Acid and Base Unit Review
4 types of problems (nearly identical to previous equilibrium problems)
1.
2.
3.
4.
Given initial concentration of acid (or base) and pH, calculate Ka (or Kb).
Given initial concentration of acid (or base) and Ka (or Kb), calculate pH.
Given pH and Ka (or Kb), calculate the initial concentration of acid (or base).
Given pH calculate % dissociation (% ionization) of a weak acid.
Types 1 & 4
1. The pH of a 0.40 M solution hypobromous (HOBr) acid is 5.05. Calculate Ka and the %
dissociation.
% of ionization is merely a percent value of how much of the initial concentration of the acid
became ionized.
acid concentration at equilibrium
% of ionization  ionizedinitial
x 100%
concentration of acid
pH = - log [H+]
5.05 = - log [H+]
10-5.05 = [H+]
8.91 x 10-6 M of H+
8.91x10 6
x100%  2.23 x10 3% or 0.00223%
1
4.0 x10
2. A 0.100 M solution of sodium carbonate (Na2CO3) has a pH of 11.618. What is Kb for the
carbonate ion (CO3-2)?
Important thing to note is the relationship between Kb, Ka, Kw.
Remember that Kw = 1.0x10-14 ([H+][OH-] = Kw = neutral water = [1.0x10-7][1.0x10-7]).
The product of the base ion constant (Kb) and the acid ion constant (Ka) equals Kw. This is
an expression of the relationship between an acid and its conjugate base. A strong acid (large
Ka) dissociates into a weak conjugate base (small Kb). Conversely, a weak acid (small Ka)
will dissociate into a strong conjugate base (large Kb). Multiply the Ka of an acid with the Kb
of its conjugate base and it will ALWAYS equal 1.0 x 10-14 (Kw).
In this question we can just calculate the Ka and divide that from Kw to get Kb.
pH = 11.618
10-11.618 = [H+] = 2.4 x 10-12
1
Ka 
[CO3 ][ Na2 ]
( this is a salt acting as an acid in a solution; no H  )
[ Na2CO3 ]
[2.4 x10 12 ][2.4 x10 12 ]
Ka 
 5.76 x10  23
1
[1.0 x10 ]
Kw 1.0 x10 14
Kb 

 1.736 x108
 23
Ka 5.76 x10
Types 2 & 4
3. Calculate the pH and % dissociation of a 1.00 M solution of hydrofluoric acid (HF).
Ka = 3.50 x 10-4. Repeat the calculation for a 1.00 x 10-2 M solution. Compare the two values
for % dissociation.
[ H  ][ F  ]
Ka 
[ HF ]
x2
3.5 x10 
1 x
0  x 2  3.5 x10  4 x  3.5 x10  4
4
 b  b 2  4ac
x
2a
 3.5 x10  4  (3.5 x10  4 ) 2  4(1)(3.5 x10  4 )
x
2(1)
 3.5 x10  4  3.74182 x10  2
x
2
x  0.0185341
pH = -log [0.0185341] = 1.732
% of ionization = ([H+]/[HF]) x 100%
= (0.0185341/1) x 100% = 1.85341 %
2
[ H  ][ F  ]
Ka 
[ HF ]
x2
3.5 x10 
0.01  x
0  x 2  3.5 x10  4 x  3.5 x10  2
4
 b  b 2  4ac
x
2a
 3.5 x10  4  (3.5 x10  4 ) 2  4(1)(3.5 x10  2 )
x
2(1)
 3.5 x10  4  0.3741657854
x
2
x  0.1869
pH = - log(0.1869)
pH = 0.7
% of ionization = ([H+]/[HF]) x 100%
= (0.1869/1) x 100% = 18.69 %
As the pH decreased by 1, the % of ionization increased by 10%
4. Calculate the [OH-] and the pH of a 0.15 M solution of sodium fluoride (NaF). Treat NaF
as if it is a STRONG ACID.
All of the acid ionizes completely. So, the concentration of all the +ions is 0.15 M.
pH = -log [0.15 M] = 0.82
pOH = 14 - 0.82 = 13.18
13.18 = - log[OH-]
10-13.18 = [OH-] = 6.6069 x 10-14 M
3
Type 3
5. A solution of ammonia has a pH of 10.50. What was the initial molarity of ammonia
(NH3)? Kb for ammonia is 1.8 x 10-5.
This question has multiple steps. To find the initial molarity of ammonia, you need to
calculate the Ka and the ion concentration of [H+] and [NH2-] first.
( Ka )( Kb)  Kw
Ka (1.8 x10 5 )  1.0 x10 14
Ka  5.5556 x10 10
Find the concentration of H 
pH   log[ H  ]
10.50   log[ H  ]
10 10.5  [ H  ]
[ H  ]  3.16228 x10 11
Now, solve for the initial concentration of NH 3

[ H  ][ NH 2 ]
Ka 
[ NH 3 ]
5.5556 x10
10
[3.16228 x10 11 ][3.16228 x10 11 ]

[ NH 3 ]
[3.16228 x10 11 ][3.16228 x10 11 ]
NH 3 
5.5556 x10 10
NH 3  1.7999856 x10 12 M
4
6. Write the balanced equation of H3PO4 reacting with water, where one proton is transferred.
H3PO4 + H2O → H2PO4- + H3O+
One proton (H+) is transferred from H3PO4 to H2O, making the conjugate base H2PO4- and
the conjugate acid H3O+
7.
NO2 is the conjugate base. It has a Kb = 1.408x10-11
Remember that Ka x Kb = Kw.
8. Complete the following (use table 15.2 on page 654, 664 and 668)
acid
acid
base
← conj. acid
base
base
acid
conj. base
→ conj. acid
← conj. acid
conj. base
conj. base
5
9. Using the list provided, determine which is the strongest acid and which has the strongest
conjugate base. Explain your answer.
STRONGEST ACID - HNO2
STRONGEST CONJUGATE BASE - CN -
10. REMOVED
11. Fill in the blank
d
__Strong__ acids ionize completely in a solution, whereas _weak_ acids only ionize partially.
Acids are known to _donate_ protons, whereas bases like to _accept_ protons.
The greater the Ka, the __stronger__ the acid.
The Kw of water is _1.00 x 10-14___. The pH of water is _7__.
6