Inorganic and Applied Chemistry

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Inorganic and Applied Chemistry

Example 5- D:

pH in a solution of polyprotic acid

We wish to determine the pH value in a 5.0 M H 3 PO 4 solution. H 3 PO 4 is a tri-protic acid being semi-strong

in the first step. In the first case we will assume that the dominating components in the solution are:

H 3 PO 4 and H 2 O,

And initially the dominating step is:

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

This equilibriums corresponding equilibrium expression is:

 

 3 4

1 7.^510324

HPO

K M H HPO

a



We will now look at the initial and equilibrium conditions analogously to other examples. The initial

concentrations are:

[H 3 PO 4 ] 0 = 5.0 M

[H 2 PO 4 - ] 0 = 0 M

[H+] 0 * 10 -7 M (from the auto protolysis of water)

and the end-concentrations are thereby:

[H 3 PO 4 ]= (5.0 – x) M

[H 2 PO 4 - ]= x M

[H+] = (10-7 + x) M

which by insertion into the expression for Ka gives:

 



x M

x

x x

HPO

H HPO

Ka 0. 19

5. 0

( 10 )

7. 510

7

3 4

(^324)
1 (^

^
The concentrations of H+ ions is 0.19 M which gives a pH of:
pH log 0. 19 0. 72
So far we have assumed that it is only the first step that participate significant to the H 3 O+ concentration
and thereby the pH value. In order to verify that the second and third step do not contribute to the H 3 O+
Acids and bases