Buffer Solutions are mixtures that contain a weak acid and its conjugate base, or a weak base and its conjugate acid. These solutions are water-soluble and resist changes in pH when small amounts of acid or alkali are added, or when they are diluted.

Buffer solutions show minimal change in pH upon the addition of a very small quantity of strong acid or strong base, making them ideal for keeping pH at a constant value. Learn more about pH of Buffer Solutions.

Table of Contents

Types of Buffer Solution

Mechanism of Buffering Action

Preparation of Buffer Solution

Handerson-Hasselbalch Equation

Solved Problems

pH Maintenance

Uses of Buffer Solutions

Answer: A buffer solution is a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid. It is used to resist changes in pH when small amounts of acid or base are added to it.

Buffer solutions are used for many purposes, such as fermentation, food preservatives, drug delivery, electroplating, printing, the activity of enzymes, and blood oxygen carrying capacity, as they are able to maintain their Hydrogen ion concentration (pH) with only minor changes upon dilution or addition of a small amount of either acid or base.

Solutions of a weak acid and its conjugate base (or weak base) and its conjugate acid are able to maintain pH, and are known as buffer solutions.

#Types of Buffer Solutions

The two primary types of buffer solutions are acidic and alkaline buffers.

Acidic Buffers

An acid buffer is a solution prepared by mixing a weak acid (such as acetic acid) and its salt (such as sodium acetate) with a strong base. This type of solution is used to maintain acidic environments, and an aqueous solution of equal concentrations of acetic acid and sodium acetate has a pH of 4.74.

The pH of these solutions is less than 7

These solutions consist of a weak acid and a salt of a weak acid.

A mixture of sodium acetate and acetic acid (pH = 4.75) is an example of an acidic buffer solution.

Alkaline Buffers

The aqueous solution of an equal concentration of ammonium hydroxide and ammonium chloride has a pH of 9.25 and is used to maintain basic conditions. This type of buffer is prepared by mixing a weak base and its salt with a strong acid.

The pH of these solutions is greater than 7

They contain a weak base and a salt of the weak base.

A mixture of ammonium hydroxide and ammonium chloride with a pH of 9.25 is an example of an alkaline buffer solution.

Also Read

Acid and Base

pH Scale and Acidity

PH and Solutions

Mechanism of Buffering Action

In solution, the salt is completely ionized and the weak acid is partly ionized.

CH3COONa <=> Na+ + CH3COO–

CH3COOH <=> H+ + CH3COO-

On the Addition of an Acid and a Base

1. The protons released from the addition of acid will be taken up by the acetate ions to form an acetic acid molecule.

H⁺ + CH₃COO^- (from added acid) ⇌ CH₃COOH (from buffer solution)

2. The hydrogen ions will remove the hydroxide released by the base upon addition of the base, resulting in the formation of water.

H⁺ + OH^- ⇌ H₂O

Preparation of Buffer Solution

If the dissociation constant of the acid and the dissociation constant of the base are known, a buffer solution can be prepared by controlling the salt-acid or the salt-base ratio.

Mixing weak bases with their corresponding conjugate acids, or mixing weak acids with their corresponding conjugate bases, are solutions that were discussed earlier.

An example of this method of preparing buffer solutions is the preparation of a phosphate buffer by mixing HPO₄^2- and H₂PO₄^-. The pH maintained by this solution is 7.4.

Handerson-Hasselbalch Equation

Preparation of Acid Buffer

Consider an acid buffer solution, containing a weak acid (HA) and its salt (KA) with a strong base (KOH). The weak acid HA ionizes, and the equilibrium can be written as:

$$HA \leftrightarrow H^+ + A^-$$

HA + H2O ⇋ H⁺ + A^-

Ka = $\frac{[H^+][A^-]}{HA}$

Taking the negative log of both the left-hand side and right-hand side:

![Buffer Solutions]()

pH = pKa + ([salt]/[acid])

The Henderson-Hasselbalch equation, also known as the Henderson equation, is a popular equation.

Preparation of Base Buffer

Consider a base buffer solution, containing a weak base (B) and its salt (BA) with a strong acid.

pOH, can be derived as above,

pOH of a basic buffer = pK_b + log([salt]/[acid])

The pH of a basic buffer can be calculated by subtracting the pKa from the logarithm of the ratio of the salt to the acid:

pH = pKa - log([salt]/[acid])

The Importance of Henderson-Hasselbalch Equation

Handerson Equation can be used to:

• Calculate the energy of a system
• Determine the thermodynamic properties of a system
• Estimate the reaction rate of a system
• Predict the equilibrium constants of a system

1. Determine the pH of the buffer created from a combination of the salt and weak acid/base.

2. Calculate the pKa value.

3. Prepare a buffer solution of the desired pH.

Limitations of the Henderson-Hasselbalch Equation

The Henderson – Hasselbalch equation cannot be used for strong acids and strong bases.

Buffering Capacity

The Buffer capacity of a buffer solution is the number of millimoles of acid or base to be added to a litre of the solution to change the pH by one unit.

Β = millimoles / (ΔpH)

Problems Regarding Buffer Solutions

Problem 1: What is the ratio of base to acid when pH = pKa in buffer solution? How about when pH = PKa + 1?

The ratio of base to acid when pH = pKa in buffer solution is 1:1. When pH = PKa + 1, the ratio of base to acid is greater than 1:1.

Given:

This is a heading

Sol:

This is a Heading

pH = -log(p*K*a) when the ratio of base to acid is 1, because log(1) = 0

When log_(10/1) = 1, then the ratio of base to acid is 10:1.

Answer: The pH of a buffered solution of 0.5 M ammonia and 0.5 M ammonium chloride when enough hydrochloric acid corresponding to make 0.15 M HCl is 9.25.

Given:

Welcome to our website

Sol:

Welcome to our website!

The pKb of ammonia is 4.75.

pKa = 9.25 + pKb = 14

0.15 M H⁺ reacts with 0.15 M ammonia to form 0.30 M ammonium.

So, the ammonium ion is 0.65 M and there is 0.35 M of remaining ammonia (base).

Using the Henderson-Hasselbalch equation

pKa = 9.25 - log(.65/.35) = 8.98

Problem 3: How many moles of sodium acetate and acetic acid must be used to prepare 1.00 L of a 0.100 mol/L buffer with a pH of 5.00?

Given:

This is a heading

Sol:

This is a heading

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

log([A−][HA]) = 5.00 - 4.74

log([A-][HA]) = 0.26

A-H<sup>+</sup> = 10.26 = 1.82