1 Titration Curve of a Weak Base with a Strong Acid

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Titration Curve of a Weak Base with a Strong Acid

Titration Curve of a Weak Base with a Strong Acid

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Titration of a Polyprotic AcidTitration of a Polyprotic Acid

if Ka1 >> Ka2, there will be two equivalence points in the titration

the closer the Ka’s are to each other, the less distinguishable the equivalence points are

if Ka1 >> Ka2, there will be two equivalence points in the titration

the closer the Ka’s are to each other, the less distinguishable the equivalence points are

titration of 25.0 mL of 0.100 M H2SO3 with 0.100 M NaOH

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Monitoring pH During a TitrationMonitoring pH During a Titration

the general method for monitoring the pH during the course of a titration is to measure the conductivity of the solution due to the [H3O+]

using a probe that specifically measures just H3O+

the endpoint of the titration is reached at the equivalence point in the titration – at the inflection point of the titration curve

if you just need to know the amount of titrant added to reach the endpoint, we often monitor the titration with an indicator

the general method for monitoring the pH during the course of a titration is to measure the conductivity of the solution due to the [H3O+]

using a probe that specifically measures just H3O+

the endpoint of the titration is reached at the equivalence point in the titration – at the inflection point of the titration curve

if you just need to know the amount of titrant added to reach the endpoint, we often monitor the titration with an indicator

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Monitoring pH During a TitrationMonitoring pH During a Titration

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IndicatorsIndicators many dyes change color depending on the pH of the solution these dyes are weak acids, establishing an equilibrium with the

H2O and H3O+ in the solution

HInd(aq) + H2O(l) Ind-(aq) + H3O+

(aq)

the color of the solution depends on the relative concentrations of Ind-:HInd ≈ 1, the color will be mix of the colors of Ind- and HInd when Ind-:HInd > 10, the color will be mix of the colors of Ind-

when Ind-:HInd < 0.1, the color will be mix of the colors of HInd

many dyes change color depending on the pH of the solution these dyes are weak acids, establishing an equilibrium with the

H2O and H3O+ in the solution

HInd(aq) + H2O(l) Ind-(aq) + H3O+

(aq)

the color of the solution depends on the relative concentrations of Ind-:HInd ≈ 1, the color will be mix of the colors of Ind- and HInd when Ind-:HInd > 10, the color will be mix of the colors of Ind-

when Ind-:HInd < 0.1, the color will be mix of the colors of HInd

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PhenolphthaleinPhenolphthalein

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Methyl RedMethyl Red

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Monitoring a Titration with an Indicator

Monitoring a Titration with an Indicator

for most titrations, the titration curve shows a very large change in pH for very small additions of base near the equivalence point

an indicator can therefore be used to determine the endpoint of the titration if it changes color within the same range as the rapid change in pH pKa of HInd ≈ pH at equivalence point

for most titrations, the titration curve shows a very large change in pH for very small additions of base near the equivalence point

an indicator can therefore be used to determine the endpoint of the titration if it changes color within the same range as the rapid change in pH pKa of HInd ≈ pH at equivalence point

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Acid-Base IndicatorsAcid-Base Indicators

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Solubility EquilibriaSolubility Equilibria

all ionic compounds dissolve in water to some degree

however, many compounds have such low solubility in water that we classify them as insoluble

we can apply the concepts of equilibrium to salts dissolving, and use the equilibrium constant for the process to measure relative solubilities in water

all ionic compounds dissolve in water to some degree

however, many compounds have such low solubility in water that we classify them as insoluble

we can apply the concepts of equilibrium to salts dissolving, and use the equilibrium constant for the process to measure relative solubilities in water

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Solubility ProductSolubility Product

the equilibrium constant for the dissociation of a solid salt into its aqueous ions is called the solubility product, Ksp

for an ionic solid MnXm, the dissociation reaction is:MnXm(s) nMm+(aq) + mXn−(aq)

the solubility product would be Ksp = [Mm+]n[Xn−]m

for example, the dissociation reaction for PbCl2 isPbCl2(s) Pb2+(aq) + 2 Cl−(aq)

and its equilibrium constant is Ksp = [Pb2+][Cl−]2

the equilibrium constant for the dissociation of a solid salt into its aqueous ions is called the solubility product, Ksp

for an ionic solid MnXm, the dissociation reaction is:MnXm(s) nMm+(aq) + mXn−(aq)

the solubility product would be Ksp = [Mm+]n[Xn−]m

for example, the dissociation reaction for PbCl2 isPbCl2(s) Pb2+(aq) + 2 Cl−(aq)

and its equilibrium constant is Ksp = [Pb2+][Cl−]2

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13

Molar SolubilityMolar Solubility

solubility is the amount of solute that will dissolve in a given amount of solution at a particular temperature

the molar solubility is the number of moles of solute that will dissolve in a liter of solution the molarity of the dissolved solute in a saturated

solution

for the general reaction MnXm(s) nMm+(aq) + mXn−

(aq)

solubility is the amount of solute that will dissolve in a given amount of solution at a particular temperature

the molar solubility is the number of moles of solute that will dissolve in a liter of solution the molarity of the dissolved solute in a saturated

solution

for the general reaction MnXm(s) nMm+(aq) + mXn−

(aq)

