CHE 1302 Tro Ch 16 Aqueous Ionic Equilibrium BB Notes

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CHE 1302

Tro 2nd Ed.

Chapter 16: Aqueous Ionic Equilibrium

Dr. Bruce E. Hodson

The Common Ion Effect

Ionization of the weak acid is suppressed: pH INCREASES

1. Consider a solution of the WEAK acid acetic acid, CH3COOH….

3. LeChâtelier’s principle: What happens to [H3O+] ?

NaC2H3O2 → Na+ + CH3COO–

2. … then add acetate ion as a second solute

CH3COOH(aq) at equilibrium: a few H3O+ ions and a

few CH3COO– ions

Add CH3COO– , reestablish

equilibrium: more acetate ions, fewer

H3O+ ions


Common ion effect : the suppression of the ionization of a weak electrolyte by the presence of a common ion from a strong electrolyte


Ionization of the weak base is suppressed

1. Consider a solution of the WEAK base ammonia, NH3 ….

NH4Cl → NH4+ + Cl–

2. … then add ammonium ion (a COMMON ION) as a second solute

3. LeChâtelier’s principle:

What happens to [OH–] ?

EXAMPLES

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16.2 Buffers: Solutions that Resist pH Change

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Pure water has no ‘buffer capacity’


In contrast, for a buffer solution the pH barely changes

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A buffer solution is one whose pH changes very little when small amounts of strong acid or strong base are added, or when diluted

Weak acid and a salt of its conjugate base

Weak base and a salt of its conjugate acid


A weak acid and a salt of it’s conjugate base…


Note that the COMMON ION EFFECT ….. is in effect …


Buffer pH higher than

pure acid and lower than pure salt

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A buffer maintains approximately constant pH because of the acid and base components which are present at all times .….

Added OH– reacts with acid present HC2H3O2 → H2O + C2H3O2–

Added H3O+ reacts with base present C2H3O2– → H2O + HC2H3O2

16.2 Buffer Solutions – Why do they work ?

weak acidHC2H3O2

conjugate base C2H3O2

–

Na+ spectator ions


A weak base and a salt of it’s conjugate acid …

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Added OH– reacts with acid present NH4+ → H2O + NH3

Added H3O+ reacts with base present NH3 → H2O + NH4

+

weak baseNH3

conjugate acid NH4

+

Cl– spectator ions

16.2 Buffer Solutions – Why do they work ?

A buffer maintains approximately constant pH because of the acid and base components which are present at all times .….

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So…. Buffers contain components which can neutralize (small amounts of) added strong acids and bases

Buffers only work because the components do not completely neutralized each another… e.g HCl + NaOH doesn’t work ….

H3O+ + OH– → 2 H2O

pH of a Buffer: Equilibrium Approach EXAMPLE


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The Henderson-Hasselbalch equation gives the pH of a ‘weak acid + salt of conjugate base’ buffer directly:

[conjugate base]pH = pKa + log –––––––––––––– [weak acid]


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Generally, a buffer is most effective if [weak acid] = [conjugate base] …..

[conjugate base]pH = pKa + log –––––––––––––– [weak acid]

pH = pKa

Under these conditions, the pH of the buffer is approx. equal to the pKa of the weak acid

16.2 Buffer Solutions – The BUFFER CONDITION

.. and within a range of 2 pH units, meaning 0.1 through 10 for the ratio base to acid

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Capacity of a buffer is the amount of strong acid, or strong base, it can consume before it no longer acts as a buffer

….the more concentrated the buffer components in a solution, the more added acid or base the solution can neutralize

16.2 Buffer Solutions - CAPACITY

More Buffer EXAMPLES

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16.4 Titration and pH Curves

4.8 Acid-Base Titrations (Stoichiometry in Solutions)

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A titration is a method for determining the concentration of a solute in a sample of known volume ………

4.8 Titrations (Stoichiometry in Solutions)

Titrations can be applied to precipitation, acid-base, or redox reactions

Known volume (L) of unknown concentration

(mol/L) reactant B (analyte)

Add a known volume (L) of a known

concentration (mol/L) reactant A (titrant)

1. Add enough A to react completely with B

2. Calculate moles of A added

3. Use a balanced equation to find moles

of B present

4. Moles B / volume B = molarity of B

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4.8 Acid – Base Titration

42.00 mL of

0.150 M NaOH(aq)

added

A TITRANT of known concentration is added until the ANALYTE solution becomes

neutral…..

