LECTURE 3: ACIDS AND BASES
DEFINITIONS OF AN ACID
Theory:
Acid=
When
Arrhenius
increases H+
1880’s
Brønsted
proton donor
1923
Lowry
proton donor
1923
Lewis
electron-pair
acceptor
1923
Who
Svante August Arrhenius
(February 19, 1859 – October 2, 1927)
Swedish chemist; Nobel Prize in Chemistry, 1903
* Arrhenius equation (activation energy)
* Greenhouse effect
Johannes Nicolaus Brønsted
(February 22, 1879-December 17, 1947)
Danish physical chemist
Thomas Martin Lowry
(October 26, 1874–November 2, 1936)
English organic chemist
Gilbert Newton Lewis
(October 23, 1875-March 23, 1946)
American physical chemist
ARRHENIUS DEFINITIONS
• Arrhenius acids and bases
– Acid: Substance that, when dissolved in water,
increases the concentration of hydrogen ions (protons,
H+).
– Base: Substance that, when dissolved in water,
increases the concentration of hydroxide ions.
BRØNSTED–LOWRY DEFINITION
• Brønsted–Lowry: must have both
1. an Acid: Proton donor
and
2. a Base:
Proton acceptor
Brønsted-Lowry acids and bases are always paired.
The Brønsted-Lowry acid donates a proton,
while the Brønsted-Lowry base accepts it.
Which is the acid and which is the base in each of these rxns?
A Brønsted–Lowry acid…
…must have a removable (acidic) proton.
HCl, H2O, H2SO4
A Brønsted–Lowry base…
…must have a pair of nonbonding electrons.
NH3, H2O
If it can be either…
...it is amphiprotic.
–
HCO3
–
HSO4
H2O
What Happens When an Acid Dissolves
in Water?
• Water acts as a Brønsted–
Lowry base and abstracts a
proton (H+) from the acid.
• As a result, the conjugate
base of the acid and a
hydronium ion are formed.
CONJUGATE ACIDS AND BASES
• From the Latin word conjugare, meaning “to join together.”
• Reactions between acids and bases always yield their
conjugate bases and acids.
ACID AND BASE STRENGTH
• Strong acids are completely
dissociated in water.
– Their conjugate bases are quite
weak.
• Weak acids only dissociate
partially in water.
– Their conjugate bases are weak
bases.
ACID BASE STRENGTH
• Substances with negligible acidity
do not dissociate in water.
– Their conjugate bases are
exceedingly strong.
ACID BASE STRENGTH
In any acid-base reaction, the equilibrium favors the
reaction that moves the proton to the stronger base.
HCl(aq) + H2O(l)  H3O+(aq) + Cl–(aq)
H2O is a much stronger base than Cl–, so the equilibrium
lies so far to the right K is not measured (K>>1).
ACID BASE STRENGTH
Acetate is a stronger base than
H2O, so the equilibrium favors the
left side (K<1).
The stronger base “wins” the
proton.
HC2H3O2(aq) + H2O
H3O+(aq) + C2H3O2–(aq)
AUTOIONIZATION OF WATER
As we have seen, water is amphoteric.
• In pure water, a few molecules act as bases and a few act
as acids.
This process is called autoionization.
EQUILIBRIUM CONSTANT FOR WATER
• The equilibrium expression for this process is
Kc = [H3O+] [OH–]
• This special equilibrium constant is referred to as the
ion-product constant for water, Kw.
• At 25°C, Kw = 1.0  10-14
pH
pH is defined as the negative base-10 logarithm of the
hydronium ion concentration.
pH = –log [H3O+]
pH
• In pure water,
Kw = [H3O+] [OH–] = 1.0  10-14
• Because in pure water [H3O+] = [OH-],
[H3O+] = (1.0  10-14)1/2 = 1.0  10-7
pH
• Therefore, in pure water,
pH = –log [H3O+]
= –log (1.0  10-7) = 7.00
• An acid has a higher [H3O+] than pure water, so its pH is <7
• A base has a lower [H3O+] than pure water, so its pH is >7.
pH
These are the pH
values for several
common
substances.
