Equilibrium Involving
ACIDS & BASES
Shane Kyle Labe 12-STEM
TABLE OF CONTENTS
1
Brønsted Acids & Bases
Strong Acids & Bases
2
Measuring Acidity with pH
3
4
Common Ion &
Buffer Solutions
Factors that Affect
Solubility
Weak Acids & Bases
5
6
7
Solubility Equilibrium
01
Brønsted
Acids & Bases
A theory proposed by Johannes Nicolaus
Brønsted(Danish Physical Chemist).
- 1923
Brønsted defined acid as a
proton(H+) donor, and a base as
a proton acceptor.
●
●
Any hydrogen-containing
compound is capable of
releasing H+ is an acid.
A base must have an atom
with a lone pair so it can
accept a proton.
Acid-Base Reaction
Example:
Involves the transfer of a
proton from an acid to a base
CONJUGATE BASE
An acid releases a proton.
CONJUGATE ACID
A base accepts a proton.
Water as both Acid & Base
01
Example
If [H₃O⁺] is 6.2 x 10⁻⁵ M, then [OH⁻]
is calculated as:
02
[OH⁻] = Kw / [H₃O⁺]
= 1.0 x 10¹⁴ / 6.2 x 10⁻⁵
= 1.6 x 10⁻¹⁰
Water can act as both as Brønsted
acid and base.
-
Kw = [H₃O⁺][OH⁻] = 1.0 x 10¹⁴
Sol’n:
SELF-IONIZATION or
AUTOIONIZATION of Water
-
[H₃O⁺] = [OH⁻]
neutral
[H₃O⁺] > [OH⁻]
acidic
[H₃O⁺] < [OH⁻]
basic
03
Between two water molecules,
one can give off a proton and
the other can accept this
proton.
Water is amphiprotic in nature.
Ions exist in equilibrium with the
water molecules.
Ionization Constant of Water (Kw)
1.0 x 10¹⁴ M at 25℃
Measuring Acidity
with pH
A concept introduced by
Søren Peter Lauritz
Sørensen(Danish Chemist)
-
1909
02
The Concept of
pH
POWER OF HYDROGEN (pH)
Indicates the acidity or
basicity of a solution.
hydrogen
power of H⁺
concentration
Mathematical Definition
The logarithm of the reciprocal
of the H⁺ molar concentration.
neutral
0
pH scale
acidic
Way of expressing the
concentration of H⁺ in
dilute solutions.
7
Neutral Solutions
[H⁺] = 1.0 x 10⁻⁷
Acidic Solutions(pH<7)
[H⁺] > 1.0 x 10⁻⁷
Alkaline Solutions(pH>7)
[H⁺] < 1.0 x 10⁻⁷
basic
14
[H⁺] = [H₃O⁺]
pH ⟶ [H⁺]
pOH ⟶ [OH⁻]
01
02
pH = –log[H⁺]
pOH = –log[OH⁻]
➔ The relationship between ph & pOH
03
can be derived from the equilibrium
expression of the autoionization of
water.
Kw = [H₃O⁺][OH⁻] = 1.0 x 10⁻¹⁴
log Kw = log[H₃O⁺][OH⁻] = log(1.0 x 10⁻¹⁴)
= log[H₃O⁺] + log[OH⁻] = -14
-log Kw = (-log[H₃O⁺]) + (-log[OH⁻]) = -(-14)
pKw = pH + pOH = 14
Formula
➔
The sum of pH & pOH for a
solution will always be equal to 14
because the product of [H₃O⁺] &
[OH⁻] is 1.0 x 10⁻¹⁴.
01
Examples
Calculate the pH of a solution with
[H⁺] = 0.010 M HCl.
Sol’n:
pH = -log (0.010) = 2.00
02
What is the concentration of H⁺
in a solution with a pH of 4.73?
Sol’n:
[H⁺] = antilog(-4.73) = 1.9 x 10⁻⁵
03
3. Calculate the pH of a 0.00015 M
NaOH solution. NaOH is a strong base
and completely dissociates in water.
Sol’n:
pOH = -log (0.00015) = 3.82
pH = 14.00-3.82 = 10.18
03
Strong
Acids & Bases
The strength of an acid or base
depends on the extent of its
dissociation or ionization in
water.
Ionization constants of
weak acids & bases are
smaller.
Strong acids & bases
ionize almost
completely.
Ionization constants of
strong acids & bases
are relatively larger.
Weak acids & bases
ionize only partially.
Note!
The stronger the acid, the
weaker its conjugate base;
the stronger the base, the
weaker its conjugate acid.
Acids as
BINARY or TERNARY
Ternary Acids
Binary Acids
“Oxoacids” or Oxyacids”
Acidic hydrogen is bonded
to an oxygen atom.
