Buffer Solutions

Definition: A buffer solution is a mixture that resists changes in pH when small amounts of acid or base are added, typically formed from a weak acid and its conjugate base or a weak base and its conjugate acid.

How It Works

A buffer needs two components present in comparable amounts: a weak acid (HA) and its conjugate base (A-), usually supplied as a salt. Both species stay in equilibrium:

HA ⇌ H+ + A-

When strong acid is added, the extra H+ reacts with the conjugate base:

A- + H+ → HA

This converts some conjugate base into weak acid instead of letting free H+ accumulate.
When strong base is added, the OH- reacts with the weak acid:

HA + OH- → A- + H2O

This absorbs the base instead of letting free OH- accumulate.
Because the disturbance is consumed by the buffer’s own equilibrium rather than directly changing [H+], pH shifts only slightly instead of sharply.

This only works within limits:

Buffer capacity depends on how much HA and A- are actually present.
Once one component is used up, the solution loses its buffering ability and behaves like an ordinary unbuffered solution.
Effective range is about one pH unit on either side of the acid’s pKa.
Outside that range, one component is too dilute relative to the other to do much work.
Buffer capacity is maximized when [HA] = [A-], exactly at pH = pKa.
At that point the mixture has equal reserves to absorb either added acid or added base.

Under the Hood

The Henderson-Hasselbalch equation relates pH directly to the ratio of conjugate base to acid:

pH = pKa + log([A-]/[HA])

Derived directly from the Ka expression by taking -log of both sides and rearranging:

Ka = [H+][A-]/[HA]

When [A-] = [HA], the log term is zero and pH = pKa exactly.

Worked example: preparing a buffer.

Given: 0.30 M acetic acid (Ka = 1.8 × 10⁻⁵, pKa = 4.74) and 0.20 M sodium acetate
Step 1: apply Henderson-Hasselbalch

pH = 4.74 + log(0.20/0.30)
pH = 4.74 + log(0.667)
pH = 4.74 + (-0.176)

Answer: pH = 4.56

Worked example: response to added strong base.

Given: 1.0 L of the buffer above, 0.30 mol acetic acid, 0.20 mol acetate, add 0.050 mol NaOH
Step 1: OH- consumes acetic acid

HA + OH- → A- + H2O

Step 2: find new amounts

HA = 0.30 - 0.05 = 0.25 mol
A- = 0.20 + 0.05 = 0.25 mol

Step 3: apply Henderson-Hasselbalch

pH = 4.74 + log(0.25/0.25) = 4.74 + 0

Answer: pH = 4.74
The pH moved only from 4.56 to 4.74, a change of 0.18 units.
Adding that same 0.050 mol NaOH to 1.0 L of pure water would jump the pH to nearly 12.7.

Worked example: buffer capacity limit.

Given: the same buffer, but add 0.35 mol NaOH instead, more than the 0.30 mol acetic acid present
Step 1: all 0.30 mol HA is consumed
Step 2: 0.05 mol NaOH remains unreacted
Answer: buffering capacity is exhausted, the solution now behaves like a simple strong base solution

Why It Matters

Blood is buffered by the carbonic acid/bicarbonate system:

H2CO3 ⇌ H+ + HCO3-

This holds blood pH between 7.35 and 7.45.
A shift outside that narrow range, acidosis or alkalosis, impairs enzyme function and can be fatal.
Labs rely on buffers like phosphate-buffered saline (PBS) or Tris to keep cell cultures and reactions at a controlled pH.
Pharmaceutical formulations use buffers to keep injectable drugs stable.
Buffers also control how quickly a drug ionizes and gets absorbed in the gut.

Common Pitfalls

Using the Henderson-Hasselbalch equation far outside its valid range, roughly pH = pKa ± 1.
At that point the approximation that equilibrium concentrations equal initial concentrations breaks down.
Assuming any acid-base mixture is a buffer. A strong acid and its conjugate base isn’t a buffer.
The conjugate base of a strong acid is too weak to meaningfully react with added H+.
Forgetting buffer capacity is finite: enough added acid or base will exhaust one component.
Once exhausted, pH swings sharply, exactly like an unbuffered solution.
Confusing pKa with pH: pKa is a fixed property of the acid; pH depends on the ratio actually present.
Diluting a buffer and assuming pH changes proportionally; diluting both components equally leaves pH nearly unchanged.
Buffer capacity does drop on dilution, even though pH doesn’t shift much.
Picking a buffer system whose pKa is far from the target pH just because it’s convenient or cheap.

Comparison

Response	Buffered Solution	Unbuffered Solution
pH change on added acid	Small, damped by conjugate base	Large, direct
pH change on added base	Small, damped by weak acid	Large, direct
Mechanism against acid	A- consumes added H+	No consuming species
Mechanism against base	HA consumes added OH-	No consuming species
Capacity limit	Works until a component is depleted	No inherent limit, changes immediately

Buffer System	Components	Typical pH Range	Common Use
Acetate	CH3COOH / CH3COO-	3.7-5.6	Lab buffers, food preservation
Phosphate	H2PO4- / HPO4²-	6.2-8.2	Cell biology (PBS), blood (secondary)
Bicarbonate	H2CO3 / HCO3-	6.1-8.1	Blood pH regulation
Tris	Tris-H+ / Tris	7.0-9.0	Molecular biology buffers

Real-World Application

Blood pH regulation links buffer chemistry directly to breathing rate.

The bicarbonate equilibrium:

CO2 + H2O ⇌ H2CO3 ⇌ H+ + HCO3-

During intense exercise, muscles produce excess CO2 and lactic acid.
This pushes the equilibrium right and lowers blood pH.
The body responds by increasing breathing rate, exhaling CO2 faster than it’s produced.
That shifts the equilibrium left, consuming H+ and restoring pH.
Respiratory compensation typically restores blood pH within minutes.
The kidneys’ compensation, excreting or retaining bicarbonate, is far slower, hours to days.

Example

Phosphate-buffered saline (PBS) is a staple of cell biology labs, formulated near pH 7.4 to match physiological conditions. Cells or tissue samples are rinsed or stored in PBS specifically because the phosphate buffer pair holds pH steady even as cells release small amounts of acidic or basic metabolic byproducts during handling.

FAQ

Can a buffer be made from a strong acid?

Not in the usual sense.
Strong acids dissociate essentially completely, leaving almost no undissociated HA to neutralize added base.
There’s no reservoir on the acid side of the equilibrium.

Does adding water to a buffer break it?

Diluting a buffer doesn’t change its pH much, the ratio of components stays the same.
It does reduce buffer capacity, since there’s less HA and A- present in total to absorb a disturbance.

Why is pKa = pH the “sweet spot” for buffer choice?

At that point [HA] = [A-].
This gives the buffer equal reserves against both added acid and added base.
That maximizes its capacity to resist change in either direction.

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