Buffer solutions

In previous posts you have learnt that dissociation constant (K_a and K_b) tells us the strength of acid or base while the strength of solution is determined by pH. Do you know that milk, water, blood, fruit juices all have a particular pH. Can you guess what happens if it gets disturbed? It will spoil the nature and taste. And it can be fatal if pH change occurs to blood. Blood transports oxygen to all the cells of the body and its normal pH is 7.35 to 7.45. If its pH decreases or increases even for few seconds, it will cause death of that person. But blood has a system to deal with any pH change and this system maintains the pH of blood. This system is called Buffer.

A buffer solution maintains the pH of a solution even if you add H⁺ or OH^- ions to it. In this post we will see how it works? 

Buffer solution is made up of weak acid and its conjugate base. If we add OH^- to the solution, weak acid gives proton to neutralize them and if we add H⁺ to the solution, conjugate base accept those protons to nullify the excess H⁺concentration. Thus, the addition of small amount of H⁺ or OH^- ions doesn’t affect the pH of the solution. Mostly buffer solutions can maintain the pH by ±1. Let’s take an example of Brönsted acid-base couple to understand the working of a buffer solution:

HA ⇌ H⁺ + A^-

Let’s write the equation for acid dissociation constant:

K_a = [H⁺] [A^-] / [HA]

Taking log of both side of equation,

Log K_a = log [H⁺] + log [A^-] / [HA]

On multiplying by -1,

-log K_a = -log [H⁺] - log [A^-] / [HA]

Or

-log K_a = -log [H⁺] + log [HA]/ [A^-]

pK_a  = pH + log [HA]/ [A^-]

here [HA] represents the acid concentration and [A^-] represents the concentration of basic form. This is the Henderson-Hasselbalch equation. It can be used to determine the pH of buffer solution. It also tells us whether a compound will exist in its acidic form or basic form at a particular pH. 

Lets take the example of CH₃COOH, its pK_a is 4.7. Let’s find out in which form it will exist at pH 5.

CH₃COOH ⇌ H⁺ + CH₃COO^-

pK_a  = pH + log [acidic form]/ [Basic form]

4.7 = 5 + log [CH₃COOH]/ [CH₃COO^-]

-0.3 = log [CH₃COOH]/ [CH₃COO^-]

10^-0.3 = [CH₃COOH]/ [CH₃COO^-]

0.5 = [CH₃COOH]/ [CH₃COO^-]

It means that the basic form of acetic acid [CH₃COO^-] will be present in excess at pH 5 or you can say that at pH 5 acetic acid exists in its basic form. There is another way to get your answer fast:

pK_a  = pH + log [acidic form]/ [Basic form]

pK_a  - pH = log [acidic form]/ [Basic form]

pK_a  - pH = log [acidic form] - log [Basic form]

On comparing both side,

pK_a ≈ log [acidic form]

And

pH ≈ log [Basic form]

So; if (pH > pK_a) that means basic form > acidic form

If (pH < pK_a) that means basic form < acidic form.

Now you can calculate pH of a buffer solution by using Henderson-Hasselbalch equation. Buffer solution is made up of weak acid and its conjugate base. 

Let’s calculate the pH of buffer solution prepared by mixing 20ml of 0.1 M formic acid and 15ml of 0.5 M sodium formate, pK_a of formic acid is 3.75.

HCOOH ⥃ H⁺ + HCOO^-

Formic acid is a weak acid so it dissociates less. It makes the acidic part of buffer.

HCOONa ⟶ Na⁺ + HCOO^-

Sodium formate is a salt and it ionises completely in water and makes the basic part of buffer.

First we have to calculate the concentration of formic acid and sodium formate in the buffer.

The relation between M, mmol and ml is as follows and we can calculate the number of mmoles or mls by simply putting the values of all the other variables in it and solving the equation.

M = Mole/ Lit = mmol/ml

So how many moles are present in 20ml of 0.1 M formic acid?

0.1M = mmol/ 20ml = 0.1 × 20

=2mmol of formic acid

And how many moles are present in 15ml of 0.5 M sodium formate?

M = mmol/ml

0.5 = mmol/15ml = 0.5 × 15

= 7.5mmol of sodium formate

Now place the values in Henderson-Hasselbalch equation to calculate the pH of buffer solution:

pK_a  = pH + log [acidic form]/ [Basic form]

3.75 = pH + log [2]/[7.5]

3.75+0.57 = pH

pH =4.32

as pH is greater than pK_a , basic form will exist in majority.

There are three ways to prepare a buffer solution.

1.      Weak acid and its salt.
2.      Weak acid and strong base

3.      Salt of weak base and strong acid

In the next post we will discuss all these ways to prepare buffer solutions.

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