Acid-Base Practice and Info
Dr. Bilicki
Acid-Base Calculations
The Ion-Product Constant for Water, Kw
Water undergoes ionization to a small extent:
H20(l)  H+(aq) + OH–(aq)
The equilibrium constant for the reaction is the ion-product constant for water Kw:


14
K w  [H ][OH ]  1.0  10
(1)
This is a key equation in acid-base chemistry. Note that the product of [H+] and [OH–] is a
constant at a given temperature (Eq(1) value is for 25oC). Thus as the hydrogen ion
concentration of a solution increases, the hydroxide ion concentration decreases (and vice
versa).
The pH scale is widely used to report the molar concentration of hydrogen ion H+(aq) in
aqueous solution. The pH of a solution is defined as

pH   log 10[H ]
(2)
Similarly, pOH and pKw are defined as

pOH   log10 [OH ]
pKw  log10(Kw ) 14.00
(3)
(4)
If you take the log10 of both sides of Eq(1), multiply the resulting equation by (-1), and use
the definitions of pH, pOH and pKw above, the result is the very useful equation
pH + pOH = pKw = 14.00
(5)
Equations (2) and (3) above may be solved for [H+] and [OH–] respectively to give

pH
[H ]  10

 pOH
[OH ] 10
(6)
(7)
x
(Here we use the well known rule that if log 10 y  x , then y  10 .) In practice, the pH scale
is only used when [H+(aq)] is less than 1.0 M.
Acidic, basic, and neutral solutions can be distinguished as shown below:
Type of Solution
Acidic
Neutral
Basic
pH
< 7.00
= 7.00
> 7.00
[H+]
> 1.0  10 7
= 1.0  10 7
< 1.0  10 7
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Color of litmus
pink
in between
blue
Acid-Base Practice and Info
Dr. Bilicki
pH and [H+] Calculations for Strong Acids and Bases
By definition, strong acids and bases are 100% ionized in water solution. Ionization of a
strong acid gives rise to H+ ions, and ionization of a strong base produces OH– ions. The
equilibrium constant for a strong acid or strong base is undefined, since the reaction the
ionization is complete. There is no equilibrium!
In nearly all cases of practical interest the [H+] for a strong acid (or the [OH–] for a strong
base) is determined completely by the stoichiometry of the reaction. Once the [OH–] or pOH
is known for a base, the [H+] or the pH of the base may be calculated using Eq(1) and/or
Eq(5).
Exercises
1. Complete the following table:
pH
(a)
(b)
(c)
(d)
[H+]
pOH
[OH–]
Acidic,
basic, or
neutral?
5.4 x 10–4
7.8 x 10-10
10.75
5.00
Answers:
(a) pH = 3.27; pOH = 10.73; [OH–] = 1.85 x 10–11 = 1.9 x 10–11, acidic (since pH < 7).
(b) pH = 4.89, [H+] = 1.3 x 10–5, pOH = 9.11, acidic (since pH < 7).
(c) [H+] = 1.8 x 10-11, pOH = 3.25, [OH–] = 5.6 x 10–4, basic (since pH > 7).
(d) pH = 9.00, [H+] = 1.0 x 10–9, [OH–] = 1.0 x 10–5, basic (since pH > 7).
2. Calculate the pH of a 0.0430 M HNO3 solution.
Answer:
Since HNO3 is a strong acid, the nitric acid solution will be 100% ionized. Thus [H +] = [NO3–] = 0.0430 M. The
pH = 1.37 (use Eq(2)).
3. Calculate the pH of a 0.020 M Ba(OH)2(aq) solution.
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Acid-Base Practice and Info
Dr. Bilicki
Answer:
Since Ba(OH)2 is a strong base it is 100% ionized. Note that ionization gives 2 OH– ions for each mole of
Ba(OH)2. Thus [OH–] = 2 x 0.020 M = 0.040 M. Eq(3) gives pOH = 1.40. Then using Eq(5), pH = 12.60.
pH and [H+] Calculations for Weak Acids and Bases
Weak acids and bases are usually less than 5% ionized. The equilibrium constant for a weak
acid equilibrium is the acid ionization constant Ka, and for a weak base equilibrium is the
base ionization constant Kb.
A typical monoprotic weak acid equilibrium can be written in two forms, the second of which
emphasizes the Brønsted acid-base nature of the reaction:
HA  H+(aq) + A–(aq)
HA + H2O  H3O+(aq) + A–(aq)
(8)
In Eq(9) the Brønsted acid HA donates a proton H+ to the Brønsted base H2O to form H3O+
and the conjugate base A–. The acid ionization constant (using the second form) is
Ka 
[H3O  ][A ]
[HA]
(9)
A typical weak base equilibrium is
B + H2O  BH+(aq) + OH–(aq)
(10)
In Eq(10) the Brønsted base B accepts a proton H+ from the Brønsted base H2O to form the
conjugate acid BH+ and OH–. The base ionization constant is
Kb 
[BH  ][OH ]
[B]
(11)
Exercises
4. Calculate (a) the pH and (b) the percent ionization of a 0.250 M HC2H3O2 solution.
Ka(HC2H3O2) = 1.8 x 10-5. (The formula for acetic acid may also be written as CH3COOH.)
HINT: Begin by filling out the equilibrium table below.
Balanced Equation
HC2H3O2 
H+
+
C2H3O2–
Initial Concentration (M)
Change (M)
Equilibrium Concentration (M)
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Acid-Base Practice and Info
Dr. Bilicki
Answer:
HC2H3O2 
Balanced Equation
Initial Concentration (M)
Change (M)
Equilibrium Concentration (M)
0.250
-x
0.250 - x
H+
+
0
x
x
C2H3O2–
0
x
x

