15.5-15.6 Autoionization of Water and pH, and Finding the [H3O+] and pH of
Strong and Weak Acid Solutions
Autoionization of Water
• Autoionization – the process by which water acts as an acid and a base with itself.
H20(l)
H+(aq) + OH-(aq)
• The equilibrium constant for this reaction is:
Kw = [H3O+][OH-] = [H+][OH-]
Kw is known as the ion product (or dissociation) constant for water.
At 25oC (298 K), Kw = 1.0X10-14. In pure water, the number of H3O+ and OH- ions are
equal, and the solution is neutral.
In a neutral solution [H3O+] = [OH-] = Kw of 1.0X10-7
In an acidic solution [H3O+] > [OH-]
In a basic solution [H3O+] < [OH-]
In all aqueous solutions [H3O+][OH-] = 1.0X10-14 (at 25oC) Remember, equilibrium
constants are temperature dependent.
The pH Scale
• The pH identifies the hydrogen potential of a solution. A low pH
signifies an acidic solution while a high pH represents a basic
solution.
• In general, the pH scale ranges from 0-14, with 7 representing a
neutral solution. 1 pH unit corresponds to a 10-fold change in
H3O+ concentration.
• pH = -log[H3O+] = -log[H+]
• When calculating the correct number of significant figures when
performing a logarithm, the log must have the same number of
decimal places as the original numbers significant figures. For
example, if you were to take the log of 1.0X10-4, which has two sig
figs., your answer should be reported to two decimal places: 4.00
Let’s Try a Practice Problem!
Calculate [H3O+] at 25oC for the following solution and
determine if the solution is acidic, basic, or neutral.
[OH-] = 1.5x10-2
Kw = [H3O+][OH-]
Kw
1.0x10-14
[H3O+] = --------- = ------------------ = 6.7x10-13
[OH-]
1.5x10-2
[OH-] > [H3O+], so the solution is basic
Let’s Try a Few Practice Problems!
Calculate the pH of the solution and indicate whether the solution is
acidic or basic:
[H3O+] = 9.5X10-9 M
pH = -log[H3O+] = - log(9.5x10-9) = 8.02, this solution is basic!
Calculate the [H3O+] for a solution with a pH of 8.37.
pH = -log[H3O+]
8.37 = -log[H3O+]
[H3O+] = 10-8.37 = 4.27x10-9
pOH and other p Scales
• Similar to the pH scale, but with respect to [OH-]
• A low pOH represents a basic solution, while a high
pOH represents a solution that is acidic.
• pOH = -log[OH-]
• The following two relationships are also on the
reference table!
• Kw = [H+][OH-] = 1.0x10-14 at 25oC
• pH + pOH = 14
• The pKa of a weak acid can also be used to quantify
its strength. pKa = -log Ka
Strong vs. Weak Acids
• For strong acids: the concentration of H3O+ in a strong acid
solution is equal to the concentration of the strong acid.
• For example: A 0.10 M HCl solution has a H3O+ concentration of
0.10 M and a pH of 1.00.
• For weak acids: This is not the case. Remember, weak acids
only partially ionize, and different weak acids at the same
concentration, can have different pH values.
– For these types of problems, we have to solve an equilibrium
problem, using an ICE table as we did in the previous chapter.
– The initial H3O+ concentrations are approximately zero because of
the negligible small contribution of H3O+ due to the autoionization
of water.
– In many cases, we can apply the x is small approximation (in these
cases, where the ration of x to the number it is subtracted from <
5.0 %, it is valid).
Finding the [H30+] and pH of Strong
and Weak Acid Solutions
• When a solution contains either a strong or weak
acid, there are two sources of hydronium ions:
HA(aq) + H2O(l)
H2O(l) + H2O(l)
H3O+(aq) + A-(aq) Strong or Weak Acid
H3O+(aq) + OH-(aq) Kw = 1.0X10-14
*side note: HA is a strong or weak acid.
In most strong or weak acid solutions, the autoionization of
water produces such a small amount of H3O+ than in pure
water, and can be ignored.
Let’s Try a Practice Problem!
Find the H30+ concentration of a 0.250 M hydrofluoric acid solution.
(Here the Ka value is 3.5x10-4 according to table 15.5 within the
chapter)
HF(aq) + H2O(l)
H3O+(aq) + F-(aq)
[HF] M
[H3O+] M
[F-] M
Initial
0.250
~0
0
Change
-x
+x
+x
Equilibrium
0.250 - x
x
x
[H3O+][F-]
(x)(x)
Ka = ---------------- = ------------[HF]
0.250 - x
Continued on the next slide 
[H3O+][F-]
(x)(x)
Ka = ---------------- = -------------[HF]
0.250 – x  x is a small value
Now I will plug in my Ka and solve for x:
x2
3.5x10-4 = --------0.250
X2 = 8.75x10-5 so, x = 9.35x10-3
Now we can check if x is valid:
9.35x10-3
------------- X 100 = 3.74 % so x is valid.
0.250
So, the H3O+ concentration is: 9.4X10-3 M
Let’s Try Another!
A 0.175 M weak acid solution has a pH of 3.25. Find Ka for the acid.
For this problem, first we need to solve for the hydronium ion concentration:
pH = -log[H3O+]
-3.25 = log[H3O+]
[H3O+] = 10-3.25 = 5.6x10-4 M
Now set up an ICE table!
[H3O+][A-]
Ka = --------------[HA]
[HA] M
[H3O+] M
[A-] M
Initial
0.175
~0
0
Change
- 5.6x10-4
+ 5.6x10-4
+ 5.6x10-4
Equilibrium
0.175 -5.6x10-4
= 0.174
5.6x10-4
5.6x10-4
Continued on the next slide 
[H3O+][A-]
Ka = --------------[HA]
(5.6x10-4)2
Ka = -------------0.174
Ka = 1.8x10-6
Let’s Try Another!!