Hydrogen compounds of
the halogens(Group 7A
elements.
Acid Strength:
HF < HCl < HBr < HI
Their strength depends on
the bonding energy
between the hydrogen
atoms & halide.
Further classified as:
monoprotic, diprotic, or
polyprotic - depending on
the number of protons an
acid molecule can donate.
Factors that influence the strength of ternary acids ⇨
HSO₃F₄ > H₂SO₄ > H₂S₂O₃
⇦
Oxyacids with the same central
atom & oxidation states.
- The electronegativity of an atom
attached to the central atom
affects the strength of the acid.
Electronegativity
of an atom
attached to the
central atom
H₂SO₄ > H₂SO₃
⇧
Oxyacids with the
same central atom.
Higher oxidation state
(central atom)
⇩
STRONGER
⇨
Higher electronegativity
(central atom)
Oxidation State
of the Central
Atom
STRONGER
HClO₄ > HBrO₄ > HIO₄
⇧
Electronegativity
of the central
atom
Factors that Influence
the relative
⇩
Strength of Ternary Acids
Oxyacids with different
central atom but similar
oxidation state.
Higher electronegativity
(central atom)
⇩
STRONGER
pH of Strong BASES & ACIDS
Strong acids completely
dissociates in water.
➔ Produces hydronium ion(H₃O⁺)
and the conjugate base of the
acid(A⁻).
➔ Concentrations of H₃O⁺ and A⁻
are both equal to the initial
concentration of the acid.
Formula in calculating the
“pH of strong acids & bases”
⇨
pH + pOH = 14
pH
Calculate the pH of the following acid or base:
3.6 x 10⁻⁷ M HNO₃
pH = -log[HNO₃] = -log(3.6 x 10⁻⁷) = 6.44
⇦
pOH
pH = -log[H₃O⁺]
(the negative sign indicates an inverse
relationship between pH & [H₃O⁺])
1.
Described as
Strong Electrolytes.
pH = -log[OH⁻]
1.
2.
2.0 x 10⁻⁴ M NaOH
pOH = -log(NaOH) = -log(2.0 x 10⁻⁴) = 3.70
pH + pOH = 14
pH = 14 - pOH = 14 - 3.70 = 10.3
Weak
Acids & Bases
“Weak electrolytes”
➔ Ionize only partially in water until
equilibrium is attained.
04
Weak Acids
Kₐ values are always less than 1
➔
➔
backwards process is favored.
There is more undissociated
acid in solution than its ion
products.
Kₐ ⇨
Acid
Ionization Constant
⇩ Kₐ = weaker the acid
⇧ Kₐ = closer to being a strong acid
pKₐ = -log(Kₐ)
⇩
a simpler ways of expressing
the acid ionization constant of a
weak acid.
➔ Donates more than one proton
Diprotic acid like H₂CO₃ can ionize twice
1st ionization: H₂CO₃ + H₂O ⇌ H₃O⁺ + HCO₃ ⁻ Kₐ₁ = 4.4x10⁻⁷
2nd ionization: HCO₃⁻ + H₂O ⇌ H₃O⁺ + CO₃ ²⁻ Kₐ₂ = 4.7x10⁻¹¹
Monoprotic Acids
➔ Undergoes only one dissociation
➔ Donates only one proton
(“deprotonation or ionization process”)
Polyprotic Acids
Triprotic Acids
➔ Have three Kₐ values
1st ionization > 2nd ionization > 3rd third ionization
Weak bases
➔ Ionizes only partially in water.
Example:
Determine the relative strength of
each pair of acids based on their Kₐ.
Identify the stronger conjugate base
that will form from the acids.
➔ Kb is less than 1.
⇩ Kb = weaker base
Answers:
Strength of Acid
Strength of
Conjugate Base
CH₃COOH > HBrO
BrO⁻ > CH₃COO⁻
HIO₃ > HOCN
OCN⁻ > IO₃⁻
Kb
Base Ionization
Constant
Relationship between Kₐ & Kb
For a conjugate acid-base pair:
Kₐ x Kb = K
(constant)
W
● A weak acid HA with the dissociation
reaction and Kₐ expression below:
HA(aq) + H₂O(l) ⇌ H₃O⁺(aq) + A⁻(aq)
Kₐ = [H₃O⁺][A⁻]
[HA]
● With A⁻ as the conjugate base in water:
A⁻(aq) + H₂O(l) ⇌ HA(aq) + OH⁻(aq)
Kb = [HA][OH⁻]
[A⁻]
➔ The larger the Kₐ the smaller the Kb.
➔ The stronger the acid, the weaker
Relationship between pKₐ and pKb:
the conjugate base.
●
pKₐ + pKb = pK
Adding the two expressions yields the reaction for the
autoionization of water.
