ACIDS AND BASES
IONIC EQUILIBRIA
ARRHENIUS DEFINITIONS
• Arrhenius acids and bases
– Acid: Substance that contains H and when dissolved in
water, increases the concentration of hydrogen ions
(protons, H+).
– Base: Substance that contains OH, 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
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.
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
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
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].
AMPHOTHERIC HYDROXIDES
Metal Ion
Insoluble Hydroxide
Complex Ion Formed
Be+2
Be(OH)2
[Be(OH)4]2-
Al+3
Al(OH)3
[Al(OH)4]-
Cr+3
Cr(OH)3
[Cr(OH)4]-
Zn+2
Zn(OH)2
[Zn(OH)4]2-
Sn+4
Sn(OH)4
[Sn(OH)6]2-
Pb+2
Pb(OH)2
[Pb(OH)4]2-
As+3
As(OH)3
[As(OH)4]-
Co+2
Co(OH)2
[Co(OH)4]2-
Cu+2
Cu(OH)2
[Cu(OH)4]2-
ACID BASE 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?
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.
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
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.
SALTS OF WEAK ACIDS AND BASES
If parent Ka = Kb
--- Neutral aqueous solutions
If parent Ka > Kb
--- Acidic aqueous solutions
If parent Ka < Kb
--- Basic aqueous solutions
SALTS OF 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.
Solution of 0.1M NaHS acidic basic or nuetral?
H2S Ka1 = 9x 10-8 Ka2 = 1.0 x 10-17
How about NaHCO3? K2HPO4?
H2CO3 Ka1 = 4.5 x 10-7
H3PO4
Ka1 = 7.2 x 10-3
Ka2 = 4.7 x 10-11
Ka2 = 6.3 x 10-8
Ka3 = 4.2 x 10-13
SAMPLE PROBLEM
What is the pH of 0.012 M Na2CO3 solution? K1= 4.2 x 10-7 K2 = 4.8 x 10-11
Solution of 0.1M NaHS acidic basic or nuetral?
H2S Ka1 = 9x 10-8 Ka2 = 1.0 x 10-17
How about NaHCO3? K2HPO4?
H2CO3 Ka1 = 4.5 x 10-7
H3PO4
Ka1 = 7.2 x 10-3
Ka2 = 4.7 x 10-11
Ka2 = 6.3 x 10-8
Ka3 = 4.2 x 10-13
SALTS THAT CONTAIN SMALL HIGHLY CHARGED
CATIONS
Salts of highly charged cations are hydrated in aqueous solution.
Solutions of such containing such cations are acidic because these
cations hydrolyze to produce excess hydronium ions.
[Al(H2O)6]+3 + H2O
Ka = 1.2 x 10-5
[Al(H2O)5(OH)]+2 + H3O+
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 ACIDS AND 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.
Lewis Acids & Bases
The Lewis definition further expands the definitions.
A base is an electron-pair donor, and an acid is an
electron-pair acceptor. The two combine to form an
adduct.
A + :B  A-B
Lewis Acids & Bases
This definition includes the “standard” BrønstedLowry acid-base reactions:
H+(aq) + :NH3(aq)  NH4+(aq)
It also includes the reactions of metal ions or atoms
with ligands to form coordination compounds:
Ag+(aq) + 2 :NH3(aq)  Ag(NH3)2+(aq)
Lewis Acids & Bases
In addition, electron-deficient compounds such as
trivalent boron is categorized as a Lewis acid.
B(CH3)3 + :NH3  (CH3)3B←NH3
The HOMO on the Lewis base interacts with the
electron pair in the LUMO of the Lewis acid. The MOs
of the adduct are lower in energy.
ACIDS AND BASES IN NON-AQUEOUS
SOLVENT
HCl is a ‘strong acid’ when water is our solvent; but is
only partially dissociated in acetic acid.
HCl in acetic acid behaves as a weak acid.
LEVELLING AND DIFFERENTIATING EFFECTS
A weakly basic solvent has less tendency than a strongly
basic one to accept a proton. Similarly a weak acid has
less tendency to donate protons than a strong acid. As a
result a strong acid such as perchloric acid exhibits more
strongly acidic properties than a weak acid such as acetic
acid when dissolved in a weakly basic solvent. On the
other hand, all acids tend to become indistinguishable in
strength when dissolved in strongly basic solvents owing
to the greater affinity of strong bases for protons. This is
called the leveling effect. Strong bases are leveling
solvents for acids, weak bases are differentiating solvents
ACIDS IN ACIDIC SOLVENTS
In H2SO4, HClO4 is practically unionized while
HNO3 is fully ionized.
SOLVENT ORIENTED APPROACH TO ACIDBASE DEFINITION
In a self-ionizing solvent, an acid is a substance that
produces the cation characteristic of the solvent, and a
base is a substance that produces the anion characteristic
of the solvent.
SELF-IONIZING NON-AQUEOUS SOLVENTS
2 Categories
Protic: NH3, HF, H2SO4, HOSO2F
Aprotic: BrF3, N2O4
LIQUID AMMONIA
AUTO-IONIZATION
2NH3 → NH2- + NH4+ K = 5.1 x 10-27
Acid-Base Reactions:
Acid + Base → Salt +
NH4Br + KNH2
NH3
→ KBr + 2NH3
LIQUID AMMONIA
Liquid NH3 use when strong base is needed
Sulfamicacid, H2NS=(O)2OH is monobasic acid in water
with pka1= 1 but is dibasic acid in NH3
Mg2Si + 4NH4Br → SiH4+ 2MgBr2 + 2NH3
also GeH4 and AsH3
LIQUID HYDROGEN FLUORIDE
AUTOIONIZATION:
2HF ↔ H2F+ + HF2ACID-BASE:
Amines and carboxylic acids are readily soluble in HF:
CH3COOH + 2HF ↔ CH3COOH2+ + HF2-
LIQUID HYDROGEN FLUORIDE
SiF4 and CF4 are insoluble in liquid HF but BF3 AsF5
and SbF5 dissolve to give very strongly acidic
solutions. (give reactions).
Very few protic acids exhibit acidic behavior in liquid
HF. (Flurosulfuric acid and perchloric acid)
SbF5 + 2HF ===== super acid capable of
protonating even a very weak base like hydrocarbons
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.
BUFFER PREPARATION
• Mixing solutions of a weak acid/base and the salt of its
conjugate base/acid.
• Mixing solutions of a weak acid and a strong base with
the weak acid in molar excess.
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.
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
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
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 from solubility of a compound
• A saturated solution of silver chromate, Ag2CrO4, has [Ag+] = 1.3 x
10-4 M. What is the Ksp for Ag2CrO4?
Molar mass =
461.01
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)
---
---
Molar Mass =
147.63
Common Ion Effect in Solubility
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.
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
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.
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 CrO4 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
8.0 x 10-28
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+.
Selective Precipitation of Ions
One can use
differences in
solubilities of salts to
separate ions in a
mixture.