An INDICATOR is added which changes color at its END

POINT

At the EQUIVALENCE POINT, the acid and base have been

brought together in exact stoichiometric proportions

HCl(aq) ? conc. (50.00 mL)

Neutral Solution

H+

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A pH indicator is a weak acid that is one color when protonated (HIn), and another color when deprotonated (In–) (the conjugate base)

HIn(aq) + acid HIn(aq) + base

16.4 Acid-Base Indicators

HIn + H2O In– + H3O+

Adding H3O+ to HIn(aq) (pink) favors HIn (RED)Adding OH– to HIn(aq) (pink) favors In– (PURPLE)

RED PURPLE

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‘H’Phenolphthalein + H2O

Phenolphthalein– + H3O+

Adding OH– favors Phenolphthalein–

Adding H3O+ favors ‘H’Phenolphthalein

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Different indicators have different values of Ka, so they exhibit color changes at different values of pH …


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Bromthymol blue changes color at the

equivalence point (approx. 20.0 mL NaOH(aq) added)

So … Ideally we want the indicator’s color change (end point) to be centered around the equivalence point

16.4 Titrations: Strong Acid with Strong Base (MONOPROTIC)

Titration Curve

HCl + NaOH → NaCl + H2O

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pH values can be calculated at any point by assessing moles of acid and base present and using a balanced equation for the neutralization

16.4 Titrations: Strong Acid with Strong Base (MONOPROTIC)

pH << 7

pH >> 7

HCl + NaOH → NaCl + H2O

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16.4 Titrations: Strong Base with Strong Acid (MONOPROTIC)

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Equivalence-point pH is NOT 7.00

16.4 Titrations: Weak Acid with Strong Base (MONOPROTIC)

HC2H3O2 + NaOH → NaC2H3O2 + H2O

pH low

pH >> 7

pH = pKa ± 1: buffer solution (weak acid

and its conjugate base): H-H equation

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16.4 Titrations: Weak BASE with Strong Acid (MONOPROTIC)

Titration curve for the titration of 0.100 M NH3 with 0.100 M HCl

EXAMPLES

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16.4 Titrations: Weak Acid with Strong Base (POLYPROTIC)

Titration of 50.0 mL of 0.10 M H3PO3 with 0.10 M NaOH

H3PO3 + NaOH → H2PO3– + H2O

H2PO3– + NaOH → HPO3

2– + H2O

(Multiple Equivalence

points)

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16.5 Solubility Equilibria and the Solubility Product Constant

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Solid begins to dissolve

As solid dissolves, some dissolved solute begins to crystallize

Eventually, the rates of dissolving and of crystallization are equal; no more solute appears to dissolve.

Longer standing does not change the amount

of dissolved solute

Equilibria Review

For a saturated solution ……..

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Many ionic compounds do not dissolve well in aqueous solution – these are termed ‘sparingly soluble’ rather than ‘insoluble’

A Dynamic Equilibrium is established between the solid solute and the ions of that solute in solution

16.5 The Solubility Product Constant, Ksp

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The Solubility Product, Ksp , is the equilibrium constant for the dissolution of a sparingly soluble ionic compound

One mole of solid is the only reactant and the concentration of the solid is not included in the Ksp expression

FeS(s) Fe2+(aq) + S2–

(aq)

Ksp = [Fe2+] [S2–]


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Ksp values for a wide variety of sparingly soluble ionic compounds are known for a given temperature …… (25 °C in the table below)


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Molar solubility is simply the solubility of a compound in moles per liter – it can be computed directly from Ksp and vice versa

AgCl(s) Ag+(aq) + Cl –

(aq) Ksp = 1.77 × 10–10

1 mol AgCl ≡ 1 mol Ag+ ≡ 1 mol Cl –

Ksp = [Ag+] [Cl–]

Molar Solubility = S = [Ag+] = [Cl–]

Ksp = [Ag+] [Cl–] = S × S = S2

S = Ksp = 1.77 × 10–10 = 1.33 × 10–5 M

16.5 Relationship between Ksp and Molar Solubility

EXAMPLES

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16.5 The Common Ion Effect and Solubility Equilibria

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If we add one of its ions to a saturated solution of a sparingly soluble compound, we predict that precipitate will form :

This follows Le Chatelier’s principle (more product shifts equilibrium to the left)

PbCrO4(s) Pb2+(aq) + CrO42-(aq) PbCrO4(s) Pb2+(aq) + CrO4

2-(aq)

K2CrO4 added


PbCrO4(s) Pb2+(aq) + CrO42-(aq)

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When K2CrO4 is added to the

saturated solution of

PbCrO4 (aq) …

… [Pb2+] attains a new, lower equilibrium

concentration as Pb2+ reacts with

CrO42–

to produce PbCrO4 (s)

The solubility of a slightly soluble ionic compound is lowered when a second solute that furnishes a common ion is added to the solution

Pb2+

CrO42–

EXAMPLES


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16.5 Solubility and pH

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One way to dissolve a sparingly soluble compound is often to add a strong acid ……….