The initial concentration and Ka’s of several weak acid (HA) solutions
are listed here. For which of these is the x is small approximation
least likely to work in finding the pH of a solution?
a.) initial [HA] = 0.100 M; Ka = 1.0x10-5
b.) initial [HA] = 0.10 M; Ka = 1.0x10-6
c.) initial [HA] = 0.0100 M; Ka = 1.0x10-3
d.) initial [HA] = 1.0 M; Ka = 1.5x10-3
c.) initial [HA] = 0.0100 M; Ka = 1.0x10-3 (the validity of the x is small
approximation depends on both the value of the equilibrium
constant and the initial concentration – the closer these are to one
another, the less likely the approximation will be valid.
Let’s Try Another!!!
Which solution is most acidic? (That is, which one has the
lowest pH)
(a) 0.10 M HCl
(b) 0.10 M HF
(c) 0.20 M HF
(a) 0.10 M HCl (a weak acid solution will usually be less
than 5% dissociated. Since HCl is a strong acid, the
0.10 M solution is much more acidic than either a weak acid
with the same concentration or even a weak acid that is
twice as concentrated.)
Percent Ionization of a Weak Acid
• Another way to quantify the ionization of a weak acid,
is to calculate the percent ionization (the percentage of
acid molecules that actually ionize).
conc. of ionized acid [H3O+]equil
Percent ionization = --------------------------------------------------- X 100
initial concentration of acid [HA]initial
•
The equilibrium H3O+ concentration of a weak acid increases with the
increasing concentration of the acid
•
The percent ionization of a weak acid decreases with increasing the
concentration of the acid.
Let’s try a practice problem!
Find the percent ionization of a 0.250 M HC2H3O2 solution at 25oC. (Ka of acetic acid is 1.8x10-5)
[H3O+]equil
Percent ionization = --------------- X 100
[HA]initial
[HC2H3O2] M
[H30+]M
[C2H3O2-]
Initial
0.250
~0
0
Change
-x
+x
+x
Equilibrium
0.250-x
x
x
[H30+][C2H3O2-]
Ka = ---------------------- =
[HC2H3O2]
x2
1.8x10-5 = ----------0.250 –x
X2 = 4.5x10-6 so, x = 2.12x10-3
[H3O+]equil
2.13x10-3
Percent ionization = --------------- X 100 = -------------- x 100 = 0.85%
[HA]initial
0.250
Let’s Try Another!!!
Which of these weak acid solutions has the greatest percent
ionization? Which solution has the lowest (most acidic pH)?
a.) 0.100 M HC2H3O2
b.) 0.500 M HC2H3O2
c.) 0.0100 M HC2H3O2
c.) 0.0100 M HC2H3O2 would have the greatest percent
ionization because percent ionization increases with decreasing
weak acid concentration. But b.) 0.500 M HC2H3O2 would have the
lowest pH, because the equilibrium H3O+ concentration increases
with increasing weak acid concentration.
Mixture of Acids
• If we were tasked with finding the pH of a mixture of acids we would:
– 1.) See if one of the acids were strong, and if so, we
could neglect the weaker acid.
– 2.) See if we are dealing only with weak acids. If we are,
and the concentrations are similar, but the Ka’s are
sufficiently different (differ by more than a factor of
several hundred), then again, we can focus on the
stronger acid as the weaker makes an insignificant
contribution.
– If we were asked to find the concentration of a part of the weak acid, we have
to set up an ICE table, except here, we can’t place a zero in for the initial
hydronium ion concentration. Instead, the initial hydronium ion concentration
is equal to the hydronium ion concentration formed by the strong(er) acid.
Let’s Try a Practice Problem that Deals with a Mixture of Two Weak Acids!
Find the pH of a mixture that is 0.150 M in HF and 0.100 M HClO.
The following are given:
HF(aq) + H2O(l)
H3O+(aq) + F-(aq) Ka = 3.5x10-4
HClO(aq) + H2O(l)
H3O+(aq) + ClO-(aq) Ka = 2.9x10-8
H2O(l) + H2O(l)
H3O+(aq) + OH-(aq) Kw = 1.0x10-14
[HF] M
[H3O+] M
[F-] M
Initial
0.150
~0
0
Change
-x
+x
+x
Equilibrium
0.150-x
x
X
[H3O+][F-]
Ka = ---------------- =
[HF]
x2
3.5X10-4 = ------------0.150 – x
x = 7.2x10-3
Let’s check if x is valid: (7.2x10-3 / 0.150) X 100 = 4.8%  Valid, but barely
pHmix = -log(7.2x10-3) = 2.14
Let’s Try Another!!
Use the information from the previous problem, to find
the ClO- concentration of the mixture of HF and HClO.
[HClO] M
[H3O+] M
[ClO-] M
Initial
0.100
7.2x10-3
0
Change
-x
+x
+x
Equilibrium
0.100-x
7.2x10-3+x
x
[H3O+][ClO-]
Ka = ------------------[HClO]
x = 4.0x10-7
(7.2x10-3+x)(x)
2.9x10-8 = ------------------0.100-x
[ClO-] = 4.0x10-7 M
Let’s Try Another Practice Problem!!!
Which solution is most acidic (that is, which has the
lowest pH)?
a.) 1.0 M HCl
b.) 2.0 M HF
c.) A solution is 1.0 M in HF and 1.0 in HClO
a.) 1.0 M HCl, this is the only strong acid of the acids
listed above.
15.5-15.6 pgs. 746-747 #’s 48, 50, 56, 58 , 66, 70, 74, 80 (a’s only)
Read 15.7-15.8 pgs. 720-731