Designing Buffer Solution

Designing Buffer Solutions

The ability to prepare a buffer solution of a desired pH is a practical application of acid-base equilibrium. The key lies in the relationship between pH, the acid dissociation constant ($K_a$), and the ratio of the concentrations of the weak acid and its conjugate base (or weak base and its conjugate acid).

Preparation of an Acidic Buffer

An acidic buffer is prepared using a weak acid and its salt with a strong base. For example, a mixture of acetic acid ($CH_3COOH$) and sodium acetate ($CH_3COONa$).

Consider a general weak acid HA that ionises in water:

The acid dissociation constant $K_a$ is given by:

Rearranging this expression to isolate $[H_3O^+]$:

Taking the negative logarithm (base 10) of both sides:

Since $pH = -\log[H_3O^+]$ and $pK_a = -\log K_a$, we get:

This is more commonly written as:

This equation is known as the Henderson–Hasselbalch equation for an acidic buffer.

Simplifying the Concentration Ratio

The term $\frac{[A^-]}{[HA]}$ is the ratio of the concentration of the conjugate base (the anion of the weak acid) to the concentration of the undissociated weak acid in the mixture.

Two important approximations are made:

1. The weak acid HA ionises to a very small extent. Therefore, the concentration of HA in the buffer mixture is negligibly different from the initial concentration of the acid taken to form the buffer.
2. Most of the conjugate base $A^-$ comes from the complete dissociation of the salt of the weak acid (e.g., $CH_3COONa$). The contribution from the ionisation of the weak acid itself is negligible. Therefore, the concentration of $A^-$ is negligibly different from the concentration of the salt added.

Using these approximations, the Henderson–Hasselbalch equation for an acidic buffer is written in its practical form:

Choosing the Acid for a Desired pH

If the molar concentrations of the weak acid and its salt (conjugate base) are equal, then $[\text{Salt}] = [\text{Acid}]$, and the ratio $\frac{[\text{Salt}]}{[\text{Acid}]} = 1$. Since $\log 1 = 0$, the Henderson–Hasselbalch equation simplifies to:

This is a critical result. It means that when the acid and its conjugate base are present in equal concentrations, the pH of the buffer is exactly equal to the $pK_a$ of the weak acid.

Example: The $pK_a$ of acetic acid is 4.76. Therefore, a buffer solution formed by mixing equal molar concentrations of acetic acid and sodium acetate will have a pH of approximately 4.76.

Preparation of a Basic Buffer

A basic buffer is prepared using a weak base and its salt with a strong acid. For example, a mixture of ammonia ($NH_3$) and ammonium chloride ($NH_4Cl$).

A similar analysis for a buffer made with a weak base (B) and its conjugate acid ($BH^+$) leads to an analogous equation for pOH.

Consider a weak base B:

The base dissociation constant $K_b$ is:

Following the same derivation steps as for the acidic buffer, we obtain:

This is the Henderson–Hasselbalch equation for a basic buffer.

Converting to pH

The pH of the buffer solution can be calculated using the relationship $pH + pOH = 14$ (at 298 K). However, a more direct equation in terms of pH can be derived.

We know the following relationships:

Substituting $pOH = pK_w - pH$ and $pK_b = pK_w - pK_a$ into the basic buffer equation:

Cancelling $pK_w$ from both sides and rearranging:

Multiplying both sides by -1 gives the final form:

This equation is identical in form to the Henderson–Hasselbalch equation for an acidic buffer, but it uses the $pK_a$ of the conjugate acid of the weak base.

Choosing the Base for a Desired pH

If the molar concentrations of the weak base and its conjugate acid (salt) are equal, then the ratio $\frac{[\text{Base}]}{[\text{Conjugate acid}]} = 1$, and $\log 1 = 0$. The equation simplifies to:

This means that when the weak base and its conjugate acid are present in equal concentrations, the pH of the buffer is equal to the $pK_a$ of the conjugate acid of the weak base. …