Chapter 14: Acids and Bases
Example: Determine the hydroxide concentration in a solution with
[ H3O(aq ) ]  1.89 104 M .
Kw
1.0 1014

 5.29 1011 M
[ H 3O(aq ) ] 1.89 104
________________________________________________________________________
pKw derivation:
K w  [OH (aq ) ][ H 3O(aq ) ]
K w  [ H 3O(aq ) ][OH (aq ) ]  [OH (aq ) ] 
 log K w   log{[OH (aq ) ][ H 3O(aq ) ]}
 log K w   log[OH (aq ) ]  log[ H 3O(aq ) ]
pK w  pOH  pH   log(1.0 1014 )  14
________________________________________________________________________
Strong Acid/Base Examples I: Write the balanced equation for each of the following
and determine the pH.
a.) 0.5000 M HClO4(aq)


HClO4(aq)  H 2O(l)  H3O(aq)
 ClO4(aq)

[H3O(aq)
]  [HClO 4(aq) ]  0.5000 M
pH   log(0.5000 M)  0.30
b.) 0.0256 M LiOH(aq)


LiOH(aq)  Li(aq)
 OH(aq)

[OH (aq)
]  [LiOH (aq) ]  0.0256 M
pH  14  pOH  14  ( log(0.0256 M))  12.41
________________________________________________________________________
Strong Acid/Base Examples II: Determine the hydronium ion concentration for a
0.01500 M Ca(OH)2.
2

Ca(OH) 2(aq)  Ca (aq)
 2OH (aq)

 OH (aq)
  2  Ca(OH) 2(aq)   2  0.01500 M  0.03000 M
Kw
1.0  1014

[H 3O (aq)
]

 3.33 10 13 M

 OH (aq)  0.03000 M
________________________________________________________________________
Weak Acid Example: Calculate [H+] and the pOH of 0.050M of benzoic acid.
HC7 H5O2( aq)
C7 H5O2( aq)  H(aq) Ka  6.5 105
HC7 H 5O2( aq )
Initial
Change
Eq
0.050
-x
0.050 - x
C7 H5O2( aq )
0.0
+x
+x
1
H (aq )
0.0
+x
+x
Ka 
[C7 H 5O2( aq ) ][ H (aq ) ]
[ HC7 H 5O2( aq ) ]

x2
 6.5 105
0.050  x
x 2  3.25 106  6.5 105 x  x 2  6.5  105 x  3.25  106  0
x
6.5 105 
 6.5 10 
5 2
 4 1  -3.25 106 
2
5
3
6.5 10  3.6110
 x  1.77 103 M
2

[ H ]  1.77 103 M
x
14  pH  pOH  pOH  14  pH  14  log[ H  ]  11.25
________________________________________________________________________
Percent Dissocation Example: Determine the percent dissociation of 0.050M of benzoic
acid.
HC7 H5O2( aq)
C7 H5O2( aq)  H(aq) Ka  6.5 105
We already found [ H  ]  1.77 103 M therefore
[ H  ] 1.77 103 M

100%  3.54%
0.050M
 HA
It should be small since our Ka is so small
________________________________________________________________________
Polyprotic Acid Example: Calculate the [H+] of 0.050M of sulfuric acid.
H 2 SO4( aq )  H (aq )  HSO4( aq )
K a  1
HSO4( aq )
H (aq )  SO4(2aq )
K a  1.2 102
Initially all 0.050M of the H2SO4 dissociates completely into
0.050M H(aq )  HSO4( aq )
H (aq )
HSO4( aq )
SO4(2aq )
Initial
0.050
0.050
0.0
Change
-x
+x
+x
Eq
0.050 - x
0.050+x
+x

2
[ H ( aq ) ][ SO4( aq ) ] (0.050  x) x
Ka 

 1.2  102

[ HSO4( aq ) ]
(0.050  x)
0.050 x  x 2  6.0 10 4  1.2 10 2 x
x 2  0.062 x  6.0 10 4  0
x
0.062 
 0.062 
2
 4 1  -6.0 10 4 
2
2
0.062  7.90 10
 x  8.51 103 M
2
[H+] = 0.050+0.0085 = 0.059M
x
2
________________________________________________________________________
Weak Base Example: Calculate the pH of 0.050 M NH3.
NH3( aq )  H 2O
OH(aq )  NH 4( aq) Kb  1.8 105
OH(aq )
NH 4( aq )
Initial
0.050
0.0
0.0
Change
-x
+x
+x
Eq
0.050 - x
+x
+x


2
[OH ( aq ) ][ NH 4( aq ) ]
x
Ka 

 1.8 105
[ NH 3( aq ) ]
0.050  x
NH 3( aq )
x 2  9  107  1.8  105 x  x 2  1.8 105 x  9 107  0
x
1.8 105 
1.8 10 
5 2
 4 1  -9 107 
2
4
x  9.4 10 M
[OH  ]  9.4 104 M
pH  14  log[OH  ]  14  3.03  10.97
___________________________________________________________________
Conversion from Kb to Ka Example: Determine the Kb of HCN if Ka = 4.9 x 10-10.
K
11014
K w  K a  Kb  Kb  w 
 2.04 105
K a 4.9 1010
___________________________________________________________________
Salt Classification Example I: Classify the following solutions as basic, acidic, or
neutral.
a.) KBr
b.) NaNO2
c.) NH4Cl
Answer:
a.) neutral
b.) basic
c.) acidic
___________________________________________________________________
Salt Classification Example II: Calculate the Ka for the cation & the Kb for the anion in
an aqueous solution containing NH4CN. Is the solution acidic, basic, or neutral?
NH 4CN  NH 4  CN 
for NH 4 : NH 4  H 2O
NH 3  H 3O 
Ka  ?
we will not find this K a in a table BUT we can find the K b of NH 3 to it:
NH 3  H 2O
NH 4  OH 
K b  1.8 105  K a 
Kw
1014

 5.56  1010
K b 1.8 105
for CN  we will have to use the K a of HCN to get its K b
CN   H 2O
HCN  OH 
Kb  ?
3
CN   H 3O 
HCN  H 2O



K a  4.9 1010  K b 

Kw
1014

 2.04 105
10
K a 4.9 10
Kb CN   K a NH 4  the soln is basic
___________________________________________________________________
Salt Example: Calculate the pH of a 0.25M NaC2H3O2, Ka = 1.76x10-5.
Kb 
Kw
11014

 5.68 1010
K a 1.76 105
C2 H3O2( aq)  H 2O(l )
HC2 H3O( aq)  OH(aq)
HC2 H 3O( aq )
OH 
C2 H3O2( aq)
Initial
0.250
0.0
0.0
Change
-x
+x
+x
Eq
0.250 - x
+x
+x

2
[ HC2 H 3O( aq ) ][OH ]
x
Kb 

 5.68 1010

[C2 H 3O2( aq ) ]
0.250  x
because we have a large concentration of acetate and a small Kb we
will try and assume 0.250 >> x
x2
x2
 5.68 1010  x  1.19 105 M
0.250  x 0.250
1.19 105
ck :
100%  0.005%  5%
0.250
therefore our assumption is valid and [OH-] = 4.77x10-5M
pH = 14 - pOH
= 14 + log[OH] = 9.08
4