(a) K a 
[H ][C 2 H3 O2 ]
x2
x2
 1.8 105 

. This approximation is OK if the %
[HC 2 H3 O2 ]
0.250 - x 0.250
ionization is < 5%; it is in this case -- see answer to (b) below. Thus x2 = 4.5 x 10-6; x = 2.12 x 10-3 = [H+]. pH =
2.67.
(b)
% ionization =
2.12 103 
 x 
100%  0.85% .
 100%  
0.250 
 0.250 
5. Calculate the pH of a 0.600 M solution of methylamine CH3NH2. Kb = 4.4 x 10–4.
HINT: Methylamine is a weak base. First write the equation for the reaction following the
pattern of Eq(10). Then fill out the equilibrium table below.
CH3NH2  
Balanced Equation
Initial Concentration (M)
Change (M)
Equilibrium Concentration (M)
CH3NH3+ +
OH–
Answer:
Since CH3NH2 is a weak base, the balanced equation for the reaction is CH3NH2 + H2O  CH3NH3+ + OH–.
Balanced Equation
CH3NH2  
CH3NH3+ +
OH–
Initial Concentration (M)
Change (M)
Equilibrium Concentration (M)
0.600
-x
0.600 - x
0
x
x
0
x
x

[BH  ][OH ] [CH3 NH3 ][OH  ]
x2
x2
Kb 



 4.4 104 . Thus x = 1.62 x 10-2 =
[B]
[CH 3NH 2 ]
0.600  x 0.600
[OH–], and pOH = 1.79. It follows from Eq(5) that pH = 12.21. NOTE: The approximation used is OK since the
% ionization is 2.7% (i.e., less than 5 %).
6. The pH of a 0.10 M solution of a weak base is 9.67. What is the Kb of the base?
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Acid-Base Practice and Info
Dr. Bilicki
Answer:
The balanced equation for a weak base B is given in Eq(10). The equilibrium table required is given below.
Balanced Equation
B
BH+
+
OH–
 
Initial Concentration (M)
Change (M)
Equilibrium Concentration (M)
0.10
-x
0.10 - x
0
x
x
0
x
x
At equilibrium, [OH–] = [BH+] = x. Use the pH to calculate the [OH–] at equilibrium (which is the value of x).

pOH
4.33
5
Here pOH = 14.00 – pH = 14.00 – 9.67 = 4.33. Thus [OH ] 10
10
 4.68 10  x .
Kb 
[BH  ][OH  ]
x2
x2
(4.68 105 ) 2



 2.2 108 . The approximation is OK since
[B]
0.10  x 0.10
0.10
the % ionization is well under 5%.
Relationship between Ka for a Weak Acid and Kb for its Conjugate Base
The relationship between Ka for a weak acid HA and Kb for its conjugate base A– is
Ka(HA)  Kb(A–) = Kw = 1.0 x 10-14
(12)
If we define pKa = - log10(Ka) and pKb = - log10(Kb), the logarithmic form of Eq(12) is
The stronger the acid, the larger the Ka and the smaller the pKa. Likewise the stronger the
base, the larger the Kb and the smaller the pKb. Eqs(12) and (3) show that as the Ka increases
(and the pKa decreases), the Kb decreases (and the pKb increases). These equations give
quantitative support to the statement “the stronger the acid, the weaker the conjugate base.”
The justification for Eq(12) follows from the equations below. Recall that if Eq(1) + Eq(2) =
Eq(3), then K1  K2 = K3.
Eq(1), Weak Acid:
HA + H2O  H3O+(aq) + A–(aq)
Eq(2), Conjugate Base:
A–(aq) + H2O  HA(aq) + OH–(aq)
Eq(3) = Eq(1) + Eq(2)
___________________________
2 H20(l)  H3O+(aq) + OH–(aq)
[H 3O  ][A ]
[HA]
[HA][OH  ]

K b (A ) 

[A ]
K a (HA) 

Relationship between Kb for a Weak Base and Ka for its Conjugate Acid
Analogous equations to Eqs(12) and (13) above can be written the relationship between Kb
for a weak base B and Ka for its conjugate acid HB+:
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
K w  [H 3O ][OH ]
Acid-Base Practice and Info
Dr. Bilicki
Kb(B)  Ka(BH+) = Kw = 1.0 x 10-14
(14)
pKb(B) + pKa(BH+) = pKw = 14.00
(15)
The equations below provide justification for these results:
Eq(1), Weak Base:
B + H2O  BH+(aq) + OH–(aq)
Eq(2), Conjugate Acid:
BH+(aq) + H2O  H3O+(aq) + B(aq)
Eq(3) = Eq(1) + Eq(2)
___________________________
2 H20(l)  H3O+(aq) + OH–(aq)
[BH  ][OH ]
[B]
[H3 O ][B]
+
K a (BH ) 
+
[BH ]
K b (B) 


K w  [H 3O ][OH ]
___________________________________________________________________________
Exercise
7. Use the following acidity constants to help answer the questions below:
Ka(HC2H3O2) = 1.8 x 10 – 5 ;
Ka(HCN) = 4.9 x 10 – 10 ;
Ka(HCOOH) = 1.7 x 10 - 4
(a)
(b)
(c)
(d)
Which of the three acids is the weakest? ________________
Which of the following bases is the strongest: C2H3O2-, CN - , or HCOO- ? __________
What is the pKa of HCN? _____________
What is the Kb for CN- ?____________
Answers:
(a) HCN; (b) CN - ; (c) 9.31; (d) 2.04 x 10-5 2.0 x 10-5.
___________________________________________________________________________
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