5.2.1- Strong acids and bases
5.2.2- Ionization constants
5.2.3- Calculating pH
5.2.4- The pH scale
5.2.5- Indicators
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Strong acids produce many H+ ions
(or H3O+ ions)
weak acids produce few H+ ions
Strong bases produce many OH- ions
weak bases produce few OH- ions

Strong acids and bases are essentially one-way
reactions - the acid or base breaks down
completely to produce ions. At equilibrium there
are only products (the ions) left.
Strong Acids
HCl(aq) → H+(aq) + Cl-(aq)
H2SO4 (aq) → H+(aq) + HSO41-(aq)
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
Strong Bases
NaOH(aq) →Na+(aq) + OH- (aq)
Mg(OH)2 (aq) → Mg2+(aq) + 2 OH-(aq)

Weak acids and bases do not ionize completely. For
weak electrolytes, equilibrium lies to the reactant side
of the equation where there will be few ions present.
Some examples:
Weak Acids
HC2H3O2 (aq) ↔ H+(aq) + C2H3O2-(aq)
HCHO2 (aq) ↔ H+(aq) + CHO2-(aq)
Weak Bases
NH3 (aq) + H2O(l) ↔ NH4+(aq) + OH-(aq)
NH2CH3(aq) + H2O(l) ↔ NH3CH3+(aq) + OH-(aq)
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Ka and KbPay attention to the physical states and
whether or not a particular substance is
included in the equilibrium
Note: Ka for acids and Kb for bases.
HCl(aq) → H+(aq) + Cl-(aq)
Ka = [H+] [Cl-] =1.3×106
[HCl]
HCl(g) + H2O(l) → H3O+(aq) + Cl-(aq)
Ka = [H3O+] [Cl-] =1.3×106
[HCl]
HCHO2 (aq) ↔ H+(aq) + CHO2-(aq)
Ka = [H+] [CHO2-] = 1.8 × 10-4
[HCHO2]
Mg(OH)2 (aq) → Mg2+(aq) + 2 OH-(aq)
Kb = [Mg2+] [OH-]2
[Mg(OH)2]
NH3 (aq) + H2O(l) ↔ NH4+(aq) + OH-(aq)
Kb = [NH4+] [OH-] = 1.8×10-5
[NH3]
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A large value of Ka
means there are many H+ ions in solution in other words, a strong acid
A large Kb indicates many OH- ions a strong base
The Table of Acid and Base Strengths gives
Ka and Kb values for a number of acids and
bases. Kavalues for very strong acids are not
given
We usually do not think of water as producing ions, but
water does ionize, but not very well.
 We can write the ionization equation for water in two
ways. The equilibrium constant, Kw, can also be written
for both equations
 Notice that H2O does not appear in the Kw expression
because it is a liquid

H2O(l) → H+(aq) + OH-(aq)
Kw = [H+] [OH-] = 1.0 × 10-14
2 H2O(l) → H3O+(aq) + OH-(aq)
Kw = [H3O+] [OH-] = 1.0 ×10-14
Values for Kw are given for 25°C.

In pure water, the balanced equation tells us
that the concentrations of H+ and OH- will be
equal to one another. We find that [H+] = [OH] = 1.0×10-7

The value of Kw is very small, meaning that
very few ions are present.
As long as temperature remains constant
Kw is a constant and it's value will not change.


Ka and Kb are used to calculate the
concentrations of ions in acids and bases. This
will be crucial to determining pH

How you calculate ion concentrations
depends on whether you have a strong acid
(or base) or a weak acid (or base).

For strong acids and bases, the concentration
of the ions can be readily calculated from the
balanced equation.
1.Calculate the hydrogen ion concentration in a 0.050 M
solution of hydrochloric acid.
Solution:
We know that HCl is a strong acid that ionizes
completely in water (you should memorize the list of
strong acids).
Begin by writing the balanced reaction:
HCl(aq) → H+(aq) + Cl-(aq)
From the balanced equation we see that 1 mole of HCl
produces 1 mole of H+ (a 1:1 ratio), therefore the
concentration of H+ will equal that of HCl.
Answer: [H+ ] = 0.050 M
Also [Cl-] = 0.050 M
2.Calculate the hydroxide ion concentration in
a 0.010 M solution of barium hydroxide,
Ba(OH)2. Barium hydroxide is a strong base.
Always begin by writing a balanced equation:
Ba(OH)2 (aq) → Ba2+(aq) + 2 OH-(aq)
Since 2 moles of OH- are produced for every 1
mole of Ba(OH)2 , the concentration of OHwill be twice the concentration of Ba(OH)2 .
Answer:
[OH-] = 2 × 0.010 = 0.020 M
Also [Ba2+] = 0.010 M
Weak acids and bases require a much different
approach to finding ion concentrations. Once
you know you have a weak acid or base,
follow these steps:
1. Write a balanced equation
2. You will need to know the value of Ka or Kb,
look it up in a Table of Acid and Base
Strengths.
3. Set up the equilibrium constant expression.
1.Calculate the hydrogen ion
concentration in a 0.10 M acetic acid
solution, C2H3O2.
Ka for acetic acid, a weak acid, is
1.8 ×10-5.
Begin by writing the balanced reaction:
HC2H3O2 (aq) ↔ H+(aq) + C2H3O2-(aq)
Concentration of the acid, HC2H3O2 is 0.10 M.
We need to find the concentration of H+, which
will also equal the concentration of C2H3O2Because ionization is NOT complete because
this is a weak acid, [H+] will NOT equal
[HC2H3O2]. Instead we must calculate it using
the equilibrium constant expression.
Ka = [H+] [ C2H3O2-]
[HC2H3O2]
Substitute values into the equation. Let x equal the
unknowns
1.8 ×10-5 = (x) (x)
0.10
Rearrange the equation
X2 = (1.8 ×10-5)(0.10)
X2 = 1.8 ×10-6 Take the square root
X = 1.3×10-3
ANSWER: [H+] = 1.3×10-3
Also [C2H3O2-] = 1.3×10-3
2.Calculate the hydroxide ion concentration,
[OH-], in a 0.025 M solution of analine,
C6H5NH2, a weak base with Kb = 4.3×10-10
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

