How to Prepare Buffer Solutions?

In the last post we have learnt that a weak acid and its conjugate base are the main constituents of a buffer solution . 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

Each of these ways will give us a pair of weak acid and its conjugate base. Let us see how.

Method I: Prepare a Buffer solution by weak acid and salt: for example we will take acetic acid (pK_{a ­­­}= 4.7) and salt sodium acetate.

CH₃COOH  ⟶  H⁺ + CH₃COO^-

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

CH₃COONa  ⟶  Na⁺ + CH₃COO^-

Sodium acetate is a salt and it completely ionizes in water and makes the basic part of buffer which is conjugate base of acetic acid.

Method II: Prepare a Buffer solution by Weak acid and strong base: for example we will take acetic acid (pK_{a ­­­}= 4.7) and NaOH.

As you have seen in the previous example, acetic acid makes the acidic part of buffer. NaOH is a strong base and it dissociates completely.

NaOH  ⟶  Na⁺ + OH^-

When it reacts with acetic acid it neutralised it and produces water and the salt sodium acetate.

CH₃COOH + Na⁺ + OH^-  ⟶  CH₃COONa + H₂O

Salt sodium acetate ionises completely in water and makes the basic part of buffer which is conjugate base of acetic acid.

Method III: Prepare a Buffer solution by Salt of weak base and strong acid: if we take sodium acetate salt and HCl let’s see what happens.

CH₃COONa  ⟶  Na⁺ + CH₃COO^-

Sodium acetate is a salt and it ionises completely in water and makes conjugate base of acetic acid.

CH₃COO^- + HCl  ⟶  CH₃COOH + Cl^-

And when HCl reacts with the conjugate base, the base accepts proton given by HCl and produces acetic acid. This makes the acidic part of buffer and so on.

You have seen that how every way reaches to the one common point. Let’s try to prepare 1M buffer of acetic acid/ sodium acetate with pH 4, pK_a of acetic acid is 4.76.

To prepare this buffer we have to find out the desired concentration of acidic form - acetic acid, and basic form - acetate ions.

By using Henderson-Hasselbalch equation:

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

4.76 = 4 + log [CH₃COOH]/ [CH₃COO^-]

0.76 = log [CH₃COOH]/ [CH₃COO^-]

10^0.76 = [CH₃COOH]/ [CH₃COO^-]

5.75 = [CH₃COOH]/ [CH₃COO^-]

5.75[CH₃COO^-] = [CH₃COOH]

Concentration of acetic acid/sodium acetate buffer solution is 1M, that means:

[CH₃COOH] + [CH₃COO^-] = 1M

5.75[CH₃COO^-] + [CH₃COO^-] = 1M

6.75[CH₃COO^-] =1M

[CH₃COO^-] = 0.15M = 15mmol

So,

[CH₃COOH] = 1-0.15 = 0.85M = 85mmol

Now we have the desired concentrations of both constituents. Lets see how to prepare the buffer by each of the methods:

Methods to Prepare Buffer Solutions

I. Weak acid and its salt: For this method, we will use 85mmol of acetic acid and 15 mmol of sodium acetate to prepare buffer with pH 4.

II. Weak acid and strong base: For this method, we will need acetic acid and NaOH. When acetic acid reacts with NaOH, it gets neutralized by it and produces conjugate base

CH₃COOH + NaOH  ⟶  CH₃COONa + H₂O

CH₃COONa  ⟶  CH₃COO^- + Na⁺

That means here NaOH acts as a source of conjugate base so, the concentration of NaOH will be equal to the desired concentration of conjugate base (acetate ion). We will need 15mmol NaOH and 85mmol acetic acid to prepare this buffer.

III. Salt of weak base and strong acid: For this method, we will need sodium acetate and HCl. When they reacts:

CH₃COONa  ⟶  CH₃COO^- + Na⁺

CH₃COO^- + HCl  ⟶  CH₃COOH + Cl^-

HCl gives proton to acetate ion and produces acetic acid. Thus HCl acts as the source of acidic part of the buffer. That means we will need HCl equal to the desired concentration of acetic acid. To prepare this buffer we will need 85mmol HCl and 15mmol sodium acetate.

Now you have learnt how to prepare buffer solution by different methods. Let’s solve some problems to build better understanding.

How would you make 100 ml buffer solution with a pH 4 that is 0.3M in acetic acid and 0.2M sodium acetate using a 1M acetic acid solution and 2M CH₃COONa solution.

First we have to calculate how many moles are present in 100ml buffer solution, when concentration of acetic acid is 0.3M

M = mmol/ ml

0.3M = mmol/ 100ml

= 30mmol

Now we know that we need 30 mmol of acetic acid to prepare desired buffer. Now calculate the amount of 1M acetic acid solution to get 30mmol.

M = mmol/ ml

1M = 30mmol/ ml

= 30ml

Similarly we have to do calculations for sodium acetate. Find out the number of mmol present in 100ml buffer, when its concentration is 0.2M

M = mmol/ ml

0.2M = mmol/ 100ml

= 20mmol

Now calculate the amount of 2M sodium acetate solution to get 20mmol.

M = mmol/ ml

2M = 20mmol/ ml

= 10ml

That means to prepare desired buffer we need to mix 30ml acetic acid solution and 10 ml sodium acetate solution and make it up to 100 ml by mixing remaining 60ml water.

I hope these posts have helped you to understand the mystery of buffers but if you have any doubts, please feel free to leave a comment. 

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