Other “p” Scales
• The “p” in pH tells us to take the negative log of the
quantity (in this case, hydronium ions).
• Some similar examples are
– pOH –log [OH-]
– pKw –log Kw
Watch This!
Because
[H3O+] [OH−] = Kw = 1.0  10-14,
we know that
–log [H3O+] + – log [OH−] = – log Kw = 14.00
or, in other words,
pH + pOH = pKw = 14.00
If you know one, you know them all:
[H+]
[OH-]
pH
pOH
How Do We Measure pH?
– Litmus paper
• “Red” paper turns blue
above ~pH = 8
• “Blue” paper turns red
below ~pH = 5
– An indicator
• Compound that changes color in
solution.
How Do We Measure pH?
pH meters
measure the voltage in the
solution
Strong Acids
• You will recall that the seven
strong acids are HCl, HBr, HI,
HNO3, H2SO4, HClO3, and HClO4.
• These are strong electrolytes and
exist totally as ions in aqueous
solution.
• For the monoprotic strong acids,
[H3O+] = [acid].
Strong Bases
• Strong bases are the soluble hydroxides, which are the alkali metal
(NaOH, KOH)and heavier alkaline earth metal hydroxides (Ca(OH)2,
Sr(OH)2, and Ba(OH)2).
• Again, these substances dissociate completely in aqueous solution.
[OH-] = [hydroxide added].
Dissociation Constants
• For a generalized acid dissociation,
the equilibrium expression is
• This equilibrium constant is called the aciddissociation constant, Ka.
DISSOCIATION CONSTANTS
The greater the value of Ka, the stronger the acid.
CALCULATING Ka FROM THE pH
• The pH of a 0.10 M solution of formic acid, HCOOH, at
25°C is 2.38. Calculate Ka for formic acid at this
temperature.
• We know that
Calculating Ka from the pH
The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C
is 2.38. Calculate Ka for formic acid at this temperature.
To calculate Ka, we need all equilibrium concentrations.
We can find [H3O+], which is the same as [HCOO−], from the
pH.
Calculating Ka from the pH
pH = –log [H3O+]
– 2.38 = log [H3O+]
10-2.38 = 10log [H3O+] = [H3O+]
4.2  10-3 = [H3O+] = [HCOO–]
Calculating Ka from pH
In table form:
[HCOOH], M
[H3O+], M
[HCOO−], M
Initially
0.10
0
0
Change
–4.2  10-3
+4.2  10-3
+4.2  10-3
0.10 – 4.2  10-3
= 0.0958 = 0.10
4.2  10-3
4.2  10 - 3
At Equilibrium
Calculating Ka from pH
Ka =
[4.2  10-3] [4.2  10-3]
[0.10]
= 1.8  10-4
CALCULATING PERCENT IONIZATION
In the example:
[A-]eq = [H3O+]eq = 4.2  10-3 M
[A-]eq + [HCOOH]eq = [HCOOH]initial = 0.10 M
Calculating Percent Ionization
4.2  10-3
Percent Ionization =
 100
0.10
= 4.2%
Calculating pH from Ka
Calculate the pH of a 0.30 M solution of acetic acid,
C2H3O2H, at 25°C.
Ka for acetic acid at 25°C is 1.8  10-5.
Is acetic acid more or less ionized than formic acid (Ka=1.8
x 10-4)?
Calculating pH from Ka
The equilibrium constant expression is:
Calculating pH from Ka
Use the ICE table:
Initial
Change
Equilibrium
[C2H3O2], M
[H3O+], M
[C2H3O2−], M
0.30
0
0
–x
+x
+x
0.30 – x
x
x
Calculating pH from Ka
Use the ICE table:
Initial
Change
Equilibrium
[C2H3O2], M
[H3O+], M
[C2H3O2−], M
0.30
0
0
–x
+x
+x
0.30 – x
x
x
Simplify: how big is x relative to 0.30?
