
A: pH = 7. Both are strong, so the equivalence point produces only water and a neutral salt (NaNO₃).
Q: Calculate the ratio [CH₃COO⁻]/[CH₃COOH] needed for a buffer at pH 5.25 (Ka = 1.8 × 10⁻⁵).
A: pKa = 4.74. 5.25 = 4.74 + log(ratio). log(ratio) = 0.51. Ratio = 10⁰·⁵¹ ≈ 3.24.
Q: What is the ideal pH range for a benzoic acid/calcium benzoate buffer (Ka = 6.5 × 10⁻⁵)?
A: pKa = 4.19. Range = 4.19 ± 1 = 3.19 to 5.19.
Q: How many grams of NaF must be added to 85 mL of 0.75 M HF to make a buffer at pH 3.45 (Ka = 7.1 × 10⁻⁴)?
A: pKa = 3.15. Ratio = 10^(3.45 − 3.15) = 2.00. Moles HF = 0.06375. Moles NaF = 0.1275. Mass = 0.1275 × 42.0 ≈ 5.35 g.
Q: If 20 mL of 0.50 M HCl is titrated with 16 mL NaOH to reach the equivalence point, what is [NaOH]?
A: Moles HCl = 0.010 mol = moles NaOH. [NaOH] = 0.010/0.016 = 0.625 M.
Q: From a titration curve of a weak acid with NaOH, the equivalence point is at 13 mL and pH ≈ 8.5. What is the pKa of the acid?
A: The half-equivalence point is at 6.5 mL. Read the pH at 6.5 mL from the curve (approximately 4 to 5). That pH equals the pKa.
Q: In a weak acid/strong base titration, which diagram represents the solution at the equivalence point?
A: The diagram showing all conjugate base (A⁻) and no undissociated acid (HA). All the weak acid has been converted to its conjugate base by the added strong base.
Buffers connect back to the Ka/Kb material in the acids and bases notes: the Henderson-Hasselbalch equation is simply a rearrangement of the Ka expression. Titration stoichiometry relies on the same moles-based reasoning used in earlier stoichiometry and solution concentration problems. Understanding the equivalence point pH connects to salt hydrolysis: at the equivalence point of a weak acid/strong base titration, you effectively have a solution of the conjugate base salt, so the pH depends on that salt's basicity.
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