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Some Buffer Solutions

 

S.No.

Buffer pair

PH

Nature

1

2


3.



4.

5.

6.

7.

CH3COOH + CH3COO-

HCOOH + HCOO-

COOH      COO-
|       +   |
COOH      COO-
H3BO3 + BO33-

NH4+ + NH3

HCO3- + CO32-

H2PO4- + HPO42-

4.74

3.7

1.42

 

7.0 – 9.1

9.25

6.38

6.2 – 8.2

Acidic

Acidic

Strongly acidic

 

Basic

Basic

Basic

Basic


Calculation of pH and pOH for buffer solutions:
Acidic Buffer solution:
CH3COOH(aq) CH3COO-(aq) + H+ (aq)

Dissociation constant, i.e. Ka= 1.8 ´ 10-5 (Known)

[H=] = Ka ×
pH of the solution can be calculated as below:
pH = -log[H+] = -[log Ka + log [Salt]/[Acid]
pH = -log Ka + log
or pH = P Ka + log
pH = P Ka + log

This pH of the above mentioned buffer will not be affected much by dilution or by adding H= ions from outside to it. If some water is added, the concentrations of CH3COOH and CH3COONa giving CH3COO- ion will be affected equally such that the ratio [CH3COOH] / [CH3COO]- is once again unity and hence pH will remain unchanged.

Now, (i) if a small amount of any acid, i.e. H+ ion is added, CH3COO- ions will react to produce CH3COOH, but as CH3COO- is taken away, more of CH3COONa, will ionize. So, again the ratio [CH3COOH] / [CH3COO-] will become 1.

Similarly, (ii) if a small amount of a base, i.e. O H ion is added, the CH3COOH in equilibrium will ionize to furnish a little more CH3COO- ions. But the change is very negligible and the ratio [CH3COOH]/ [ CH3COO-] again remains nearly 1.

Hence, the buffer maintains its pH (4.74) even on dilution or adding any acid or base to it. Similarly, the working of any other buffer can also be explained.

Alkaline Buffer solution(NH4Cl + NH4OH)


We can treat a mixture of ammonium chloride and ammonium hydroxide in two ways. We can consider either the reaction.
NH4OH(aq) NH4+(aq) + OH-(aq) Or

we take the conjugate acid NH4+ and consider the reaction:

NH4+(aq) + H2O(I) NH3(aq) + H3O+(aq)
Or NH4+(aq) + H2O(1) NH3(aq) + H+(aq)

To calculate the pH, we arrange this equation in the form
[H+(aq)] = Ka

Taking negative logarithm on both sides, we get
H = pKa – log

It may be noted that pKa + PKb = 14 and pKa for NH4+ = 9.25.
Therefore, pH = 9.25 – log Or pH = 9.25 + log   Or pH = 9.25 + log

It is to be noted that the highest buffer capacity should be around 9.25 and furthermore concentration of ammonia is actually the concentration of ammonium hydroxide.




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