• It can be shown either by calculation or using pH papers that pH of chemically pure water can be changed by 3 units by adding 1 cm³ of 0.1 mol dm^-3 HCl or NaOH to 1.0 dm³ of water. This shows that a small amount of an acid or an alkali can cause a drastic change in pH. But there are solutions/systems that can resist such changes.

• A buffer solution is a one which resists the change in pH on addition of a small amount of H⁺ or OH^– or water.

• The following table shows how pH changes when the given volumes of a 0.1 mol dm^-3 NaOH solution are added to 50.00 cm³ of a 0.1 mol dm^-3 CH₃COOH solution.

• According to the above table, the change in pH in between the addition of 10.00 cm³ and 25.00 cm³of NaOH solution is very small showing that in this range, the system is resistant to the change in pH brought about by the addition of NaOH. This action is called buffer action.

Buffer systems

• Solutions containing a weak acid and its conjugate base act as buffers.
eg. CH₃COOH and CH₃COONa.

• Solutions containing a weak base and its conjugate acid also act as buffers.
eg. NH₄OH and NH₄Cl.

The buffer action of the CH₃COOH and CH₃COONa system.

CH₃COONa(aq) → CH_{3-(aq) + Na⁺(aq) (Complete dissociation)
CH3COOH(aq) + H2O(l) ⇌ CH3COO^–(aq) + H3O⁺(aq) (Incomplete dissociation)}

When H₃O⁺ ions are added to the system, they are removed by the CH₃COO^– ion forming weakly dissociated CH₃COOH. Therefore, pH almost remains constant

When a small amount of OH^– ions are added to the system, they are immediately removed forming almost unionised water.

OH^–(aq) + H₃O⁺(aq) ⇌ 2H2O(l)

More CH₃COOH ionises to replenish the lost H₃O⁺, so pH remains almost constant.

The buffer action of the NH₄OH(aq) and NH₄Cl(aq) system.

NH₄OH(aq)   ⇌  NH₄⁺(aq) + OH^–(aq)
NH₄Cl(aq)  →   NH₄⁺(aq)  + Cl^–(aq)

When a small amount of an acid is added, H⁺ ions are removed by the OH^– forming water and more NH₄OH ionises restoring the OH^– ions. Hence, pH does not change widely. When an alkali is added, OH^– ions combine with NH₄ ⁺ to form NH₄OH bringing down the OH^– concentration. Hence, pH approximately remains constant.

Henderson equation

Here is the dissociation equation for HA:

HA ⇌ H⁺ + A^–

From which, we write the K_a expression:

take the negative log of each of the three terms in the above equation. They become:

1) -log [H⁺]
2) – log K_a
3) -log ([HA] / [A¯])

However,

1) this is the pH
2) this is the pK_a
3) to get rid of the negative sign  flip the log term to get this: + log ([A¯] / [HA])

Inserting these last three items (the pH, the pK_a and the rearranged log term), we arrive at the Henderson-Hasselbalch Equation:

Here is a common way the HH equation is presented in a textbook explanation:

Remember that, in a buffer, the two substances differ by only a proton. The substance with the proton is the acid and the substance without the proton is the salt.

However, remember that the salt of a weak acid is a base (and the salt of a weak base is an acid).

Consequently, another common way to write the Henderson equation is to substitute “base” for “salt form” (sometimes you will see “conjugate base” or “base form”). This is probably the most useful way to decribe the interactions between the acidic form (the HA) and the basic form (the A¯).

Here it is:

The alternate form starts from the ionization equation for a generic base called B:

B + H₂O ⇌ HB⁺ + OH¯

By the way, here is an example of the above generic equation, using ammonia:

NH₃ + H₂O ⇌ NH₄⁺ + OH¯

The B simply represents the entire base and HB⁺ represents the substances with an additional H⁺.

Next, we write the K_b expression for this reaction:

Next, we isolate the [OH¯] on the left-hand side of the equation:

We negative log each of the three terms of the above equation to give pOH, pKb and we flip the third term so as to make it an addition, not a subtraction.

Here is the alternate form of the Henderson-Hasselbalch Equation, expressed in terms of pOH and pK_b: