Download free books at BookBooN.comInorganic and Applied Chemistry
Which by insertion into the expression for Ka gives:
x M
xx x
CHCOOHK M HO CHCOO
a5
3(^5331). 7 10
( 0. 49 )
- 810
^ (^0.^51 )(
The concentration of H 3 O+ ions is thereby 1.7 · 10-5 M which is why pH becomes:
pH log 1. 7 10
5 4. 76
This pH value could also be found with the use of the buffer equation as we still have a buffer system with
a weak corresponding acid-base pair of the same order (x is very small):
[CH 3 COOH] = (0.49 – x) M 0.49 M
[CH 3 COO-] = (0.51 + x) M 0.51 M
- 76
- 51
log 4. 74 log^0.^49
3
(^3) "
$
%
&
'
(^) M
M
CHCOO
CHCOOH
pH pKa
pH has thereby changed from 4.74 to 4.76 (0.02 pH units) by the addition of 0.010 moles solid NaOH to 1
litre of the buffer solution. It shows clearly that the buffer solution is capable of damping the external pH
action.
As a comparison we will calculate how much pH would have changed if we had added the same amount of
NaOH to 1.0 L of pure water. In this case the concentration of OH- ions will be 0.01 M. the concentration
of H 3 O+ ions may now be calculated knowing the conditions connected with the autoprotolysis of water:
M
M
K M HO OH HO M
w
12
14 2
3 3
(^1421). 010
- 01
- 0 10
( 1.^0 10
which is why pH becomes:
pH log 1. 0 10
12 12. 0
which thereby is a change of 5 pH units as pure water has a pH value of 7.0? compared to the buffer
solution this is a very large change in pH which clearly shows the pH damping abilities of the buffer
solutions.
Similarly the buffer capacity may be illustrated by the addition of an equivalent amount of strong acid to the
same type of buffer solution just with a higher concentration of buffer components.
Acids and bases
- 0 10