CALCULATING pH FROM Ka
Use the ICE table:
Initial
Change
Equilibrium
[C2H3O2], M
[H3O+], M
[C2H3O2−], M
0.30
0
0
–x
+x
+x
0.30 – x ≈ 0.30
x
x
Simplify: how big is x relative to 0.30?
Calculating pH from Ka
Now,
(1.8  10-5) (0.30) = x2
5.4  10-6 = x2
2.3  10-3 = x
Check: is approximation ok?
CALCULATING Ka from pH
The pH of a 0.01M hypochlorous acid (HClO) is 4.76.
Calculate its Ka.
SAMPLE PROBLEM
Calculate the pH of a 0.02M Hydroflouric acid solution.
Ka (HF) = 6.8 x 10-4
POLYPROTIC ACIDS
Have more than one acidic proton.
If the difference between the Ka for the first dissociation and
subsequent Ka values is 103 or more, the pH generally depends
only on the first dissociation.
SAMPLE PROBLEM
Calculate the pH of a 0.1 M H3PO4. Ka1 = 7.5 x 10-3
Ka2 = 6.8 x 10-8 Ka3 = 4.2 x 10-13
WEAK BASES
Bases react with water to produce hydroxide ion.
WEAK BASES
The equilibrium constant expression for this reaction
is
where Kb is the base-dissociation constant.
WEAK BASES
Kb can be used to find [OH–] and, through it, pH.
pH of Basic Solutions
What is the pH of a 0.15 M solution of NH3?
Kb =
[NH4+] [OH−]
[NH3]
= 1.8  10-5
pH OF BASIC SOLUTIONS
Tabulate the data.
Initial
Equilibrium
[NH3], M
[NH4+], M
[OH−], M
0.15
0
0
0.15 - x  0.15
x
x
Simplify: how big is x relative to 0.15?
pH OF BASIC SOLUTIONS
1.8  10-5 =
(x)2
(0.15)
(1.8  10-5) (0.15) = x2
2.7  10-6 = x2
1.6  10-3 = x2
Check: is approximation ok?
pH of Basic Solutions
Therefore,
[OH–] = 1.6  10-3 M
pOH = –log (1.6  10-3)
pOH = 2.80
pH = 14.00 – 2.80
pH = 11.20
SAMPLE PROBLEM
A 0.01M solution of caffeine, a weak organic base, has a
pH of 11.3. Calculate its dissociation constant.
Ka and dissociation constant of a conjugate base
HCN
CN-
+
+
H2O
CN-
H 2O
HCN
KaKb = Kw
+
+
H3O+
OH-
SALT HYDROLYSIS
Salts of strong acids and bases
Salts of strong base and a weak acid
Salt of strong acid and a weak base.
Ka and Kb are linked:
Combined reaction = ?
Ka and Kb are linked:
Combined reaction = ?
Ka and Kb
Ka and Kb are related in this way:
Ka  Kb = Kw
Therefore, if you know one of them, you can calculate the other.
SAMPLE PROBLEM
Calculate the pH of 0.10 M NH4Cl solution. Kb(NH3)=1.8 x
10-5
Calculate the % hydrolysis of a 0.36M CH3COONa.
Ka=1.75 x 10-5
ACID – BASE TITRATIONS
A. Strong Acid – Strong Base Titration.
NaOH + HCl  H2O + NaCl
B. Weak acid with a strong Base.
𝑝𝐻 = 𝑝𝐾𝑎 +
[𝑐𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒 𝑏𝑎𝑠𝑒]
log
[𝑎𝑐𝑖𝑑]
Polyprotic acids
H3PO4 + H2O
H2PO4- + H2O
HPO42-
+
H 2O
H2PO4- + H3O+
𝐾𝑎1 = 7.5 x 10 -3
HPO42- + H3O+
𝐾𝑎2 = 6.2 x 10-8
PO43- + H3O+
𝐾𝑎3 = 4.8 x 10-13
Polyprotic Acids
Have more than one acidic proton.
