Buffer Solutions

A buffer solution is a solution which resists a change in its pH when such a change is caused by the addition of a small amount of acid or base. This does not mean that the pH of the buffer solution does not change (we make this assumption while doing numerical problems). It only means that the change in pH would be less than the pH that would have changed for a solution that is not a buffer.

There are three types of buffer solutions:

(i) weak acid–salt buffer

(ii) weak base–salt buffer and

(iii) salt buffer

Buffer of a Weak Acid and its Salt with a Strong Base :

It is possible to prepare a buffer solution by the addition of a weak acid and a salt of the acid with a strong base.

We shall explain the buffer action by the following example. Let us consider a buffer solution made up of CH3COOH and CH3COONa. The weak acid dissociates to a very small extent more so due to the common ion effect of its salt.

CH3COOH <—-> CH3COO + H+

CH3COONa —> CH3COO + Na+

Now let us assume that a buffer solution contains 20 moles of CH3COOH and 20 moles of CH3COONa. The salt being a strong electrolyte would completely dissociate while the acid would be hardly dissociated. We assume that the amount of CH3COO is 20 moles as the contribution from CH3COOH would negligible. We also assume that the amount of CH3COOH to be 20 moles as it would be very weakly dissociated. Therefore the solution contains 20 moles of CH3COOH and 20 moles of CH3COO– ions.

Let us now add 10 moles of H+ ions to this solution. These 10 moles of H+ ions would react with the 20 moles of CH3COO ions to produce 10 moles of CH3COOH. (This is because the reaction of CH3COOH to give CH3COO ions and H+ ions has an equilibrium constant value of approximately 10–5. Therefore, the reverse reaction, that is the reaction of CH3COO ions to combine with H+ ions to give CH3COOH would have an equilibrium constant of approximately 105. So this reaction can be assumed to be complete). These 10 moles of CH3COOH formed would ionize weakly because it is a weak acid. Moreover due to the common ion effect of CH3COO ion (10 moles, left over), it would ionize even less. So the amount of H+ ion produced back is much less than the 10 moles that were added. So effectively, 10 moles of H+ ions were consumed and an amount much less than that is produced back which causes an insignificant change in pH of the buffer solution.

Let us add 10 moles of OH ions to the same buffer solution containing 20 moles of CH3COOH and 20 moles of CH3COO ions. These 10 moles of OH ions added would be consumed by the 10 moles of CH3COOH to produce 10 moles of CH3COO ions (CH3COOH + OH —-> CH3COO + H2O.
This is because the reverse reaction is the hydrolysis of CH3COO– ions which has an equilibrium constant of Kw/Ka ≈ 10–9.
So the forward reaction’s equilibrium constant would be 109, which implies that the reaction is practically complete). The 10 moles of CH3COO ions produced would be hydrolysed weakly (due to a very low K value of 10–9). On top of that, the presence of 10 moles of CH3COOH would further hamper the hydrolysis process and thereby the OH ion produced is much less than the amount that was added. This causes a minor change in pH.

For a buffer to act as a good buffer the amount of CH3COOH and CH3COO ion should be high.

The pH of a buffer can be calculated as follows.

Since acetic acid is in equilibrium,

$\large \frac{[H^+] [CH_3 COO^-]}{[CH_3 COOH]} = K_a$

$\large \frac{[H^+] [Conjugate \; base]}{[Acid]} = K_a$

$\large  [H^+]  = K_a \frac{[Acid]}{[Conjugate \; base]}$

$\large  log[H^+]  = log K_a  + log\frac{[Acid]}{[Conjugate \; base]}$

$\large  -log[H^+]  = -log K_a  + log\frac{[Conjugate \; base]}{[Acid]}$

$\large  p_H  = p_{K_a} + log\frac{[Conjugate \; base]}{[Acid]}$

This equation is called Henderson – Hasselbalch equation.

