352
Acid-Base Equilibria
concentrations, and because the two are the same, we represent this value by x
This gives a concentration of (0 0500 - t) mole/liter of undissociated HC 2 H 3 O 2
molecules, because 0 0500 mole of acid is put into solution and v moles
dissociate
0 0500 - v x
K
HC 2 H 3 O 2 ?± H
+ + C 2 H 3 O 2 -
Substitute the molar concentrations into the K equation
0 0500 - x
= 1 74 x 10To solve an equation of this type, we usually first assume v to be so small
that 0 0500 - x may be considered as 0 0500 (in other words, subtraction of \
from 0 0500 does not appreciably change the value) This gives, as the simplified
equation,
X
= 1 74 x
00500
^ = 8 70 x 10~
7
x = [H
+ ] = 9 33 x 10~
4 M
pH = -log (9 33 x 104
) = -(0 97 - 4 00) - 3 03
You can see that x is much smaller than the original concentration of 0 0500 M,
and that we were justified in neglecting it compared to 0 0500 In general, we
shall say that, >f the calculated value of x is less than 10 0% of the number from
which it is subtracted or to which it is added, it is permissible to make the
approximation as we did in this problem
Polyprotic Acids
When a solution contains a weak acid that can lose more than one proton
(such as H 2 S, H 2 SO 3 , or H 3 AsO 4 ) the question arises as to whether we should
write chemical equations and K, expressions to show the loss of all these
protons The answer is NO As you can tell from Table 23-1, each successive
proton comes off with very much greater difficulty than the one before, and
even the first one doesn't contribute much in the way of [H
+ ] In addition,
the H
+ ions that come from the first dissociation tend to repress second and
third dissociations, just as would H
+ ions from some other source The net
result is that second and third dissociations can be neglected, and the equilibrium expressions are always written for the loss of only one proton
Acid-Base Equilibria
concentrations, and because the two are the same, we represent this value by x
This gives a concentration of (0 0500 - t) mole/liter of undissociated HC 2 H 3 O 2
molecules, because 0 0500 mole of acid is put into solution and v moles
dissociate
0 0500 - v x
K
HC 2 H 3 O 2 ?± H
+ + C 2 H 3 O 2 -
Substitute the molar concentrations into the K equation
0 0500 - x
= 1 74 x 10To solve an equation of this type, we usually first assume v to be so small
that 0 0500 - x may be considered as 0 0500 (in other words, subtraction of \
from 0 0500 does not appreciably change the value) This gives, as the simplified
equation,
X
= 1 74 x
00500
^ = 8 70 x 10~
7
x = [H
+ ] = 9 33 x 10~
4 M
pH = -log (9 33 x 104
) = -(0 97 - 4 00) - 3 03
You can see that x is much smaller than the original concentration of 0 0500 M,
and that we were justified in neglecting it compared to 0 0500 In general, we
shall say that, >f the calculated value of x is less than 10 0% of the number from
which it is subtracted or to which it is added, it is permissible to make the
approximation as we did in this problem
Polyprotic Acids
When a solution contains a weak acid that can lose more than one proton
(such as H 2 S, H 2 SO 3 , or H 3 AsO 4 ) the question arises as to whether we should
write chemical equations and K, expressions to show the loss of all these
protons The answer is NO As you can tell from Table 23-1, each successive
proton comes off with very much greater difficulty than the one before, and
even the first one doesn't contribute much in the way of [H
+ ] In addition,
the H
+ ions that come from the first dissociation tend to repress second and
third dissociations, just as would H
+ ions from some other source The net
result is that second and third dissociations can be neglected, and the equilibrium expressions are always written for the loss of only one proton