Chapter 15
Acids and Bases
• First defined by Svante Arrhenius
1903 Nobel Prize winner who
proposed that:
Acids - produce H⁺ ions in aqueous
solution.
HCl(g)
H2O
H⁺(aq) + Cl⁻ (aq)
Arrhenius also proposed a useful
definition for bases.
Bases - produce OH⁻ ions in aqueous
solution.
H2O
NaOH(s)
Na⁺(aq) + OH⁻ (aq)
Acids and bases using according to the
Bronsted-Lowry model.
• Acid - proton donor
• Base – proton acceptor
• According to this model water, due to its
polar nature, aids in removing the proton.
In solution, the following reaction
occurs
HA (aq) + H₂O₍ι₎ → H₃O⁺ (aq) + A⁻ (aq)
• In this example: HA donates a proton to water
which would make HA the acid and water
accepts the proton which make it a base.
• The conjugate acid is H₃O⁺ the hydronium ion.
• The conjugate base is A⁻, which is everything
that remains after the acid donates its proton
Using a specific acid
• HBr (aq) + H₂O₍ι₎ → H₃O⁺ (aq) + Br⁻ (aq)
•
Acid
base
conj. acid
conj. Base
• H₂SO₄ (aq) + H₂O₍ι₎ → H₃O⁺ (aq) + HSO₄⁻ (aq)
•
Acid
Base
conj. Acid
conj. Base
• H₂S (aq) + H₂O₍ι₎ → H₃O⁺ (aq) + HS⁻ (aq)
•
Acid
base
conj. acid
conj. Base
• With Bronsted-Lowry acid base theory some themes persist.
• Water is the base, the conjugate acid is H₃O⁺ and the conjugate
base is simply the acid without a proton (H⁺ ion)
Differences between the 2 theories
• With Arrhenius, acidic aqueous solutions
contain the H⁺ ion.
• With Bronsted-Lowry H₂O is included in the
product, so a Hydronium ion (H₃O⁺) is found in
acidic solutions.
The Hydronium ion, H₃O⁺
• In water, H⁺ combines with water to
form the hydronium ion
H⁺ + H₂O → H₃O⁺
• For our purposes H⁺ and H₃O⁺ can be
used interchangeably.
Including H₂O as a reactant
• The ionization of an acid would look slightly
different if water is included as a reactant.
• Let’s use HBr as an example
• HBr + H₂O → H₃O⁺ + Br⁻
• Presence of either H⁺ or H₃O⁺ make a solution
acidic the difference is only how we choose to
write the ionization.
Water is amphoteric
• Amphoteric - a substance that can behave as
either an acid or a base.
• Equation for the ionization of water
H₂0₍ι₎ + H₂0₍ι₎ → H₃O⁺(acid) + OH⁻(base)
Or
H₂O → H⁺(acid) + OH⁻(base)
[
#
]
• Placing a number in brackets means that that
number then represents the a concentration
in “moles per liter” aka Molarity
• [OH⁻] = 1.6
• Means that the hydroxide ion concentration is
1.6 M
Comparing [H⁺] to [OH⁻]
• In neutral solution
[H⁺] = [OH⁻]
• In acidic solution
[H⁺] > [OH⁻]
• In basic solution
[H⁺] < [OH⁻]
Neutralization Reaction
• This is usually a double replacement reaction with an
acid as a reactant and a base as the other reactant.
The products are salt and water.
• Ex.
• HCl (aq) + NaOH (aq) → NaCl (aq) + HOH₍ι₎
•
Acid
base
salt
water
Other salts are possible
H₂CO₃ + Mg(OH)₂ → MgCO₃ + HOH
Acid
base
salt
water
HNO₃ + LiOH → LiNO₃ + HOH
Acid
base
salt
water
The pH scale
This is a convenient way to express relatively
small [H⁺]
Generally, pH results should range from 0-14.
To calculate pH you will use the log function on
your calculator.
pH values of common materials
Calculating pH
•pH = - log [H⁺]
Remember, [H⁺] is the hydrogen ion
concentration expressed in mol/L
a.k.a. MOLARITY
Calculating pH
An aqueous solution has a [H⁺] =2.97x10⁻⁶ what is
the pH of the solution?
pH = - log [H⁺]
pH = - log [2.97x10⁻⁶]
3 sig figs
pH = 5.527
3 sig figs
Calculating [H⁺] from pH
• This operation involves the inv log function or
the antilog so we will have an activity sheet
designed to help you to enter this correctly in
your individual calculator.
• A solution has a pH of 4.58 what is the [H⁺]?
• [H⁺] = 2.6 x 10⁻⁵ M
An HCl solution has a [H⁺] of 1.4 x 10⁻⁴ M, what
is the pH?
pH = - log [H⁺]
pH = - log [1.4 x 10⁻⁴]
pH = 3.85
Calculating pOH
• The calculation is identical to the pH
calculation except it involves the [OH⁻].
• pOH = - log [OH⁻]
Calculating pOH
What is the pOH for a solution whose
[OH⁻] = 1.89 x10⁻⁶ ?
pOH = - log [OH⁻]
pOH = - log [1.89 x 10⁻⁶]
pOH = 5.724
What is the pOH of a solution with an
[OH⁻] = 3.33 x 10⁻⁴ ?
pOH = - log [OH⁻]
pOH = - log 3. 33 x 10⁻⁴
pOH = 3.478
Calculating [OH⁺] from pH
• This operation involves the inv log function or
the antilog so we will have an activity sheet
designed to help you to enter this correctly in
your individual calculator.
• A solution has a pOH of 5.55 what is the
[OH⁺]?
• [OH⁺] = 2.8 x 10⁻⁶ M
Calculating:
pOH from pH, pOH from pH
Due to experiment/observation we know that
pH + pOH = 14.00
So we can convert between pH + pOH with a
simple subtraction problem.
pH and pOH conversions
The pH of a solution is 2.29, what is the pOH?
pOH = 14.00 – pH
pOH = 14.00 – 2.29
pOH = 11.71
The pOH of a solution is 6.65, what is the pH?
pH = 14.00 – pOH
pH = 14.00 - 6.65
pH = 7.35
Reactions of Acids with Metals
Many acids will react with the more reactive
metals to produce the salt of that acid and
hydrogen gas.
Ex.
2Al + 6HCl → 2AlCl₃ + 3H₂
Solid metal
aqueous acid
salt
hydrogen gas
Mg + H₂SO₄ → MgSO₄ + H₂
Solid metal
aqueous acid
salt
hydrogen gas
Acidity of Basicity on pH Scale
Summary:
pH = 7 Neutral
pH < 7 Acidic
pH > 7 Basic
Ion Product Constant
In aqueous solution
[OH⁻] [H⁺] = 1.00 x 10⁻¹⁴
Ex.
If an aqueous solution has a [H⁺]
= 1.83 x 10⁻⁵ what is the [OH⁻] ?
[OH⁻] = 1.00 x 10⁻¹⁴ / 1.83 x 10⁻⁵
[OH⁻] = 5.46 x 10⁻¹⁰
If an aqueous solution has an [OH⁻] of
5.43 x 10⁻⁴, what is the [H⁺] ?
[H⁺] = 1.00 x 10⁻¹⁴ / [OH⁻]
[H⁺] = 1.00 x 10⁻¹⁴ / 5.43 x 10⁻⁴
[H⁺] = 1.84 x 10⁻¹¹