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Autoionization of Water
 Although water is a molecular substance, very
low concentrations of hydronium ions and
hydroxides ions are formed by autoionization.
H
O
H
+
H
O
H
H
O
-
+
H
O+ H
H
Autoinization
Autoionization of Water
 Autoionization:
 the process in which water spontaneously forms
low concentrations of H+ and OH - ions by proton
transfer from one water molecule to another
2 H2O (l)
H3O+ (aq) + OH- (aq)
 At any given time only a very small number of
water molecules are ionized.
• If every letter in our text represented a water
molecule, you would have to look through about 50
texts to find one H3O+!
Autoionization of Water
 An equilibrium constant expression can be
written for the autoionization of water:
2 H2O (l)
H3O+ (aq) + OH - (aq)
Kw = [H3O+] [OH -] = [H+ ] [OH- ]
where Kw = ionization constant for water
= ion-product constant
= 1.00 x 10-14 at 25oC
Autoionization of Water
 The value of Kw (and all other equilibrium
constants) varies with temperature:
 Kw = 1.14 x 10-15 at 0oC
 Kw = 1.00 x 10-14 at 25oC
 Kw = 9.61 x 10-14 at 60oC
 You should assume 25oC unless otherwise
stated in a problem.
Autoionization of Water
By definition:
 Neutral solution:
[H+ ] = [OH- ]
 Acidic solution: [H+ ] > [OH- ]
 Basic solution: [H+ ] < [OH- ]
Autoionization of Water
Example: Calculate the [H+ ] and [OH- ] in a
neutral solution at 25oC.
Autoionization of Water
Example: What is the [H+ ] at 25oC for a
solution in which [OH- ] = 0.010 M.
Autoionization of Water
Example: What is the [OH- ] at 25oC in a
solution in which [H+ ] = 2.5 x 10-6 M.
pH
 Since the [H+] is usually very small in aqueous
solutions, we normally express the [H+] in
terms of pH.
pH = - log10 [H+]
 Sig Figs and logs: only the digits after the
decimal point are significant.
pH
Example: Calculate the pH of a solution with
[H+] = 2.52 x 10-5.
pH
Example: Calculate the pH of a solution with
[OH-] = 6.5 x 10-5.
pH
 The negative log is also used to express the
magnitude of other small quantities:
 pOH = - log [OH- ]
 pH and pOH are related by the following
equation that is derived by taking the negative
log of the expression for Kw
pH + pOH = 14.00 at 25oC
pH
Example: Calculate the pOH of a solution with
[OH - ] = 2.5 x 10-3 M.
pH
Example: Calculate the pH of a solution with
[OH - ] = 2.5 x 10-3 M.
Approach 1:
pH
Approach # 2:
pH
 Given the pH of a solution, you can also find
the [H+] and the [OH-].
 Since pH = - log [H+],
[H+] = 10
–pH
 Since pOH = - log [OH-],
[OH-] = 10
-pOH
pH
Example: What are the [H+] and [OH-] for a
solution with a pH of 2.50 at 25oC?
Strong Acids
 Strong acid:
 Strong electrolyte
 Ionizes completely in aqueous solution
HNO3 (aq)
H
+ (aq)
+ NO3 – (aq)
 The only significant source of H+ ion in an
aqueous solution of a strong acid is usually the
strong acid.
Strong Acids
 Consequently, the [H+] in a solution of a strong
monoprotic acid can be determined easily using
the concentration of the strong acid itself.
HNO3 (aq)
H
+ (aq)
+ NO3 – (aq)
 In a 0.05 M HNO3 (aq) solution,
[H+] = 0.05 mol HNO3 x 1 mol H+
L
1 mol HNO3
= 0.05 M
Strong Acids
Example: What is the pH of a 0.25 M HCl (aq)
solution?
Strong Bases
 Strong Base
 strong electrolyte
 ionizes completely in aqueous solution
NaOH (aq)
Na+ (aq) + OH- (aq)
 Common strong bases
 alkali metal hydroxides
 hydroxides of Ca, Sr, and Ba
Strong Bases
 The pH of an aqueous solution of a strong
base can be determined using the
concentration of the strong base
NaOH (aq)
Na+ (aq) + OH- (aq)
 A 0.25 M solution of NaOH has an [OH-] of
0.25 M:
0.25 mol NaOH x 1 mol OHL
1 mol NaOH
= 0.25 M
Strong Bases
 The pH of the base solution can then be found
in two ways:
 Calculate pOH
use pH + pOH = 14.00 to determine pH
 Calculate [H+]
use [H+] [OH-] = 1.00 x 10-14
Then calculate pH
Strong Bases
Example: Calculate the pH of a 0.25 M Ca(OH)2
(aq) solution.
