ACIDS AND BASESIONIC EQUILIBRIA

• 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.

ARRHENIUS DEFINITIONS

• Brønsted–Lowry: must have both1. an Acid: Proton donor

and2. a Base: Proton acceptor

BRØNSTED–LOWRY DEFINITION

The Brønsted-Lowry acid donates a proton,

while the Brønsted-Lowry base accepts it.

Brønsted-Lowry acids and bases are always paired.

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

• From the Latin word conjugare, meaning “to join together.”• Reactions between acids and bases always yield their

conjugate bases and acids.

CONJUGATE ACIDS AND BASES

• 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 AND 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)

ACID BASE STRENGTH

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.

AUTOIONIZATION OF WATER

• The equilibrium expression for this process isKc = [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

EQUILIBRIUM CONSTANT FOR WATER

pH is defined as the negative base-10 logarithm of the hydronium ion concentration.

pH = –log [H3O+]

pH

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 HYDROXIDESMetal 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 acid-dissociation 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

At Equilibrium 0.10 – 4.2 10-3

= 0.0958 = 0.104.2 10-3 4.2 10 - 3

Calculating Ka from pH

[4.2 10-3] [4.2 10-3][0.10]

Ka =

= 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

Percent Ionization = 1004.2 10-3

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:[C2H3O2], M [H3O+], M [C2H3O2

−], M

Initial 0.30 0 0

Change –x +x +x

Equilibrium 0.30 – x x x

Calculating pH from Ka

Use the ICE table:[C2H3O2], M [H3O+], M [C2H3O2

−], M

Initial 0.30 0 0

Change –x +x +x

Equilibrium 0.30 – x x x

Simplify: how big is x relative to 0.30?

CALCULATING pH FROM Ka

Use the ICE table:[C2H3O2], M [H3O+], M [C2H3O2

−], M

Initial 0.30 0 0

Change –x +x +x

Equilibrium 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?

[NH4+] [OH−]

[NH3]Kb = = 1.8 10-5

pH OF BASIC SOLUTIONS

Tabulate the data.

[NH3], M [NH4+], M [OH−], M

Initial 0.15 0 0

Equilibrium 0.15 - x 0.15 x x

Simplify: how big is x relative to 0.15?

pH OF BASIC SOLUTIONS

(1.8 10-5) (0.15) = x2

2.7 10-6 = x2

1.6 10-3 = x2

(x)2

(0.15)1.8 10-5 =

Check: is approximation ok?

pH of Basic Solutions

Therefore,[OH–] = 1.6 10-3 MpOH = –log (1.6 10-3)pOH = 2.80pH = 14.00 – 2.80pH = 11.20

Ka and dissociation constant of a conjugate base

HCN + H2O CN- + H3O+

CN- + H2O HCN + OH-

KaKb = Kw

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 Ka2 = 4.7 x 10-11

H3PO4 Ka1 = 7.2 x 10-3 Ka2 = 6.3 x 10-8 Ka3 = 4.2 x 10-13

SAMPLE PROBLEMWhat 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 Ka2 = 4.7 x 10-11

H3PO4 Ka1 = 7.2 x 10-3 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 [Al(H2O)5(OH)]+2 + H3O+

Ka = 1.2 x 10-5

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 StrengthResonance 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. (B-

L 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 & BasesThis definition includes the “standard” Brønsted-

Lowry 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 & BasesIn 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 for acids.

ACIDS IN ACIDIC SOLVENTS

In H2SO4, HClO4 is practically unionized while HNO3 is fully ionized.

SOLVENT ORIENTED APPROACH TO ACID-BASE 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-IONIZATION2NH3 → NH2

- + NH4+ K = 5.1 x 10-27

Acid-Base Reactions:

Acid + Base → Salt + NH3

NH4Br + KNH2 → 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+ + HF2

-

ACID-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:

• If acetate ion is added to the solution, Le Châtelier says the equilibrium will shift to the left.

HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2−(aq)

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.

[H3O+] [F−][HF]

Ka = = 6.8 10-4

THE COMMON-ION EFFECT

Because HCl, a strong acid, is also present, the initial [H3O+] is not 0, but rather 0.10 M.

[HF], M [H3O+], M [F−], M

Initially 0.20 0.10 0

Change −x +x +xAt Equilibrium 0.20 − x 0.20 0.10 + x 0.10 x

HF(aq) + H2O(l) H3O+(aq) + F−(aq)

The Common-Ion Effect

= x

1.4 10−3 = x

(0.10) (x)(0.20)6.8 10−4 =

(0.20) (6.8 10−4)(0.10)

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]

Ka =

HA + H2O H3O+ + A−

BUFFER CALCULATIONS

Rearranging slightly, this becomes

[A−][HA]Ka = [H3O+]

Taking the negative log of both side, we get

[A−][HA]−log Ka = −log [H3O+] + −log

pKapH

acid

base

BUFFER CALCULATIONS

• SopKa = 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 BuffersThe 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 BuffersNow use the Henderson–Hasselbalch equation to calculate the new pH:

pH = 4.74 + log (0.320)(0. 200)

pH = 4.74 + 0.06

pH = 4.80

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 maybe 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 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?

The Common Ion Effect on Solubility

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 Ksp Solubility(mol/L)

CuS 8.5 x 10-45 9.2 x 10-23

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 CrO4

2- will be diluted. We have to do a dilution calculation!

Dilution: C1V1 = C2V2

[Pb2+] =

[CrO42-] =

2

2

11 Pb M 0.0080 mL) (45

mL) M)(15 (0.024 V

VC

-24

2

11 CrO M 0.020 mL) (45

mL) M)(20 (0.030 V

VC

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 the two solutions are mixed!

Note: If Qsp = Ksp, the mixture is saturatedIf Qsp < Ksp, the solution is unsaturated

Either way, no ppte will form!

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.