Inorganic and Applied Chemistry

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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

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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

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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