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Home / Find the Ph of 100 Ml of the Buffer Solution

Find the Ph of 100 Ml of the Buffer Solution

Buffer Solution is a water solvent based solution which consists of a mixture containing a weak acid and the conjugate base of the weak acid, or a weak base and the conjugate acid of the weak base. They resist a change in pH upon dilution or upon the addition of small amounts of acid/alkali to them.

The pH of Buffer Solutions shows minimal change upon the addition of a very small quantity of strong acid or strong base. They are therefore used to keep the pH at a constant value.

Table of Content

  • Types
  • Mechanism
  • Preparation
  • Handerson-Hasselbalch Equation
  • Solved Problems
  • pH Maintenance
  • Uses of Buffers

What is Buffer Solution?

The buffer solution is a solution able to maintain its Hydrogen ion concentration (pH) with only minor changes on the dilution or addition of a small amount of either acid or base. Buffer Solutions are used in fermentation, food preservatives, drug delivery, electroplating, printing, the activity of enzymes, blood oxygen carrying capacity need specific hydrogen ion concentration (pH).

Solutions of a weak acid and its conjugate base or weak base and its conjugate acid are able to maintain pH and are buffer solutions.

Types of Buffer Solution

The two primary types into which buffer solutions are broadly classified into are acidic and alkaline buffers.

Acidic Buffers

As the name suggests, these solutions are used to maintain acidic environments. Acid buffer has acidic pH and is prepared by mixing a weak acid and its salt with a strong base. An aqueous solution of an equal concentration of acetic acid and sodium acetate has a pH of 4.74.

  • pH of these solutions is below seven
  • These solutions consist of a weak acid and a salt of a weak acid.
  • An example of an acidic buffer solution is a mixture of sodium acetate and acetic acid (pH = 4.75).

Alkaline Buffers

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

  • The pH of these solutions is above seven
  • They contain a weak base and a salt of the weak base.
  • An example of an alkaline buffer solution is a mixture of ammonium hydroxide and ammonium chloride (pH = 9.25).


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 Addition of Acid and Base

1. On addition of acid, the released protons of acid will be removed by the acetate ions to form an acetic acid molecule.

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

2. On addition of the base, the hydroxide released by the base will be removed by the hydrogen ions to form water.

HO– + H+ (from added base) ⇌ H2O  (from buffer solution)

Preparation of Buffer Solution

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

As discussed earlier, these solutions are prepared by mixing the weak bases with their corresponding conjugate acids, or by mixing weak acids with their corresponding conjugate bases.

An example of this method of preparing buffer solutions can be given by 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). Weak acid HA ionizes, and the equilibrium can be written as-

HA + H2O ⇋ H+ + A−

Acid dissociation constant = Ka = [H+] [A–]/HA

Taking, negative log of RHS and LHS:

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

The equation is the Henderson-Hasselbalch equation, popularly known as the Henderson equation.

Preparation of Base Buffer

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

pOH, can be derived as above,

  • pOH of a basic buffer = pKb +  log ([salt]/[acid])
  • pH of a basic buffer = pKa – log ([salt]/[acid])

Significance of Handerson Equation

Handerson Equation can be used to:

  1. Calculate the pH of the buffer prepared from a mixture of the salt and weak acid/base.
  2. Calculate the pKa value.
  3. Prepare buffer solution of needed pH.

Limitations of Henderson-Hasselbalch Equation

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

Buffering Capacity

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

Β = millimoles /(ΔpH)

Problems on Buffer Solution

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

Sol:

pH = pK a when the ratio of base to acid is 1 because log 1 = 0.

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

What is 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? The pK b of ammonia is 4.75.

pK a = 14 – pK b. = 9.25

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

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

Using Henderson-Hasselbalch equation,

pKa – log ([salt]/[acid]) = 9.25 – log (.65/.35) = 9.25 – .269 = 8.98

Problem 2: How many moles of sodium acetate and acetic acid must you use to prepare 1.00 L of a 0.100 mol/L buffer with pH 5.00.

Sol:

pH = pKa + log([A−][HA])

5.00 = 4.74 + log([A−][HA])

log([A−][HA]) = 0.26

[A−][HA]=10.26 = 1.82 [A⁻] = 1.82[HA]

Also, [A⁻] + [HA] = 0.100 mol/L

1.82[HA] + [HA] = 0.100 mol/L

2.82[HA] = 0.100 mol/L

[HA] = 0.0355 mol/L [A⁻] = (0.100 – 0.0355) mol/L = 0.0645 mol/L

0.0355 mol of acetic acid and 0.0645 mol of sodium acetate is required to prepare 1 L of the buffer solution.

pH Maintenance

In order to understand how buffer solutions maintain a constant pH, let us consider the example of a buffer solution containing sodium acetate and acetic acid.

In this example, it can be noted that the sodium acetate almost completely undergoes ionization whereas the acetic acid is only weakly ionized. These equilibrium reactions can be written as:

  • CH3COOH ⇌ H+ + CH3COO–
  • CH3COONa ⇌ Na++ CH3COO–

When strong acids are added, the H+ions combine with the CH3COO– ions to give a weakly ionized acetic acid, resulting in a negligible change in the pH of the environment.

When strongly alkaline substances are introduced to this buffer solution, the hydroxide ions react with the acids which are free in the solution to yield water molecules as shown in the reaction given below.

CH3COOH + OH– ⇌ CH3COO– + H2O

Therefore, the hydroxide ions react with the acid to form water and the pH remains the same.

Uses of Buffer Solutions

  • There exists a few alternate names that are used to refer buffer solutions, such as pH buffers or hydrogen ion buffers.
  • An example of the use of buffers in pH regulation is the use of bicarbonate and carbonic acid buffer system in order to regulate the pH of animal blood.
  • Buffer solutions are also used to maintain an optimum pH for enzyme activity in many organisms.
  • The absence of these buffers may lead to the slowing of the enzyme action, loss in enzyme properties, or even denature of the enzymes. This denaturation process can even permanently deactivate the catalytic action of the enzymes.

Find the Ph of 100 Ml of the Buffer Solution

Source: https://byjus.com/jee/buffer-solutions/