HA(aq) + H₂O(l) ⇌ H₃O⁺(aq) + A⁻(aq)
A⁻(aq) + H₂O(l) ⇌ HA(aq) + OH⁻(aq)
2H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq)
● Multiply Kₐ and Kb with the equation Kw:
Kₐ x Kb =[H₃O⁺][A⁻] x [HA⁺][OH⁻]
[HA]
[A⁻]
= [H₃O⁺][OH⁻] = KW
● The relationship between pKₐ + pKb is then obtained.
W
pH of Weak Acids & Bases
- The partial ionization is considered when
calculating the pH.
- The concentration of H⁺( or H₃O⁺) in solution
depends on the Kₐ value of the weak acid.
- Unknown equilibrium concentrations can be
represented with x.
- The remaining amount of the unionized
substance must be equal to: molarity of the
substance - x.
A 0.010 M CH₃COOH partially ionizes in water until it reaches equilibrium.
CH₃COOH(aq) + H₂O(l) ⇌ H₃O⁺(aq) + CH₃COO⁻(aq)
Equilibrium constant expression:
[H₃O⁺][CH₃COO⁻]
Kₐ =
= 1.8 x 10⁻⁵
[CH₃COOH]
=
(x)(x)
= 1.8 x 10⁻⁵
(0.010-x)
Because CH₃COOH is a weak acid with a very low ionization, x
is assumed to be much smaller than 0.010 M.
Thus, 0.010-x ≈ 0.010.
The equation is simplified to: Kₐ =
X²
0.010 = 1.8 x 10⁻⁵
Solving for x,
Kₐ = 1.8 x 10⁻⁵
CH₃COOH
H₃O⁺
CH₃COO⁻
Initial
0.010
0
0
Change
-x
+x
+x
Equilibrium
0.010-x
x
x
Percent ionization is calculated as:
% ionization =
X = √(0.010)(1.8 x 10⁻⁵) = 4.2 x 10⁻⁴ = [H₃O⁺]
Solving for pH,
pH = -log [H₃O⁺] = -log(4.2 x 10⁻⁴) = 3.37
=
[H₃O⁺]eq
[CH₃COOH]₀
x 100
(4.2 x 10⁻⁴)
x 100 = 4.2%
(0.010)
➔ Weak acids such as acetic acid give a percent ionization of less
than 100%, unlike strong acids that completely ionize(i.e., 100%).
05
Common Ion & Buffer Solutions
Common ions can influence the pH of a weak acid or base;
Buffer solutions can resist drastic changes in pH.
Effect of a Common Ion
-
Equilibrium constant expression:
[H₃O⁺][CH₃COO⁻]
Kₐ =
= 1.8 x 10⁻⁵
[CH₃COOH]
The presence of a common ion
increases the pH(decreases the
acidity) of a weak acid in solution.
=
Example:
As calculated earlier, a pH of 3.37 corresponds to 0.010 M
solution of CH₃COOH. What will be the resulting pH if
the solution initially contains 0.010 M CH₃COOH and
0.0050 M NaCH₃COO?
CH₃COOH
H₃O⁺
CH₃COO⁻
Initial
0.010
0
0.0050
Change
-x
+x
+x
Equilibrium
0.010-x
x
0.0050+x
(x)(0.0050+x)
= 1.8 x 10⁻⁵
(0.010-x)
x is negligible to be compared to 0.010 M & 0.0050 M
such that,
0.010-x ≈ 0.010.
and
0.0050+x ≈ 0.0050.
The equation is simplified to:
(x)(0.0050)
Kₐ = (0.010) = 1.8 x 10⁻⁵
Solving for x,
X = 3.6 x 10⁻⁵ = [H₃O⁺]
Solving for pH,
pH = -log [H₃O⁺] = -log(3.6 x 10⁻⁵) = 4.44
Buffer Solutions
- Have the ability to resist drastic changes in pH
when added with an acid or base.
- A mix of weak acid(or a weak base) and a salt
of its conjugate base(or conjugate acid) that
serves as a common ion.
- A buffer solution works in two ways:
HA(aq) + H₂O(l) ⇌ H₃O⁺(aq) + A⁻(aq)
● If an acid is added to the buffer solution, A⁻
will react and be consumed.
- Causes a slight decrease in pH.
● If an base is added to the buffer solution, HA
will react with it.
- Causes a slight increase in pH.
●
Henderson-Hasselbalch equation
- Suggests that the pH of a buffer solution is close to the pKₐ
of the weak acid.
- Addition of an acid or base will only cause a small change in
the pH value.
- If acid and base concentrations are equal, then pH is equal
to pKₐ; the condition when the buffer exerts its maximum
capacity.
