Inorganic and Applied Chemistry

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Inorganic and Applied Chemistry

concentration it is necessary with additional calculations of [HPO 4 2-] and [PO 4 3-] from the second and from

the third step of equilibrium.

The concentrations of HPO 4 2- may be found in terms of the following equilibrium with corresponding

expression of equilibrium:

H 2 PO 4 - (aq)  H+(aq) + HPO 4 2-(aq)

 

^



2 4

2

2 6.^21084

HPO

K M H HPO

a

where:

[H+] [H 2 PO 4 - ] = 0.19 M (The autoprotolysis of water is neglected) 

[HPO 4 2-] = Ka2 = 6.2 · 10-8 M

In order to calculate [PO 4 3-] we use the third equilibrium step with corresponding equilibrium expression.

HPO 4 2-(aq)  H+(aq) + PO 4 3-(aq)

 



  PO M

M

M PO

HPO

K M H PO

a

(^319)
8 4
3
4
2
4
3
3 13 4 1.^610
6. 210
4. 810 0.^19

(^) 
*^
It is seen from the equilibrium expressions that the concentrations of H 3 O+ and HPO 4 2- respectively simply
are taken from the calculations in the first and second step. From the calculated values of [HPO 4 2-] and
[PO 4 3-] both being very small it is reasonable to neglect these two acids contribution to the pH of the
solution. The value of pH in solution seems thereby to be excellently approximated by the pH value of pH
= 0.72.
5.4 Acid properties of salts
Some salts exhibit properties making solutions acid when the salt is dissolved in water. E.g., the following
reaction takes place when NH 4 Cl is dissolved in water:
NH 4 Cl(s)  NH 4 +(aq) + Cl-(aq)
NH 4 +(aq) +H 2 O(l)  NH 3 (aq) + H 3 O+(aq)
The last reaction is an acid-base reaction and we shall look at how pH is calculated in such a solution in the
following example:
Acids and bases