Step 1: Determine [OH-]
Strong Bases
Step 2: Calculate pOH
Step 3: Calculate pH
Strong Bases
Example: What is the pH of a solution prepared
by mixing 10.0 mL of 0.015 M Ba(OH)2 and 30.0
mL of 7.5 x 10-3 M NaOH?
Strong Bases
Step 1: Find the total [OH-]
Strong Bases
Step 2: Find pOH
Step 3: Find pH
Weak Acids
 Most acidic substances are weak acids:
 partially ionize in solution
 the solution contains an equilibrium mixture
of acid molecules and its component ions
CH3CO2H
H+ (aq) +
CH3CO2 - (aq)
Weak Acids
 The extent to which a weak acid ionizes can
be expressed using an equilibrium constant
known as the acid-dissociation constant (Ka).
 For a general reaction:
HX (aq)
H+ (aq) + X- (aq)
Ka = [H+][X-]
[HX]
Note: The rules for writing
an expression for Ka are the
same as those for Kc, Kp
and Ksp.
Weak Acids
 The magnitude of Ka indicates the tendency of
the hydrogen ion in an acid to ionize.
 The larger the value of Ka, the stronger the
acid is.
 The pH of a weak acid solution can be
calculated using the initial concentration of the
weak acid and its Ka.
Weak Acids
 To calculate the pH of a weak acid solution:
 Write the ionization equilibrium for the
acid.
 Write the equilibrium constant expression
and its numerical value.
 Set up a table showing initial concentration,
change, equilibrium concentration.
 Substitute equilibrium concentrations into
the equilibrium constant expression.
Weak Acids
 To calculate the pH of a weak acid solution
(cont):
 Solve for the change in concentration.
Assume that the change in concentration
is small (i.e. < 5%) compared to the
initial concentration of the weak acid.
 Check the validity of previous assumptions.
If x/initial concentration x 100% >5.0%, you
must use the quadratic equation to solve for
x.
 Calculate the final concentrations and pH.
Weak Acids
Example: Calculate the pH of a 0.20 M
solution of HCN. Ka = 4.9 x 10-10
Step 1: Write the equation for the
ionization.
Step 2: Write the expression for Ka.
Weak Acids
Step 3: Set up a table.
Weak Acids
Step 4: Substitute equilibrium concentrations
into the Ka expression.
Step 5: Assume that x << 0.20 M and solve
for x.
Weak Acids
Step 6: Check the validity of our assumption.
Weak Acids
 Step 7: Substitute value for x into the table
to find the [H+].
Weak Acids
Step 8: Calculate the pH using the [H+]
Weak Bases
 Many substances behave as weak bases in
water.
Weak base + H2O
conjugate acid + OH-
 The extent to which a weak base reacts with
water to form its conjugate acid and OH- ion
can be expressed using an equilibrium constant
known as the base-dissociation constant (Kb).
Weak Bases
 Kb always refers to the equilibrium in
which a base reacts with water to form its
conjugate acid and OH- ion.
 For the reaction:
NH3 (aq) + H2O (l)
Kb = [NH4+] [OH-]
[NH3]
NH4+ (aq) + OH- (aq)
Note: The rules for writing
an expression for Kb are the
same as those for Kc, Kp and
Ksp.
Weak Base
 To calculate the pH of a weak base solution:
 Write the ionization equilibrium for the
base.
 Write the equilibrium constant expression
and its numerical value.
 Set up a table showing initial concentration,
change, equilibrium concentration.
 Substitute equilibrium concentrations into
the equilibrium constant expression.
Weak Bases
 To calculate the pH of a weak base solution
(cont):
 Solve for the change in concentration.
Assume that the change in concentration
is small (i.e. < 5%) compared to the
initial concentration of the weak base.
 Check the validity of previous assumption.
 Calculate the [OH-] concentration and pOH
 Use pOH to calculate pH.
Weak Bases
Example: Calculate the pH of a 0.20 M solution
of methylamine, CH3NH2. Kb = 3.6 x 10-4.
Step 1: Write the equation for the ionization.
Step 2: Write the expression for Kb.
Weak Bases
Step 3: Set up a table.
Weak Bases
Step 4: Substitute equilibrium concentrations
into the Kb expression.
Step 5: Assume that x << 0.20 M and solve
for x.
Weak Bases
Step 6: Check the validity of our assumption.
Weak Bases
 Step 7: Substitute value for x into the table
to find the [OH-].
Weak Acids
Step 8: Calculate the pOH
Step 9: Calculate the pH