Solving for the pH of a buffer solution.
pH = pKₐ - log
[HA][OH⁻]
[A⁻]
Example:
What is the pH of a buffer solution containing 0.025 M HNO₂
(Kₐ = 7.2 x 10⁻⁴) and 0.020 M NaNO₂?
Sol’n:
pH = pKₐ - log [HNO₂] = -log(7.2 x 10⁻⁴) - log (0.025) = 3.05
[NO₂⁻]
(0.020)
Solubility
Equilibria
Exists between an insoluble
solid(“precipitate”) and its dissolved ions.
Equilibrium constant expression:
Ksp = [A⁺]a [B⁻]b
Ksp ⇨ Solubility Product Constant
(solid substance dissolving in an aqueous solution)
06
Solubility & Molar Solubility
maximum amount(of a compound) that can
dissolve a given volume of solvent at a
specific temperature.
solubility = s x MM
g/L
↑
↑
↑
mol/L
(Can be substituted with Ksp)
g/mol
Problem:
1. The Ksp of AgCl in water is 1.8 x 10⁻¹⁰. What is its molar
solubility? What is its solubility in g/L? The molar mass
of AgCl is 143.4 g/mol.
Sol’n:
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)
Ksp = [Ag⁺][Cl⁻] = s²
Solving for s,
Ksp = 1.8 x 10⁻¹⁰ = s²
s² = 1.8 x 10⁻¹⁰
s = √1.8 x 10⁻¹⁰
= 1.3 x 10⁻⁵ mol/L
Solving for solubility,
solubility = s x MM
= (1.3 x 10⁻⁵ mol/L)(143.4 g/mol)
= 1.9 x 10⁻³ g/L
acidic
neutral
basic
Predicting Precipitation
- Calculating ion product Qsp
and comparing it with Ksp.
Relative Solubility
The lower the Ksp value, the
less soluble the compound.
Note in using Ksp values
to compare solubilities:
● Ksp values can be directly used
to compare the relative molar
solubility of compounds that
give equivalent number of ions;
those that give similar ion ratio.
Ksp = 1.8 x 10⁻¹⁰ Ksp = 1.6 x 10⁸
AgCl < PbSO₄
lower Ksp
less soluble
higher Ksp
more soluble
● Ksp values cannot be directly
used to compare the solubility
of compounds that do not have
similar ion ratios when dissolved
in water. Their molar solubility
should be the basis for the
comparison.
s = 1.1 x 10⁻² s = 1.3 x 10⁻⁵
Ca(OH)₂ > AgCl
more soluble
less soluble
Qsp = Ksp
⇨ solution is saturated
(solution contains the max amount of
solute that the given solvent can dissolve)
Qsp < Ksp
⇨ solution is unsaturated
(amount of ions in the solution is less)
*equilibrium shifts to the right,
favoring the dissolution of solid*
Qsp > Ksp
⇨ solution is unsaturated
(more ions are present)
*equilibrium shifts to the left,
favoring the formation of precipitate*
07
Factors that Affect
Solubility
The solubility of ion compounds(salt)
can be affected by the presence of a
common ion and the pH of a solution.
Effect of pH
Common Ion Effect
⇩
⇩
For solubility equilibria, the presence of a common
ion decreases the solubility of ionic compounds in
water(equilibrium shifts to the left).
Ionic compounds that contain basic anions(conjugate bases
of weak acids) tend to be more soluble at low pH conditions.
● The solubility of ionic compounds that contain anions that
conjugate bases of strong acids are not affected by pH.
They are not protonated because of the high tendency to
be in their dissociated form.
1. What will be the solubility of lead(II) iodid(PbI₂) in
0.018 M sodium iodide(NaI) solution? The Ksp of
PbI₂ is 7.1 x 10⁻⁹ an its molar mass is 461.0 g/mol.
PbI₂ ⇋ Pb²⁺(aq) + 2 I⁻(aq)
Ksp = [Pb²⁺][I⁻]² = (s)(2s)² = 7.1 x 10⁻⁹
PbI₂
Pb²⁺
I⁻
Initial
[PbI₂]
0.018
0
Change
-s
+s
+s
[PbI₂]-s
0.018+s
s
Equilibrium
Ksp = (0.018+s)(2s)² = 7.1 x 10⁻⁹
Since Ksp is small, s is negligible; 0.018+s ≈ 0.018.
Ksp = (0.018+s)(4s²) = 7.1 x 10⁻⁹
Solving for s,
s² =
Ksp
4(0.018) =
7.1 x 10⁻⁹
4(0.018) = 9.9 x 10⁻⁸
s = √9.9 x 10⁻⁸ = 3.15 x 10⁻⁴ mol/L
solubility = s x MM
= (3.15 x 10⁻⁴ mol/L)(461.0 g/mol)
= 0.15 g/L ; The assumption “s is negligible” is valid.
Thank You!
(see you again next time)
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