Buffer Solutions
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
Preparation of Buffer Solution
Handerson-Hasselbalch Equation
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
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)(pKa) of the acid and the (dissociation constant)(pKb) of the base are known, a buffer solution can be prepared by controlling the (salt-acid)(pKa) or the (salt-base)(pKb) 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 HPO42- 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:
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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
- 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:
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Sol:
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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:
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Sol:
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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:
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Sol:
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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
