Acids - Base Equilibria
Part I: pH and Acidity
Jespersen Chap. 17 Sec 1 & 2
Dr. C. Yau
Spring 2013
1
Autoionization of Water
H2O + H2O
2 H 2O
H3O+ + OH
H3O+ + OH
[H 3 O + ][OH - ]
Kc =
[H 2 O]2
With H2O as a solvent, [H2O] becomes a
constant. We rearrange the equation to give
Kc [H2O]2 = [H3O+][OH]
Kw = [H3O+][OH]
For simplicity, we write Kw = [H+][OH]
At 25°C, Kw = 1.0x10-14 = [H+][OH]
Know well! Note: 2 sig. fig. for 1.0
2
Kw = 1.0x10-14 = [H+][OH]
As usual, changes in concentration does not
change Kw. They only cause shifts in the
equilibrium. Only T can change Kw.
Table 17.1
p. 775
3
Kw = 1.0x10-14 = [H+][OH]
Neutral solutions: [H+]=[OH]
Acidic solutions: [H+]>[OH]
Basic solutions: [H+]<[OH]
Note how small Kw is.
This means water is a VERY weak acid.
This is why it is considered nonelectrolytic.
It does not conduct electricity unless it is
contaminated with ionic compounds.
Tap water IS electrolytic because it is not pure
water and contains salts.
4
H2O auto-ionization occurs in any solution
• When other ions present
– [H+]  [OH]
– But Kw = [H+]·[OH] = 1.0 x 1014
• In aqueous solution,
• Product of [H+] and [OH] equals Kw
• [H+] and [OH] may not actually equal each
other
5
In a sample of blood at 25°C, [H+] is 4.6x10-8 M.
Find the molar concentration of OH-, and decide if
the sample is acidic, basic, or neutral.
[H+][OH]= 1.0x10-14
-14
1.0x10
[OH - ] 
[H + ]
-14
1.0x10
=
-8
4.6x10
-7
=2.2x10
Compare
Ans. [OH-] = 2.2x10-7 M OH(very slightly basic, more OH- than H+)
Pract Exer 17.1 & 17.2 p. 776
6
pH Scale
When we have a dilute solution of acid, the
H+ concentration is very small, such as
0.0001M, and it becomes cumbersome to
be talking about H+ concentration of
0.0001 M or 1x10-4M.
The pH scale was invented to make it easier
to communicate the acidity of a solution.
KNOW THIS WELL: pH = - log [H+]
or pH = - log [H3O+]
7
e.g. [H+] = 2.43x104 M, what is the pH?
pH =  log [H+]
=  log 2.43x104 (3 sig.fig.)
=  (3.614)
(3 decimal places)
= 3.614
e.g. pH = 8.71, what is the [H+]?
pH =  log [H+]
8.71=  log [H+]
8.71 = log [H+]
Antilog(-8.71) = antilog log [H+]
10-8.71 = [H+]
(2 decimal places)
Ans. [H+] = 1.9x10-9 M (2 sig. fig.)
REMEMBER: 10-pH = [H+]
8
We see that the pH and H+ concentration has an
inverse relationship:
As H+ concentration increased, the pH decreases.
The smaller the pH, the more acidic it is.
The larger the pH, the more basic it is.
So, if the swimming pool water has a pH that is too
high, that means it is too basic.
9
pOH =  log [OH-]
e.g. If [OH] = 4.2x103M, what is the pOH?
pOH =  log 4.2x103 (2 sig.fig.)
=  2.38 = +2.38 (2 decimal places)
e.g. If pOH = 12.3, what is the hydroxide conc?
[OH] = 1012.3
(1 decimal place)
= 5x1013 M (1 sig. fig.)
THIS IS HOW pH AND pOH ARE RELATED:
Remember that [H+] [OH] = 1.0x1014
Finding the log of both sides of the equation gives us…
log [H+] + log [OH] = log 1.0x1014
Multiply both sides of equation by 1.
( log [H+] + log [OH])
= ( log 1.0x1014)
( log [H+] )+(  log [OH]) =  (14.00)
pH
+
pOH
= 14.00
REMEMBER THIS!
10
Practice Exer. 17.4 p. 778
Because rain washes pollutants out of the
air, the lakes in many parts of the world
have undergone pH changes. In a New
England state, the water in one lake was
found to have a hydrogen ion conc of
3.2x10-5 mol L-1. What are the calculated
pH and pOH? Is it acidic or basic?
Ans. pH = 4.49 (acidic)
pOH = 9.50
11
What is the pH of a NaOH soln at 25 oC in
which the hydroxide ion concentration
equals 0.0026 M?
Ans. pH = 11.41 (basic, as expected).
Practice Exercises 17.3 & 17.5 p. 778
12
Calculate the values of pH, pOH and [OH-] for the
following solutions:
a) 0.020M HCl
b) 0.00035 M Ba(OH)2.
You must first note that HCl is a strong acid. Only
because this is so can you assume 0.020M HCl
means 0.020 M H+.
Ans. a) pH = 1.70, pOH = 12.30,
[OH-]=5.0x10-13M
b) 0.00035M Ba(OH)2 means [OH-] = ?
Ans. [OH-] = 0.00070M, pOH = 3.15, pH = 10.85
13
Acid or basic solutes suppress the ionization of water.
H2O + H2O
2 H 2O
H3O+ + OH
H3O+ + OH
Remember Le Chatelier’s Principle:
If an acidic solute (e.g. HCl) is added to water, it shifts
the equilibrium to the left, thus suppressing the
ionization of water.
This means that in an acidic solution the only source of
H+ is from the acid and essentially no contribution from
water.
In the same way, in a basic solution the only source of
OH- is from the base, none from the dissociation of
14
water.
Calculating without a calculator
Often we do not need a calculator to
calculate pH or [H+].
[H+] = 108 What is the pH?
[OH] = 106 What is the pH?
pH = 4
What is the pOH?
pOH = 3
What is the pH?
15
A soln was made by dissolving 0.837 g
Ba(OH)2 in 100 mL final volume. What is
the molar conc of OH- in the soln? What
are the pOH and the pH? What is the
hydrogen ion concentration in the
solution? MM(Ba(OH)2) = 171.3 g mol-1
Ans. [OH-] = 0.0977 M, pOH = 1.010
pH = 12.990, [H+] = 1.02x10-13M
Practice Exer 6, 7, & 8 p. 779
16
What is the pH of 0.00000001 M HCl?
[H+] =1x108M
pH = 8 (basic)
How can a solution of HCl be basic??
Answer: Its pH will NOT be 8.
At very dilute solutions, the H+ will NOT
suppress the dissociation of water.
Water will contribute to the H+ concentration
and keep it from becoming basic!
Ans. pH = 7 (stays neutral)
17