Begin by writing a balanced equation. Since analine is a base
that doesn't contain the hydroxide ion, include H2O as a
reactant.
C6H5NH2 (aq) + H2O (l) ↔ C6H5NH3+(aq) + OH-(aq)
Set up the Kb expression and solve for ion concentrations. We
see from the balanced equation that the ions have a 1:1 ratio,
therefore [OH-] will equal the [C6H5NH3+].
Set up the Kb equation, omitting liquid water:
Kb = [C6H5NH3+] [OH- ]
[C6H5NH2]
Substitute values into the equation. Let x equal the unknowns
4.3×10-10 = (x) (x)
0.025
X2 = (4.3×10-10)(0.25)
X2 = 1.1×10-11 Take the square root
X = 3.3×10-6
ANSWER:[OH-] = 3.3×10-6 M
Also [C6H5NH3+] = 3.3×10-6 M
H2O(l) → H+(aq) + OH-(aq); [H+= 0.050 M]
Kw = [H+] [OH-] = 1.0 × 10-14
remember that equilibrium constants are constant. Thus the
value of Kw will still have a value of 1.0 ×10-14 even if [H+] has
increased due to the presence of the acid. We can use this
information to calculate the concentration of hydroxide
ions present in the aqueous solution:
Kw = [H+] [OH-]
Rearrange the equation
[OH- ] = Kw______
[H+]
Substitute known values and solve for [OH- ]
[OH- ] = 1.0 ×10-14
0.05
[OH- ] = 2.0 ×10-13
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

For any acid or base you can calculate
both [H+] and [OH-]
Acids
First determine [H+] then use Kw to calculate
[OH-]
Bases
First determine [OH-] then use Kw to calculate
[H+]
Recall: When we disrupt an equilibrium system
by increasing the concentration of a reaction
participant, equilibrium will shift to minimize the
stress.
 When we increase the H+ ion concentration in
the water equilibrium, the reaction will shift to
the left to "use up" the additional H+. This will
cause the concentration of OH- to decrease.
Indeed, we see that [OH-] does decrease, from
1.0×10-7 in pure water to 2.0 ×10-13 in our acid
solution.
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
Practice set 5.2.4- do questions 1-7

pH is just another way to express the
hydrogen ion concentration ([H+]), of an
acidic or basic solution.
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
pH is defined as the negative log of hydrogen
ion concentration.
Mathematically it looks like this:
pH = - log [H+]
[H+]
1 ×10-3
2.5 ×10-11
4.7 ×10-9
5.8 ×10-4
1.0×10-7
pH
3.0
10.6
8.3
3.2
7.0
Acids
Bases
Neutral solutions
pH < 7
pH > 7
pH = 7
The lower the pH,
the stronger the acid
The higher the pH,
the stronger the base
1.Calculate the pH of a 0.01M HNO3 solution?
Begin finding pH by first finding [H+].
HNO3 is a strong acid,based on the balanced
equation, we see that there is a 1:1 ratio
between HNO3 and H+, so [HNO3] = [H+]:
HNO3 (aq) → H+(aq) + NO3-(aq)
[H+] = -log [H+] = -log (0.01)
= - (-2.0) = 2.0 answer
Find the pH of a 0.01 M solution of ammonia.
Ammonia is a weak base with Kb= 1.8 × 10-5
This is a more difficult question because it’s a
weak base (similar for a weak acid). We must
use Kb to determine ion concentrations. That
will require a balanced equation- for weak
bases (that don't contain an OH–), be sure to
include water, H2O as a reactant.
Remember that the base will gain a hydrogen
ion:
NH3 (g) + H2O(l) ↔ NH4+(aq) + OH-(aq)
First calculate [OH-]. Then use Kw to determine [H+]
Set up the Kb equation:
Kb = [NH4+] [OH- ]
[NH3]
Substitute values into the equation. Let x equal the unknowns
1.8 ×10-5 = (x) (x)
0.01
Rearrange the equation
X2 = (1.8 × 10-5) (0.01)
X2 = 1.8 × 10-7 Take the square root
X = 4.2 × 10-4
[OH-] = 4.2 × 10-4 M
Next we calculate [H+]:
Kw = [H+] [OH-]
[H+ ] = __Kw___
[OH-]
Substitute in known values and calculate [H+]
[H+] = 1.0 ×10-14
4.2 × 10-4
[H+] = 2.4 × 10-11
Finally, convert [H+] into pH:
[H+] = 2.4 × 10-11
pH= -log[H+] = -log(2.4 × 10-11)
pH= 10.6 answer
There is a way to simplify the last parts of this
operation, by using pOH:
pOH = - log [OH-]
Once we find [OH-] for a base, we can determine pOH:
[OH-] =4.2 × 10-4
pOH= -log [OH-] = -log (4.2×10-4)
pOH= 3.4 answer
Next we use of the following easy-to-memorize
relationship:
pH + pOH = 14
Once we find pOH, it is a simple matter to find pH:
pH + pOH = 14
OR
pH = 14 - pOH = 14 - 3.4 pH= 10.6