If the difference between the Ka for the first dissociation and
subsequent Ka values is 103 or more, the pH generally depends
only on the first dissociation.
SAMPLE PROBLEM
What is the pH of 0.025 M H2S solution? K1= 5.7 x 10-8 K2 = 1.2 x 10-15
SAMPLE PROBLEM
What is the pH of 0.012 M Na2CO3 solution? K1= 4.2 x 10-7 K2 = 4.8 x 10-11
SAMPLE PROBLEM
A 50 ml of 0.05M formic acid solution (Ka = 1.77 x 10-4) is titrated with 0.05 M
NaOH solution. What is the pH at equivalence point?
PRACTICE EXERCISES
1. Niacin, one of the B vitamins, has the following molecular structure:
A 0.020 M solution of niacin has a pH of 3.26. (a) What percentage of the acid is ionized in this solution?
(b) What is the acid-dissociation constant, Ka, for niacin?
2. What is the pH of (a) a 0.028 M solution of NaOH, (b) a 0.0011 M solution of Ca(OH)2? What percentage of the bases are ionized?
3. Calculate the percentage of HF molecules ionized in (a) a 0.10 M HF solution, (b) a 0.010 M HF solution. Ka for HF is 6.8 x10-4.
Reactions of Anions with Water
• Anions are bases.
• As such, they can react with water in a hydrolysis
reaction to form OH– and the conjugate acid:
X–(aq) + H2O(l)
HX(aq) + OH–(aq)
Reactions of Cations with Water
• Cations with acidic protons (like
NH4+) lower the pH of a solution by
releasing H+.
• Most metal cations (like Al3+) that are
hydrated in solution also lower the
pH of the solution; they act by
associating with H2O and making it
release H+.
Reactions of Cations with Water
• Attraction between nonbonding electrons
on oxygen and the metal causes a shift of
the electron density in water.
• This makes the O-H bond more polar and
the water more acidic.
• Greater charge and smaller size make a
cation more acidic.
Effect of Cations and Anions
1. An anion that is the conjugate base of
a strong acid will not affect the pH.
2. An anion that is the conjugate base of
a weak acid will increase the pH.
3. A cation that is the conjugate acid of
a weak base will decrease the pH.
Effect of Cations and Anions
4. Cations of the strong Arrhenius
bases will not affect the pH.
5. Other metal ions will cause a
decrease in pH.
6. When a solution contains both the
conjugate base of a weak acid and
the conjugate acid of a weak base,
the affect on pH depends on the Ka
and Kb values.
What effect on pH?
Why?
An anion that is the conjugate base of a strong
acid does not affect pH.
= very weak base
An anion that is the conjugate base of a weak
acid increases pH.
= strong base
A cation that is the conjugate acid of a weak
base decreases pH.
= strong acid
Cations of the strong Arrhenius bases (Na+,
Ca2+) do not affect pH.
= very weak acid
Other metal ions cause a decrease in pH.
Weak acid + weak base
(not really acidic at all)
= moderate bases
(cations)
Depends on Ka and Kb
Factors Affecting Acid Strength
• The more polar the H-X bond and/or the weaker the H-X bond, the
more acidic the compound.
• Acidity increases from left to right across a row and from top to
bottom down a group.
Factors Affecting Acid Strength
In oxyacids, in which an OH
is bonded to another atom,
Y,
the more electronegative Y
is, the more acidic the acid.
Factors Affecting Acid Strength
For a series of oxyacids, acidity increases with the number
of oxygens.
Factors Affecting Acid Strength
Resonance in the conjugate bases of carboxylic acids
stabilizes the base and makes the conjugate acid more acidic.
Lewis Acids
• Lewis acids are defined as electron-pair acceptors.
• Atoms with an empty valence orbital can be Lewis acids.
• A compound with no H’s can be a Lewis acid.
Lewis Bases
• Lewis bases are defined as electron-pair donors.
• Anything that is a Brønsted–Lowry base is also a Lewis base. (BL bases also have a lone pair.)
• Lewis bases can interact with things other than protons.