If the anion of the compound is the conjugate base of a weak acid, the H3O+ added will react with it lower its concentration

Added H3O+ reacts with, and removes F–; LeChâtelier’s principle says more F– forms

CaF2(s) Ca2+(aq) + 2 F–

(aq)

H3O+(aq) + F–

(aq) HF(aq) + H2O(l)


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Qsp = [Ca2+] [F–]2 < Ksp

equilibrium shifts to the rightSome CaF2 (s) dissolves (increase [F–] and [Ca2+]), so that Qsp becomes equal to Ksp. Continues until all acid is consumed or all solid dissolves.


CaF2(s) Ca2+(aq) + 2 F–

(aq)

Ksp = [Ca2+] [F–]2 = 5.3 × 10–9

What about Ksp and Qsp ? Adding H3O+ reduces the [F–], hence Qsp < Ksp

Remove product

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AgCl(s) Ag+ (aq) + Cl–

(aq)

This method will not work when the anion of the sparingly soluble compound is the conjugate base of a strong acid ……..

…….. because the anion is therefore a very weak base

H3O+(aq) + Cl–

(aq)

BaSO4(s) Ba2+ (aq) + SO4

2– (aq)

H3O+(aq) + SO4

2– (aq)

NO REACTION

NO REACTION


EXAMPLES

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Limitations of the Ksp Concept

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Ksp can only be used for sparingly soluble compounds if we wish to use molarities in the equilibrium constant expression …….

…… in more concentrated solutions, stoichiometric and effective concentrations (activities) are not equal


Ksp = [A+] [B–]

… if [AB] = 1.0 × 10–5 M … is [A+] = [B–] = 1.0 × 10–5 M ???

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Ions that are not common to the precipitate can affect solubility (and hence Ksp values) :

CaF2 is more soluble in 0.010 M Na2SO4 than it is in waterdue to inter-ionic attractions

Ca2+

Na+

SO42–

F–

Na+

SO42–

Na+

Ca2+

F–

H2O Na2SO4 (aq)


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16.6 Precipitation

16.6 Criteria for Precipitation and its completeness

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To determine if precipitation will occur, we calculate the Ion Product Reaction Quotient, Qsp, and compare it to Ksp

…. the Qsp expression takes the same form of the Ksp expression, but concentrations are not equilibrium values

FeS(s) Fe2+(aq) + S2–

(aq)

Ksp = [Fe2+] [S2–]

precipitation should occur (reaction shifts left)Qsp > Ksp

Qsp < Ksp precipitation will not occur (reaction shifts right)

Qsp = Kspsolution is just saturated (equilibrium)

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We generally consider precipitation to be “complete” if about 99.9% of the target ion is precipitated (0.1% or less left in solution)

In calculations, assume as much ppt. has formed as possible, calculate concentrations of remaining ions and then equilibrium

is reached from that point

16.6 Criteria for Precipitation and its completeness

EXAMPLES

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16.6 Selective (Fractional) Precipitation

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We can often completely precipitate one ion while leaving another ion in solution – Selective Precipitation

AgNO3 added to a mixture containing Cl– and I–

16.6 Selective (Fractional) Precipitation

EXAMPLES

… for this to work, there must be a fairly significant difference in the solubilities (Ksp values) of the substances being separated

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16.8 Equilibria Involving Complex Ions

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Complex ions have a more complex structure than the ions we have previously considered

Ligands are LEWIS BASES, the central metal atom/ion is a LEWIS ACID. Coordinate covalent bonds are formed.

Attached groups (LIGANDS) are ions or molecules with at

least one lone pair of electrons

3+


Common ligands:Cl–, OH–, H2O, NH3

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The formula of a complex ion is written in square [!!!!!] brackets, any charge is placed superscript right ……

A coordination compound is one which contains a complex ion, e.g. Na[AgCl2] or [Cu(NH3)4]Cl2

[AgCl2]–

–

[Cu(NH3)4]2+

2+


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The production of a complex ion can be described by an equilibrium reaction……..

