Inorganic and Applied Chemistry

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Inorganic and Applied Chemistry

Example 5- E:

pH in a solution of aluminium chloride

We wish to determine pH in a 0.01 M solution of AlCl 3. It has been observed that when aluminium

chloride is dissolved in water an acid solution is the result. Even though the Al3+ ion is not an acid itself as

it cannot produce a H+ ion, it is the so-called hydrate compound Al(H 2 O) 6 3+ that is a weak acid reacting

with water in the following manner:

Al(H 2 O) 6 3+(aq) + H 2 O(l)  Al(OH)(H 2 O) 5 2+(aq) + H 3 O+(aq)

The aluminium-ion will thereby in aqueous solution be surrounded by 6 water molecules. The Ka value for

Al(H 2 O) 6 3+ is 1.4 · 10-5 M. Again we will start by looking at the main components in the solution. These

are in this case:

Al(H 2 O) 6 3+, Cl- and H 2 O

On the basis of the reaction scheme above we have the following equilibrium expression:

 



 

3

2 6

2

(^5325)
( )

1. 4 10 ( )( )
   AlHO
   K M HO AlOH HO
   a
   We will now look at start and equilibrium conditions completely similar to prior examples. The initial
   concentrations are:
   [Al(H 2 O) 6 3+] 0 = 0.01 M
   [Al(OH)(H 2 O) 5 2+] 0 = 0 M
   [H 3 O+] 0 = 10-7 M (from the autoprotolysis of water)
   and the end concentrations thereby become:
   [Al(H 2 O) 6 3+] = (0.01 – x) M
   [Al(OH)(H 2 O) 5 2+] = x M
   [H 3 O+] = (10-7 + x) M
   which by insertion in the expression for Ka gives:
    
   
   x M
   x
   x x
   AlHO
   K M HO AlOH HO
   a
   (^74)
   3
   2 6
   2
   (^53253). 7 10


2. 01
   ( 10 )
   ( )


3. 410 ( )( )^
   
    
   (^) (
   ^
   Thus the concentrations of H 3 O+ ions is 3.7 · 10-4 M by which pH becomes:
   Acids and bases