When a weak dibasic acid is taken then the pH at half neutralisation is given by a special formula which is derived by balancing charge and mass of reactants and products. For example, at half neutralisation of H2CO3 by NaOH, the pH is given by

$\large p_H = \frac{p_{K_a1} + p_{K_a2}}{2}$

Where Ka1 and Ka2 are 1st and 2nd dissociation constant of H2CO3

Buffer of a Weak Base and its Salt with a Strong Acid :

This combination works on the principle as stated in the previous section. If we take the example of NH4OH and NH4Cl, then the OH– ion added would react with (from the salt) to produce NH4OH. The H+ ion added would react with NH4OH to produce

The Henderson – Hasselbalch equation appears as

$\large p_{OH} = p_{K_b} + log \frac{[Conjugate \; base]}{[Acid]}$

Exercise : (a) Calculate the pH after the addition of 90 ml and 100ml respectively of 0.1N NaOH to 100ml 0.1N CH3COOH (Given pKa for CH3COOH = 4.74)

(b) What is pH of 1M CH3COOH solution? To what volume must one litre of this solution be diluted so that the pH of resulting solution will be twice the original value. Given : Ka = 1.8 × 10–5

Illustration : 100 mL 1 M CH3COOH is treated with 0.8 gm NaOH. Calculate pH of solution. How much NaOH needs to be added further so as to increase its pH to 4.74? Will the dilution of the mixture will change the pH of solution
(Ka = 1.8 × 10–5).

Solution: 100 mL 1M CH3COOH ≡ 100 × 1 i.e. 100 millimoles of CH3COOH

0.8 gm i.e. 800 mg NaOH ≡ 800/40 i.e. 20 millimoles of NaOH

20 millimoles NaOH will react with 20 millimoles CH3COOH leaving behind 80 millimoles of free CH3COOH. The mixture will be thus, an acid buffer mixture whose pH can be calculated using the equation.

$\large p_H = p_{K_a} + log\frac{[salt]}{[acid]}$

= – log (1.8 × 10–5) + log(20÷80)

= 4.74 – 0.60 = 4.14

To make the pH of solution equal to 4.74 , $\frac{[salt]}{[acid]}$  must be equal to 1. The acid present in the original acid solution is 100 millimoles. So we have to add 50 millimoles i.e. (50 ×40)/1000 (=2g) of NaOH. Thus the amount of NaOH to be added further is 2 – 0.8 i.e. 1.2g.

Upon dilution the ratio $\frac{[salt]}{[acid]}$   and also pKa will remain unchanged and hence pH of solution will remain unchanged.


Ostwald Theory of Indicator: An indicator, generally organic weak acids or weak bases, is a substance which is used to determine the end point in a titration. They change their colour within certain pH range. Generally, the colour change is due to shifting indicator equilibrium.

Illustration : An acid type indicator HIn differs in colour from its conjugate base (In). Human eye is sensitive to colour differences only when the ratio [[In]/[HIn] is greater than 10 or smaller than 0.1. What should be the minimum change in the pH of the solution to observe a complete colour change (Ka = 1.0 × 10–5)?

Solution: The two conditions when colour of indictor will be visible.

$\large p_H = pK_{in}+ log\frac{[I_n^-]}{[HI_n]}$

$\large (i) p_H = -log_{10} 1 \times 10^{-5} + log_{10} (10) = 6 $

$\large (ii) p_H = -log_{10} 1 \times 10^{-5} + log_{10} (0.1) = 4 $

Thus minimum change in pH = 6 – 4 = 2

Salt Buffer :

The fundamental principle behind a buffer action is the fact that on adding an acid the system consumes the H+ ion added to produce a weak acid and on adding a base, it consume the OH ion added to produce a weak base. This ensures that H+ or OH ion added is consumed and the weak acid or the weak base produced gives less H+ or OH– ion as they are weak.

Based on this principle, a solution of a salt of a weak acid and weak base is also called as a buffer. Let us take the example of CH3COONH4. It dissociates as,

CH3COONH4 —> CH3COO + NH4+ . When H+ ion is added, CH3COO ion consumes it to give CH3COOH. When OH ion is added, NH4+ ion consumes it to give NH4OH. Hence it acts as a buffer.

$\large [H^+] = \sqrt{\frac{K_w K_a}{K_b}}$

(derived earlier).

Next Page →

←Back Page

Leave a Reply