The Common-Ion Effect
• Consider a solution of acetic acid:
HC2H3O2(aq) + H2O(l)
H3O+(aq) + C2H3O2−(aq)
• If acetate ion is added to the solution, Le Châtelier says
the equilibrium will shift to the left.
The Common-Ion Effect
“The extent of ionization of a weak electrolyte is
decreased by adding to the solution a strong electrolyte
that has an ion in common with the weak electrolyte.”
The Common-Ion Effect
Calculate the fluoride ion concentration and pH of a solution
that is 0.20 M in HF and 0.10 M in HCl.
Ka for HF is 6.8  10−4.
Ka =
[H3O+] [F−]
[HF]
= 6.8  10-4
The Common-Ion Effect
H3O+(aq) + F−(aq)
HF(aq) + H2O(l)
Because HCl, a strong acid, is also present, the initial
[H3O+] is not 0, but rather 0.10 M.
Initially
Change
At Equilibrium
[HF], M
[H3O+], M
[F−], M
0.20
0.10
0
−x
+x
0.20 − x  0.20
0.10 + x  0.10
+x
x
The Common-Ion Effect
6.8 
10−4
(0.20) (6.8  10−4)
(0.10)
(0.10) (x)
(0.20)
=
=x
1.4  10−3 = x
The Common-Ion Effect
• Therefore, [F−] = x = 1.4  10−3
[H3O+] = 0.10 + x = 0.10 + 1.4  10−3 = 0.10 M
• So,
pH = −log (0.10)
pH = 1.00
Buffers:
• Solutions of a weak
conjugate acid-base pair.
• They are particularly
resistant to pH changes,
even when strong acid or
base is added.
Buffers
If a small amount of hydroxide is added to an equimolar solution of
HF in NaF, for example, the HF reacts with the OH− to make F− and
water.
Buffers
If acid is added, the F− reacts to form HF and water.
Buffer Calculations
Consider the equilibrium constant expression for the
dissociation of a generic acid, HA:
H3O+ + A−
HA + H2O
Ka =
[H3O+] [A−]
[HA]
Buffer Calculations
Rearranging slightly, this becomes
Ka =
[H3O+]
[A−]
[HA]
Taking the negative log of both side, we get
−log Ka =
pKa
pH
−log [H3O+] + −log
base
[A−]
[HA]
acid
Buffer Calculations
• So
pKa = pH − log
[base]
[acid]
• Rearranging, this becomes
pH = pKa + log
[base]
[acid]
• This is the Henderson–Hasselbalch equation.
Henderson–Hasselbalch Equation
What is the pH of a buffer that is 0.12 M in lactic
acid, HC3H5O3, and 0.10 M in sodium lactate? Ka
for lactic acid is
1.4  10−4.
Henderson–Hasselbalch Equation
[base]
[acid]
pH = pKa + log
pH = −log (1.4 
10−4)
pH = 3.85 + (−0.08)
pH = 3.77
+ log
(0.10)
(0.12)
pH Range
• The pH range is the range of pH values over which a
buffer system works effectively.
• It is best to choose an acid with a pKa close to the
desired pH.
When Strong Acids or Bases Are Added
to a Buffer…
…it is safe to assume that all of the strong acid or base is
consumed in the reaction.
Addition of Strong Acid or Base to a
Buffer
1. Determine how the neutralization reaction
affects the amounts of the weak acid and its
conjugate base in solution.
2. Use the Henderson–Hasselbalch equation
to determine the new pH of the solution.
Calculating pH Changes in Buffers
A buffer is made by adding 0.300 mol HC2H3O2 and
0.300 mol NaC2H3O2 to enough water to make 1.00 L
of solution. a) Calculate the pH of this solution after
0.020 mol of NaOH is added. Ka = 1.8 x 10-5
b) calculate the pH after 0.020 mole HCl is added.