The equilibrium constant is the FORMATION CONSTANT, Kf

Ag+(aq) + 2 NH3 (aq) [Ag(NH3)2]+

(aq)

Kf = [ [Ag(NH3)2]+ ]

Pb2+(aq) + 4 I–

(aq) [PbI4]2– (aq)

[Ag+] [NH3]2

Kf = [ [PbI4]2– ]

[Pb2+] [I–]4


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At a molecular level, the actual process (mechanism) is stepwise ……

M(H2O)42+ M(H2O)3(NH3)2+ M(NH3)4

2+

NH3

3NH3

The product of individual formation constants for each step gives the overall formation constant, Kf


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Sparingly soluble compounds can be dissolved by treatment with a complexing agent, which contains a ligand with great affinity for the metal ion

The above process is similar to dissolving sparingly soluble compound in strong acid

AgI (s) Ag+ (aq) + I–

(aq)

Ag+(aq) + 2 S2O3

2– (aq) [Ag(S2O3)2]3–

(aq)

KSP = 5.3 × 10–9

Kf = 1.7 × 1013

AgI(s) + 2 S2O32–

(aq) [Ag(S2O3)2]3– (aq) + I–

(aq)

Kc = Ksp × Kf = 1.4 × 10–3


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1. AgCl is insoluble in

water.…

2…but if the concentration of NH3 is made high

enough …

3 … the AgCl forms the soluble

[Ag(NH3)2]+ ion.

Similarly for AgCl(s) and NH3 …….


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CaSO4 will not dissolve in strong acid, however adding EDTA …….

The complexation reaction competes successfully with the precipitation reaction and CaSO4 dissolves

CaSO4 (s) Ca2+ (aq) + SO4

2– (aq)

Ca2+ (aq) + EDTA 4–

(aq) [CaEDTA]2– (aq)

KSP = 9.1 × 10–6

Kf = 1.0 × 1011


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EXAMPLES

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A hydrated cation (a complex ion) can be a weak acid ….

High charge denisty of the metal ion tends to withdraw electron density from, and hence weaken, the O–H bonds e.g. Fe2+

(aq), Zn2+(aq), Al3+

(aq)

[Fe(H2O)6]3+ (aq) + H2O (l) [Fe(H2O)5OH]2+

(aq) + H3O+ (aq)

Ka = 9.0 × 10–4


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Certain metal hydroxides, insoluble in water, are amphoteric; they will react with both strong acids or strong bases

Al(OH)3(H2O)3 (s)[Al(H2O)6]3+ (aq) [Al(OH)4(H2O)2]–

(aq)

NaOH (aq)HCl (aq)


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The amphoteric behavior is a result of the balance between acidic and basic behavior of the Al–O–H unit:

+ OH–

(OH)2Al O H

(OH)2Al O+ H2O

–

In an amphoteric substance, M (Al above) is not strongly electronegative as it is for the oxoacids, nor is it highly electropositive as it is in strong bases

As an acid, O–H breaks

+ H3O+

(OH)2Al O H

(OH)2Al+ 2 H2O

As a base, M–O breaks

+


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16.7 Qualitative Cation Analysis

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QIA is used to identify the cations and anions present in a sample

It has largely been replaced by modern instrumental methods

Add precipitating

ionC

entri

fuge

Add precipitating

ion

Cen

trifu

ge


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Step 1 Add

NH3(aq)

Cen

trifu

ge

Cen

trifu

ge

Step 2 Add HCl

Step 3 Add NaOH

Cen

trifu

geStep 4

Add HCl, Na2HPO4

Step 5 Dissolve in

HCl and add KSCN

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Chapter 16 Summaries

The Solubility Product Constant, Ksp, represents an equilibrium between a slightly soluble ionic compound and its ions in a saturated aqueous solution

The Common Ion Effect is responsible for the reduction in solubility of a slightly soluble ionic compound according to LeChatelier’s principle

Precipitation is assumed to be complete if no more than 0.1% of the target ion remains in solutionThe Solubilities of some slightly soluble compounds depends strongly on pH

Certain solutes become more soluble in the presence of species that canserve as ligands (Lewis Bases) in Complex Ions

The ability of water to donate protons when acting as a ligand accounts for theacidic character of some complex ions

Precipitation, acid–base, redox, complex-ion formation, and amphotericbehavior are all used for the Qualitative Analysis of common cations

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Chapter 16 Summaries

A strong electrolyte that produces an ion common to the ionization equilibrium of a weak acid or a weak base suppresses the ionization of the weak electrolyte – this is called the COMMON ION EFFECT

Acid–base indicators are weak acids for which the acid and its conjugate base have different colors

CHE 1302 Tro Ch 16 Aqueous Ionic Equilibrium BB Notes

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