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Calculate the molar solubility of PbCl2 in pure water at 25°CCalculate the molar solubility of PbCl2 in pure water at 25°C

Write the dissociation reaction and Ksp expression

Create an ICE table defining the change in terms of the solubility of the solid

[Pb2+] [Cl−]

Initial 0 0

Change +S +2S

Equilibrium S 2S

PbCl2(s) Pb2+(aq) + 2 Cl−(aq)

Ksp = [Pb2+][Cl−]2

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Calculate the molar solubility of PbCl2 in pure water at 25°CCalculate the molar solubility of PbCl2 in pure water at 25°C

Substitute into the Ksp expression

Find the value of Ksp from Table 16.2, plug into the equation and solve for S

[Pb2+] [Cl−]

Initial 0 0

Change +S +2S

Equilibrium S 2S

Ksp = [Pb2+][Cl−]2

Ksp = (S)(2S)2

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Determine the Ksp of PbBr2 if its molar solubility in water at 25°C is 1.05 x 10-2 M

Determine the Ksp of PbBr2 if its molar solubility in water at 25°C is 1.05 x 10-2 M

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Determine the Ksp of PbBr2 if its molar solubility in water at 25°C is 1.05 x 10-2 M

Determine the Ksp of PbBr2 if its molar solubility in water at 25°C is 1.05 x 10-2 M

Write the dissociation reaction and Ksp expression

Create an ICE table defining the change in terms of the solubility of the solid

[Pb2+] [Br−]

Initial 0 0

Change +(1.05 x 10-2) +2(1.05 x 10-2)

Equilibrium (1.05 x 10-2) (2.10 x 10-2)

PbBr2(s) Pb2+(aq) + 2 Br−(aq)

Ksp = [Pb2+][Br−]2

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Determine the Ksp of PbBr2 if its molar solubility in water at 25°C is 1.05 x 10-2 M

Determine the Ksp of PbBr2 if its molar solubility in water at 25°C is 1.05 x 10-2 M

Substitute into the Ksp expression

plug into the equation and solve

Ksp = [Pb2+][Br−]2

Ksp = (1.05 x 10-2)(2.10 x 10-2)2

[Pb2+] [Br−]

Initial 0 0

Change +(1.05 x 10-2) +2(1.05 x 10-2)

Equilibrium (1.05 x 10-2) (2.10 x 10-2)

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Ksp and Relative SolubilityKsp and Relative Solubility

molar solubility is related to Ksp

but you cannot always compare solubilities of compounds by comparing their Ksps

in order to compare Ksps, the compounds must have the same dissociation stoichiometry

molar solubility is related to Ksp

but you cannot always compare solubilities of compounds by comparing their Ksps

in order to compare Ksps, the compounds must have the same dissociation stoichiometry

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The Effect of Common Ion on SolubilityThe Effect of Common Ion on Solubility

addition of a soluble salt that contains one of the ions of the “insoluble” salt, decreases the solubility of the “insoluble” salt

for example, addition of NaCl to the solubility equilibrium of solid PbCl2 decreases the solubility of PbCl2

PbCl2(s) Pb2+(aq) + 2 Cl−(aq)

addition of a soluble salt that contains one of the ions of the “insoluble” salt, decreases the solubility of the “insoluble” salt

for example, addition of NaCl to the solubility equilibrium of solid PbCl2 decreases the solubility of PbCl2

PbCl2(s) Pb2+(aq) + 2 Cl−(aq)

addition of Cl− shifts the equilibrium to the left

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Calculate the molar solubility of CaF2 in 0.100 M NaF at 25°CCalculate the molar solubility of CaF2 in 0.100 M NaF at 25°C

Write the dissociation reaction and Ksp expression

Create an ICE table defining the change in terms of the solubility of the solid

[Ca2+] [F−]

Initial 0 0.100

Change +S +2S

Equilibrium S 0.100 + 2S

CaF2(s) Ca2+(aq) + 2 F−(aq)

Ksp = [Ca2+][F−]2

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Calculate the molar solubility of CaF2 in 0.100 M NaF at 25°CCalculate the molar solubility of CaF2 in 0.100 M NaF at 25°C

Substitute into the Ksp expressionassume S is small

Find the value of Ksp from Table 16.2, plug into the equation and solve for S

[Ca2+] [F−]

Initial 0 0.100

Change +S +2S

Equilibrium S 0.100 + 2S

Ksp = [Ca2+][F−]2

Ksp = (S)(0.100 + 2S)2

Ksp = (S)(0.100)2

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The Effect of pH on SolubilityThe Effect of pH on Solubility

for insoluble ionic hydroxides, the higher the pH, the lower the solubility of the ionic hydroxide and the lower the pH, the higher the solubility higher pH = increased [OH−]