Calculating pH Changes in Buffers
Before the reaction, since
mol HC2H3O2 = mol C2H3O2−
pH = pKa = −log (1.8  10−5) = 4.74
Calculating pH Changes in Buffers
The 0.020 mol NaOH will react with 0.020 mol of the acetic acid:
HC2H3O2(aq) + OH−(aq)  C2H3O2−(aq) + H2O(l)
HC2H3O2
C2H3O2−
OH−
Before reaction
0.300 mol
0.300 mol
0.020 mol
After reaction
0.280 mol
0.320 mol
0.000 mol
Calculating pH Changes in Buffers
Now use the Henderson–Hasselbalch equation to calculate the
new pH:
pH = 4.74 + log
pH = 4.74 + 0.06
pH = 4.80
(0.320)
(0. 200)
Titration
A known concentration of
base (or acid) is slowly added
to a solution of acid (or base).
Titration
A pH meter or indicators are
used to determine when the
solution has reached the
equivalence point, at which
the stoichiometric amount of
acid equals that of base.
Titration of a Strong Acid with a Strong
Base
From the start of the titration
to near the equivalence
point, the pH goes up slowly.
Titration of a Strong Acid with a Strong
Base
Just before and after the
equivalence point, the pH
increases rapidly.
Titration of a Strong Acid with a Strong
Base
At the equivalence point,
moles acid = moles base, and
the solution contains only
water and the salt from the
cation of the base and the
anion of the acid.
Titration of a Strong Acid with a Strong
Base
As more base is added, the
increase in pH again levels
off.
Titration of a Weak Acid with a Strong
Base
• Unlike in the previous case, the
conjugate base of the acid affects
the pH when it is formed.
• The pH at the equivalence point
will be >7.
• Phenolphthalein is commonly used
as an indicator in these titrations.
Titration of a Weak Acid with a Strong
Base
At each point below the equivalence point, the pH of the
solution during titration is determined from the amounts of
the acid and its conjugate base present at that particular
time.
Titration of a Weak Acid with a Strong
Base
With weaker acids, the initial
pH is higher and pH changes
near the equivalence point
are more subtle.
Titration of a Weak Base with a Strong
Acid
• The pH at the equivalence
point in these titrations is <
7.
• Methyl red is the indicator of
choice.
Titrations of Polyprotic Acids
In these cases there
is an equivalence
point for each
dissociation.
Solubility Equilibria
Solubility Rules
• Salts are generally more soluble in HOT water
(Gases are more
soluble in COLD water)
• Alkali Metal salts are very soluble in water.
NaCl, KOH, Li3PO4, Na2SO4 etc...
• Ammonium salts are very soluble in water.
NH4Br, (NH4)2CO3 etc…
• Salts containing the nitrate ion, NO3-, are very soluble in water.
• Most salts of Cl-, Br- and I- are very soluble in water - exceptions are
salts containing Ag+ and Pb2+.
soluble salts: FeCl2, AlBr3, MgI2 etc...
“insoluble” salts: AgCl, PbBr2 etc...
Dissolving a salt...
• A salt is an ionic compound - usually a
metal cation bonded to a non-metal anion.
• The dissolving of a salt is an example of
equilibrium.
• The cations and anions are attracted to
each other in the salt.
• They are also attracted to the water
molecules.
• The water molecules will start to pull out
some of the ions from the salt crystal.
• At first, the only process occurring is the
dissolving of the salt - the dissociation of
the salt into its ions.
• However, soon the ions floating in the
water begin to collide with the salt crystal
and are “pulled back in” to the salt.
(precipitation)
• Eventually the rate of dissociation is equal
to the rate of precipitation.
• The solution is now “saturated”. It has
reached equilibrium.
Solubility Equilibrium:
Dissociation = Precipitation
Na+ and Cl - ions
surrounded by
water molecules
In a saturated solution, there is
no change in amount of solid
precipitate at the bottom of the
beaker.
Concentration of the solution
is constant.
NaCl Crystal
Dissolving NaCl in water
The rate at which the salt is
dissolving into solution equals
the rate of precipitation.
