Buffer Solution
Do you know that our blood maintains a constant pH, irrespective of a number of cellular
acid – base reactions. Is it possible to maintain a constant hydronium ion concentration in such reactions? Yes, it is possible due to buffer action.
Buffer is a solution which consists of a mixture of a weak acid and its conjugate base (or) a weak base and its conjugate acid. This buffer solution resists drastic changes in its pH upon addition of a small quantities of acids (or) bases, and this ability is called buffer action. The buffer containing carbonic acid (H CO2 3 ) and its conjugate base HCO3
- is present in our blood. There are two types of buffer solutions.
1. Acidic buffer solution : a solution containing a weak acid and its salt.
Example : solution containing acetic acid and sodium acetate
2. Basic buffer solution : a solution containing a weak base and its salt.
Example : Solution containing NH OH4 and NH Cl 4
Buffer action
To resist changes in its pH on the addition of an acid (or) a base, the buffer solution should contain both acidic as well as basic components so as to neutralize the effect of added acid (or) base and at the same time, these components should not consume each other.
Let us explain the buffer action in a solution containing CH COOH3 and CH COONa 3
. The dissociation of the buffer components occurs as below.
CH3COOH (aq) CH3 - COO(aq) + H3O+(aq)
-
If an acid is added to this mixture, it will be consumed by the conjugate base CH COO3 - to
form the undissociated weak acid i.e, the increase in the concentration of H+ does not reduce the pH significantly.
+ H CH3COOH (aq)(aq)(aq)CH3COO
If a base is added, it will be neutralized by H3O +, and the acetic acid is dissociated to
maintain the equlibrium. Hence the pH is not significantly altered.
OH (aq) + CH3COOH (aq) CH3COO (aq) + H2O (l)
CH3COOH (aq) CH3COO (aq) + H2O (l)
OH (aq) + (aq) H2O (l)H3O +
(aq)H3O +
These neutralization reactions are identical to those reactions that we have already discussed in common ion effect.
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Les us analyse the effect of the addition of 0.01 mol of solid sodium hydroxide to one litre of a buffer solution containing 0.8 M CH COOH3 and 0.8 M CH COONa3 . Assume that the volume change due to the addition of NaOH is negligible. (Given: Ka for CH COOH
3 is 1.8 10
-5× )
CH -COOH(aq) CH COO (aq) + H (aq) 3
H O
3
- +
0.8-
2
α α α →←
2H O - + 3 3
0.80.8 0.8 CH COONa(aq) CH COO (aq)+Na (aq)→
The dissociation constant for CH COOH3 is given by
K = [CH COO ][H ]
[CH COOH] ;
[H ]=K [CH COOH]
[CH COO ]
a 3
3
+ a
3
3
- +
-
The above expression shows that the concentration of H+ is directly proportional to
[CH COOH]
[CH COO ] 3
3 -
.
Let the degree of dissociation of CH COOH3 be α then,
[CH COOH]=0.8-3 α and [CH COO ]= +0.83 - α
∴[H ]=K (0.8- ) (0.8+ )
«0.8,
+ a
α α
α 0.8- 0.8 and 0.8+ ∴ α α 0.8
[H ]= K
[H K + a + a
( . ) ( . )
] 0 8
0 8 ⇒ =
Given that
K for CH COOH is 1.8 10 [H pH = - log (1. a 3
-5
× ∴ = × −] . ;1 8 10 5 8 10 = 5 - log 1
-5× ) .8
= 5 - 0.26 pH = 4.74
Calculation of pH after adding 0.01 mol NaOH to 1 litre of buffer.
Given that the volume change due to the addition of NaOH is negligible ∴ = [OH 0.01M-] . The consumption of OH- are expressed by the following equations.
CH COOH (aq) CH COO (aq) + H (aq)3 3 +
0.8 - α α α
-
CH COONa(aq) CH COO (aq)+Na (aq)3 3 - +
0.8 0.8 0.8 →
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18
CH COOH + OH (aq) CH COO (aq) + H O (l) [CH COOH] = 0.8
3 3 2
3
- -→ ∴ - - 0.01 = 0.79 -
[CH COO ]= +0.8+0.01=0.81+ 3
α α
α α- «0.8; 0.79 - 0.79 and 0.81 + 0.81
α α α
∴ = × ×
= × ∴
−
−
[H
H pH = - log ( 1.76
] ( . ) . .
[ ] .
