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Inorganic and Applied Chemistry
pH logH 3 O log( 3. 110
3 ) 2. 51
On the way to the first point of equivalence we have a buffer system consisting of the weak acid H 2 A and
the corresponding weak base HA- which is why pH does not change very much. Halfway towards the first
point of equivalence pH I calculated form the buffer equation:
1 log^2 "^1 4.^00
#
% $
&
'
a HA pKa
pH pK H A
We note that half way towards the point of equivalence the amounts of H 2 A are HA- equally large why the
logarithmic relation becomes 0. First point of equivalence is reached when the amounts of H 2 A is equal to
the amount of added NaOH exactly when 20.0 mL of 0,100 M NaOH is added. We there have a solution
containing HA- which may function both as an acid and as a base. Such a specie is called an amfolyte and
pH in such solutions may be determined from the following equation:
pH ½ pKa(acid)pKa(amfolyte) (5- 4)
As both Ka values are known pH is:
pH ½ 4. 0 9. 0 6. 50
Between the first and the second point of equivalence we have a buffer system in this case consisting of a
weak acid HA- and its corresponding weak base A2-. Halfway towards the second point of equivalence we
have from the buffer equation:
2 log 2 ""^2 9.^00
#
$
%%
&
'
(^)
a A pKa
pH pK HA
The second and last point of equivalence is reached when the amount of NaOH is the double of the initial
amount of H 2 A which is when 40.0 mL of 0,100 M NaOH is added. All the acid is now on the base form
A2- and the volume of the solution is increased to totally 60.0 mL. Now the problem is to determine pH in
a solution of the base A2- if the “initial” concentration is:
mol L
L
A L mol L 3. 310 /
0. 0600
2
0.^02000.^100 /
2
Now we write in normal fashion the aqueous equilibrium with corresponding equilibrium expression.
A2- (aq) + H 2 O HA-(aq) + OH- (aq)
Acids and bases