Dissolving silver sulfate, Ag2SO4, in water
• When silver sulfate dissolves it dissociates into ions. When the
solution is saturated, the following equilibrium exists:
Ag2SO4 (s)  2 Ag+ (aq) + SO42- (aq)
• Since this is an equilibrium, we can write an equilibrium expression
for the reaction:
Ksp = [Ag+]2[SO42-]
Notice that the Ag2SO4 is left out of the expression! Why?
Since K is always calculated by just multiplying concentrations, it is called a “solubility
product” constant - Ksp.
Writing solubility product expressions...
• For each salt below, write a balanced equation showing its
dissociation in water.
• Then write the Ksp expression for the salt.
Iron (III) hydroxide, Fe(OH)3
Nickel sulfide, NiS
Silver chromate, Ag2CrO4
Zinc carbonate, ZnCO3
Calcium fluoride, CaF2
Some Ksp Values
Note:
These are experimentally determined, and may
be slightly different on a different Ksp table.
Calculating Ksp of Silver Chromate
• A saturated solution of silver chromate, Ag2CrO4, has [Ag+] = 1.3 x
10-4 M. What is the Ksp for Ag2CrO4?
Ag2CrO4 (s)  2 Ag+ (aq) + CrO42- (aq)
----
1.3 x 10-4 M
----
Calculating the Ksp of silver sulfate
• The solubility of silver sulfate is 0.014 mol/L. This means that 0.0144
mol of Ag2SO4 will dissolve to make 1.0 L of saturated solution.
Calculate the value of the equilibrium constant, Ksp for this salt.
Ag2SO4 (s)  2 Ag+ (aq) + SO42- (aq)
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---
Calculating solubility, given Ksp
• The Ksp of NiCO3 is 1.4 x 10-7 at 25°C. Calculate its molar solubility.
NiCO3 (s)  Ni2+ (aq) + CO32- (aq)
---
---
Other ways to express solubility...
• We just saw that the solubility of nickel (II) carbonate is 3.7 x 10-4
mol/L. What mass of NiCO3 is needed to prepare 500 mL of
saturated solution?
3.7 x 10 4 mol NiCO3 0.500 L
118.72 g
x
x
 0.022 g
1L
1 mol NiCO3
0.022 g of NiCO3 will dissolve to make 500 mL solution.
Calculate the solubility of MgF2 in water. What mass will
dissolve in 2.0 L of water? Ksp = 7.4 x 10-11
MgF2 (s)  Mg2+ (aq) + 2 F- (aq)
Solubility and pH
• Calculate the pH of a saturated solution of silver hydroxide, AgOH.
Ksp = 2.0 x 10-8.
AgOH (s)  Ag+ (aq) + OH- (aq)
The Common Ion Effect on Solubility
The solubility of MgF2 in pure water is 2.6 x 10-4 mol/L. What happens to
the solubility if we dissolve the MgF2 in a solution of NaF, instead of pure
water?
Calculate the solubility of MgF2 in a solution of 0.080 M
NaF.
MgF2 (s)  Mg2+ (aq) + 2 F- (aq)
Explaining the Common Ion Effect
The presence of a common ion in a solution will lower the
solubility of a salt.
• LeChatelier’s Principle:
The addition of the common ion will shift the solubility
equilibrium backwards. This means that there is more solid
salt in the solution and therefore the solubility is lower!
Ksp and Solubility
• Generally, it is fair to say that salts with very small solubility product
constants (Ksp) are only sparingly soluble in water.
• When comparing the solubilities of two salts, however, you can
sometimes simply compare the relative sizes of their Ksp values.
• This works if the salts have the same number of ions!
• For example… CuI has Ksp = 5.0 x 10-12 and CaSO4 has Ksp = 6.1 x
10-5. Since the Ksp for calcium sulfate is larger than that for the
copper (I) iodide, we can say that calcium sulfate is more soluble.
But be careful...
Salt
Solubility
(mol/L)
Ksp
-45
9.2 x 10
-23
CuS
8.5 x 10
Ag2S
1.6 x 10-49
3.4 x 10-17
Bi2S3
1.1 x 10-73
1.0 x 10-15
Mixing Solutions - Will a Precipitate Form?