It doesn't matter which method you use to
find [H+] and pH for a base - both will give you
the same answer. Choose whichever method
works best for you.

Do questions 8,9,11 from the practice
problems 4-2-4
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
You know the pH of a solution and need to
find [H+], or the concentration of the acid
solution. How do you do that?
To convert pH into [H+] involves taking
the antilog of the negative value of pH .[H+] =
antilog (-pH)
different calculators work slightly differently make sure you can do the following
calculations using your calculator. Practice as
we go along . . .
We have a solution with a pH = 8.3. What is
[H+]?
1.
2.
3.
Enter 8.3 as a negative number
Use your calculator's 2nd or Shift or INV
key to type in the symbol found above
the LOG key. The shifted function
should be 10x.
You should get the answer 5.0 × 10-9
Other calculators require you to enter keys in the
order they appear in the equation.
1. Use the Shift or second function to key in the
10x function.
2. Use the +/- key to type in a negative number,
then type in 8.3
3. You should get the answer 5.0 × 10-9
If neither of these methods work, try rearranging
the order in which you type in the keys.

1.Find the hydronium ion concentration
in a solution with a pH of 12.6. Is this
solution an acid or a base? How do you
know?
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
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The solution is a base because pH > 7.
To find hydronium ion concentration, [H3O+],
which will be the same as [H+].
To convert pH into [H3O+]:
[H3O+] = antilog (-pH)
= antilog (-12.6)
[H3O+] = 2.5 × 10-13 answer
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In order to determine Ka we need to know the concentrations of
several things: [H+], [HCO3-], and [H2CO3]
Begin with a balanced equation for the acid:
H2CO3 (aq) ↔ H+(aq) + HCO3-(aq)
Next set up the equilibrium expression, which will be needed to
find Ka:
Ka = [H+][HCO3-]
[H2CO3]
Next find [H+] from pH.[H+] = antilog (-pH) = antilog (-3.49)
[H+] = 3.2 × 10-4
The balanced equation tells us that [H+] = [HCO3-]. Substitute
values into our Ka expression and solve:
Ka = [H+][HCO3-]
[H2CO3]
The question gave us the concentration of the acid, H2CO3, as
0.24 M
=[3.2 × 10-4] [3.2 × 10-4]
[0.24]
Ka=4.3 ×10-7 answer
A 0.24M solution of the weak acid, H2CO3,
has a pH of 3.49. Determine Ka for
H2CO3 (carbonic acid).

Assignment 5.2.4

Indicators are dyes that change colour under
varying conditions of acidity. Litmus is an
indictor that changes colour from red to blue
in the pH range of 5.5 to 8.0. Other indicators
and their colours are listed in the table of
Acid - Base Indicators
Indicators may be in solution form or paper form. pH
paper is prepared by treating the paper with the
indictor solution. When the paper is then dipped into
the solution you are testing, it will change colour
depending on the acidity of the solution.
 Here are some questions to try. A short acid-base
indicator table is given here:

Indicator
pH range
Colour change
methyl orange
3.2 - 4.4
red to yellow
Litmus
5.8 - 8.0
red to blue
Phenolphthalein 8.2 - 10.0
colourless to pink

A given solution turns methyl orange yellow,
litmus blue, and phenolphthalein red. What is
the approximate pH of the solution?


Methyl orange in yellow when pH is above 4.4
Litmus is blue when pH is above 8.0, and
Phenolphthalein is red when pH is above
10.0.
Therefore the solution would have to have a
pH above 10.0


What color would methyl orange, litmus, and
phenolphthalein turn when testing:
a. vinegar (pH = 3)
b. sea water (pH = 8)
methyl
orange
litmus
phenolphthal
ein
vinegar
red
red
colourless
sea water
yellow
red-blue
colourless