1 8 10 0 79 0 81
1 76 10
5
5
×10 = 5 - log 1.76 = 5 - 0.25
-5 )
pH = 4.75
The addition of a strong base (0.01 M NaOH) increased the pH only slightly ie., from 4.74 to 4.75 . So, the buffer action is verified.
Evaluate yourself - 8
a) Explain the buffer action in a basic buffer containing equimolar ammonium hydroxide and ammonium chloride.
b) Calculate the pH of a buffer solution consisting of 0.4M CH COOH3 and 0.4M CH COONa3 . What is the change in the pH after adding 0.01 mol of HCl to 500ml of the above buffer solution. Assume that the addition of HCl causes negligible change in the volume. Given: ( Ka = × −1 8 10 5. . )
Buffer capacity and buffer index
The buffering ability of a solution can be measured in terms of buffer capacity. Vanslyke
introduced a quantity called buffer index, β , as a quantitative measure of the buffer capacity. It is defined as the number of gram equivalents of acid or base added to 1 litre of the buffer solution to change its pH by unity.
β= dB d(pH)
….(8.19) Here,
dB = number of gram equivalents of acid / base added to one litre of buffer solution.
d(pH) = The change in the pH after the addition of acid / base.
Henderson – Hasselbalch equation
We have already learnt that the concentration of hydronium ion in an acidic buffer solution
depends on the ratio of the concentration of the weak acid to the concentration of its conjugate base present in the solution i.e.,
H O =K
[acid] [base]3
+ a
eq
eq
….(8.20)
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The weak acid is dissociated only to a small extent. Moreover, due to common ion effect, the dissociation is further suppressed and hence the equilibrium concentration of the acid is nearly equal to the initial concentration of the unionised acid. Similarly, the concentration of the conjugate base is nearly equal to the initial concentration of the added salt.
H O =K [acid]
[salt]3 +
a ….(8.21)
Here [acid] and [salt] represent the initial concentration of the acid and salt, respectively used to prepare the buffer solution
Taking logarithm on both sides of the equation
log [H O ]=logK +log [acid]
[salt]3 +
a
….(8.22) reverse the sign on both sides
-log [H O ]=-logK -log [acid] [salt]3
+ a
….(8.23) We know that pH = log [H O ] and pK = logK
pH = pK log [acid] [salt
3 +
a a
a
- - -⇒
] ….(8.24)
⇒ +pH = pK log [salt] [acid]a
….(8.25)
Similarly for a basic buffer, pOH = pK +log [salt] [base]b ….(8.26)
Example 8.6
1. Find the pH of a buffer solution containing 0.20 mole per litre sodium acetate and 0.18 mole per litre acetic acid. Ka for acetic acid is 1.8 10-5× .
pH = pK +log [salt]
[acid] a
Given that K a = 1.8 10
-5× ∴ = − × = −− pK
a log( . ) log .1 8 10 5 1 85
= 5 -0.26 = 4.74
∴pH = 4.74 + log 0.20 0.18
= 4.74 + log 10 9 = 4.74 + log 10 - log 9
= 4.74 + 1 - 0.95 = 5.74 - 0.95 = 4.79
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Example 8.7
What is the pH of an aqueous solution obtained by mixing 6 gram of acetic acid and 8.2 gram of sodium acetate making the volume equal to 500 ml. (Given: Ka for acetic acid is 1.8 10
-5× )
According to Henderson – Hasselbalch equation,
pH = pK +log [salt] [acid]
P =-logK =-log(1.8 10 )= 4.74
a
K a
-5a × (Refer previous example)
Salt = Number of moles of sodium acetate Volume of the solu
[ ] tion ( litre )
Number of moles of sodium acetate = mass of sodium acetate molar mass of sodium acetate
= 8.2 82
Salt mole Litre
M
=
∴[ ]= =
0 1
0 1 1
2 0 2
.
. .
acid =
mass of CH COOH molar mass of CH COOH Volume of
3
3[ ]
solution in litre
=
6 60 1
2 = 0
.2M
∴pH = 4.74 + log (0.2) (0.2)
pH = 4.74 + log 1 pH = 4.74 + 0 = 4.74
Evaluate yourself - 9
a) How can you prepare a buffer solution of pH 9. You are provided with 0.1M NH OH4 solution and ammonium chloride crystals. (Given: pKb for NH OH4 is 4.7 at 25 C .
b) What volume of 0.6M sodium formate solution is required to prepare a buffer solution of pH 4.0 by mixing it with 100ml of 0.8M formic acid. (Given: pK
a for formic
acid is 3.75. )
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