If 15 mL of 0.024-M lead nitrate is mixed with 30 mL
of 0.030-M potassium chromate - will a precipitate form?
Pb(NO3)2 (aq) + K2CrO4 (aq)  PbCrO4 (s) + 2 KNO3 (aq)
Pb(NO3)2 (aq) + K2CrO4 (aq)  PbCrO4 (s) + 2 KNO3 (aq)
Step 1: Is a sparingly soluble salt formed?
We can see that a double replacement reaction can occur and
produce PbCrO4. Since this salt has a very small Ksp, it may
precipitate from the mixture. The solubility equilibrium is:
PbCrO4 (s)  Pb2+ (aq) + CrO42- (aq)
Ksp = 2 x 10-16 = [Pb2+][CrO42-]
If a precipitate forms, it means the solubility equilibrium has shifted
BACKWARDS.
This will happen only if Qsp > Ksp in our mixture.
Step 2: Find the concentrations of the ions that form the sparingly
soluble salt.
Since we are mixing two solutions in this example, the concentrations
of the Pb2+ and CrO42- will be diluted. We have to do a dilution
calculation!
Dilution: C1V1 = C2V2
[Pb2+] =
[CrO42-] =
C1V1 (0.024 M)(15 mL)

 0.0080 M Pb 2
V2
(45 mL)
C1V1 (0.030 M)(20 mL)

 0.020 M CrO 4 2V2
(45 mL)
Step 3: Calculate Qsp for the mixture.
Qsp = [Pb2+][CrO42-] = (0.0080 M)(0.020 M)
Qsp = 1.6 x 10-4
Step 4: Compare Qsp to Ksp.
Since Qsp >> Ksp, a precipitate will form when
solutions are mixed!
Note: If Qsp = Ksp, the mixture is saturated
If Qsp < Ksp, the solution is unsaturated
Either way, no ppte will form!
the two
A common laboratory method for preparing a precipitate
is to mix solutions of the component ions. Does a
precipitate form when 0.100 L of 0.30 M Ca(NO3)2 is
mixed with 0.200 L of 0.060 M NaF?
Solubility Products
Consider the equilibrium that exists in a saturated
solution of BaSO4 in water:
BaSO4(s)
Ba2+(aq) + SO42−(aq)
Solubility Products
The equilibrium constant expression for this equilibrium is
Ksp = [Ba2+] [SO42−]
where the equilibrium constant, Ksp, is called the solubility
product.
Solubility Products
• Ksp is not the same as solubility.
• Solubility is generally expressed as the mass of solute
dissolved in 1 L (g/L) or 100 mL (g/mL) of solution, or in mol/L
(M).
Factors Affecting Solubility
• The Common-Ion Effect
– If one of the ions in a solution equilibrium is already
dissolved in the solution, the equilibrium will shift to the
left and the solubility of the salt will decrease.
BaSO4(s)
Ba2+(aq) + SO42−(aq)
Factors Affecting Solubility
• pH
– If a substance has a basic
anion, it will be more soluble in
an acidic solution.
– Substances with acidic cations
are more soluble in basic
solutions.
Factors Affecting Solubility
• Complex Ions
– Metal ions can act as Lewis acids and form complex ions
with Lewis bases in the solvent.
Factors Affecting Solubility
• Complex Ions
– The formation of
these complex ions
increases the
solubility of these
salts.
Factors Affecting Solubility
• Amphoterism
– Amphoteric metal oxides and
hydroxides are soluble in strong
acid or base, because they can
act either as acids or bases.
– Examples of such cations are
Al3+, Zn2+, and Sn2+.
Will a Precipitate Form?
• In a solution,
– If Q = Ksp, the system is at equilibrium and the
solution is saturated.
– If Q < Ksp, more solid will dissolve until Q = Ksp.
– If Q > Ksp, the salt will precipitate until Q = Ksp.
Selective Precipitation of Ions
One can use
differences in
solubilities of salts to
separate ions in a
mixture.