354
Acid-Base Equilibria
SOLUTION:
First, write the chemical equation:
NH 3 + H 2 O ?± NH 4
+ + OHSecond, write the K, expression based on the chemical equation, obtaining the
needed value of K, from Table 23- 1 .
-'•»"•Third, write what you know and do not know. You are asked for the pH of a
basic solution, so you will first have to find the [OH~]. Let [OH~] = x. Because
NHJ and OH~ ions are formed in equal amounts, [NH^] also equals x. Of the
original 0. 100 mole/liter of NH 3 , x moles/liter will dissociate and leave (0. 100 - x)
mole/liter at equilibrium. Associate these concentrations with the chemical equation, and substitute them into the K, expression.
0.100-*
x
x
NH 3 + H 2 O <=* NH 4
+ + OHWe try to simplify the solution by neglecting x compared to 0.100, obtaining
x
2 = (0.100)(1.74 x 10~
5
)
x = [QH-J = 1.32 x 10~
3 M
We see that the assumption about neglecting x was sound.
pOH = -log (1.32 x 1C3
) = -(0.12 - 3.00) = 2.88
pH = 14.00 - 2.88 = 11.12
COMMON-ION EFFECT
If we add some sodium acetate or other source of C 2 H 3 O^ to a solution of
HC 2 H 3 O 2 , we shift the equilibrium
HC 2 H 3 0 2 ?± H
+ + C 2 H 3 0 2 -
to the left, and use up some of the H
+ ions to form more undissociated
HC 2 H 3 O 2 . We may use the regular K, expression for acetic acid to compute
the pH of such a mixture of acetic acid and sodium acetate.
Acid-Base Equilibria
SOLUTION:
First, write the chemical equation:
NH 3 + H 2 O ?± NH 4
+ + OHSecond, write the K, expression based on the chemical equation, obtaining the
needed value of K, from Table 23- 1 .
-'•»"•Third, write what you know and do not know. You are asked for the pH of a
basic solution, so you will first have to find the [OH~]. Let [OH~] = x. Because
NHJ and OH~ ions are formed in equal amounts, [NH^] also equals x. Of the
original 0. 100 mole/liter of NH 3 , x moles/liter will dissociate and leave (0. 100 - x)
mole/liter at equilibrium. Associate these concentrations with the chemical equation, and substitute them into the K, expression.
0.100-*
x
x
NH 3 + H 2 O <=* NH 4
+ + OHWe try to simplify the solution by neglecting x compared to 0.100, obtaining
x
2 = (0.100)(1.74 x 10~
5
)
x = [QH-J = 1.32 x 10~
3 M
We see that the assumption about neglecting x was sound.
pOH = -log (1.32 x 1C3
) = -(0.12 - 3.00) = 2.88
pH = 14.00 - 2.88 = 11.12
COMMON-ION EFFECT
If we add some sodium acetate or other source of C 2 H 3 O^ to a solution of
HC 2 H 3 O 2 , we shift the equilibrium
HC 2 H 3 0 2 ?± H
+ + C 2 H 3 0 2 -
to the left, and use up some of the H
+ ions to form more undissociated
HC 2 H 3 O 2 . We may use the regular K, expression for acetic acid to compute
the pH of such a mixture of acetic acid and sodium acetate.