If a buffer solution is 0.120 M in a weak acid (Ka = 3.5 × 10-5)...
Homework Answers
Concentration of a weak acid is related to its 
and pH of a solution as shown in the following Henderson -Haselbach equation.
![[salt]
pH pklogSalt]
[acid]](http://img.homeworklib.com/questions/f4984170-78ce-11ea-b644-79d782ec57af.png?x-oss-process=image/resize,w_560)
Here, 
is the acid dissociation constant; ![salt]](http://img.homeworklib.com/questions/f53d8080-78ce-11ea-b8e6-53a29ccf002f.png?x-oss-process=image/resize,w_560)
, and![[acid]](http://img.homeworklib.com/questions/f58ca4b0-78ce-11ea-b572-413f1c0945d6.png?x-oss-process=image/resize,w_560)
are the concentration terms of the salt and the acid respectively. This equation is applicable to a system that contains weak acids or weak bases only.
Because the acid is a weak acid, we can use the above equation to calculate the pH.
The value of pH for the solution can calculate by substituting the concentration values in the formula.
An aqueous solution that contains a mixture of a weak acid and its conjugate base or vice versa is called as buffer solution.
Conjugate base differs with a proton
to its acid.
Consider the weak acid as HA. It is dissociated as shown below.
The Henderson -Haselbach equation can be written as follows:
![[A]
pH pKalog
[HA](http://img.homeworklib.com/questions/f68b7e90-78ce-11ea-b684-491e13a8ba4a.png?x-oss-process=image/resize,w_560)
Calculate the value of 
and then pH of the solution by substituting the concentrations in the equation.
Now calculate the pH as follows:
![[A]
pH pK,logHA
[HA
=4.45+log0.500
0.120
4.45 0.6197
= 5.06](http://img.homeworklib.com/questions/f78e5f30-78ce-11ea-a3cb-2366eabaf84b.png?x-oss-process=image/resize,w_560)
Thus, the pH of the solution is 
.
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