M(OH)n(s) Mn+(aq) + nOH−(aq)

for insoluble ionic compounds that contain anions of weak acids, the lower the pH, the higher the solubility

M2(CO3)n(s) 2 Mn+(aq) + nCO32−(aq)

H3O+(aq) + CO32− (aq) HCO3

− (aq) + H2O(l)

for insoluble ionic hydroxides, the higher the pH, the lower the solubility of the ionic hydroxide and the lower the pH, the higher the solubility higher pH = increased [OH−]

M(OH)n(s) Mn+(aq) + nOH−(aq)

for insoluble ionic compounds that contain anions of weak acids, the lower the pH, the higher the solubility

M2(CO3)n(s) 2 Mn+(aq) + nCO32−(aq)

H3O+(aq) + CO32− (aq) HCO3

− (aq) + H2O(l)

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PrecipitationPrecipitation

precipitation will occur when the concentrations of the ions exceed the solubility of the ionic compound

if we compare the reaction quotient, Q, for the current solution concentrations to the value of Ksp, we can determine if precipitation will occur Q = Ksp, the solution is saturated, no precipitation

Q < Ksp, the solution is unsaturated, no precipitation

Q > Ksp, the solution would be above saturation, the salt above saturation will precipitate

some solutions with Q > Ksp will not precipitate unless disturbed – these are called supersaturated solutions

precipitation will occur when the concentrations of the ions exceed the solubility of the ionic compound

if we compare the reaction quotient, Q, for the current solution concentrations to the value of Ksp, we can determine if precipitation will occur Q = Ksp, the solution is saturated, no precipitation

Q < Ksp, the solution is unsaturated, no precipitation

Q > Ksp, the solution would be above saturation, the salt above saturation will precipitate

some solutions with Q > Ksp will not precipitate unless disturbed – these are called supersaturated solutions

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precipitation occurs if Q > Ksp

a supersaturated solution will precipitate if a seed crystal is

added

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Selective PrecipitationSelective Precipitation

a solution containing several different cations can often be separated by addition of a reagent that will form an insoluble salt with one of the ions, but not the others

a successful reagent can precipitate with more than one of the cations, as long as their Ksp values are

significantly different

a solution containing several different cations can often be separated by addition of a reagent that will form an insoluble salt with one of the ions, but not the others

a successful reagent can precipitate with more than one of the cations, as long as their Ksp values are

significantly different

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What is the minimum [OH−] necessary to just begin to precipitate Mg2+ (with [0.059]) from seawater assuming

Ksp=2.06x10-13)?

What is the minimum [OH−] necessary to just begin to precipitate Mg2+ (with [0.059]) from seawater assuming

Ksp=2.06x10-13)?

precipitating may just occur when Q = Ksp

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What is the [Mg2+] when Ca2+ (with [0.011]) just begins to precipitate from seawater?

What is the [Mg2+] when Ca2+ (with [0.011]) just begins to precipitate from seawater?

precipitating Mg2+ begins when [OH−] = 1.9 x 10-6 M

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What is the [Mg2+] when Ca2+ (with [0.011]) just begins to precipitate from seawater?

What is the [Mg2+] when Ca2+ (with [0.011]) just begins to precipitate from seawater?

precipitating Mg2+ begins when [OH−] = 1.9 x 10-6 M

precipitating Ca2+ begins when [OH−] = 2.06 x 10-2 M

when Ca2+ just begins to precipitate out, the [Mg2+] has dropped from 0.059 M to 4.8 x 10-10 M

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Qualitative AnalysisQualitative Analysis

an analytical scheme that utilizes selective precipitation to identify the ions present in a solution is called a qualitative analysis scheme wet chemistry

a sample containing several ions is subjected to the addition of several precipitating agents

addition of each reagent causes one of the ions present to precipitate out

an analytical scheme that utilizes selective precipitation to identify the ions present in a solution is called a qualitative analysis scheme wet chemistry

a sample containing several ions is subjected to the addition of several precipitating agents

addition of each reagent causes one of the ions present to precipitate out

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Qualitative AnalysisQualitative Analysis

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Group 1Group 1

group one cations are Ag+, Pb2+, and Hg22+

all these cations form compounds with Cl− that are insoluble in water as long as the concentration is large enough PbCl2 may be borderline

molar solubility of PbCl2 = 1.43 x 10-2 M

precipitated by the addition of HCl

group one cations are Ag+, Pb2+, and Hg22+

all these cations form compounds with Cl− that are insoluble in water as long as the concentration is large enough PbCl2 may be borderline

molar solubility of PbCl2 = 1.43 x 10-2 M

precipitated by the addition of HCl

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Group 2Group 2

group two cations are Cd2+, Cu2+, Bi3+, Sn4+, As3+, Pb2+, Sb3+, and Hg2+

all these cations form compounds with HS− and S2− that are insoluble in water at low pH

precipitated by the addition of H2S in HCl

group two cations are Cd2+, Cu2+, Bi3+, Sn4+, As3+, Pb2+, Sb3+, and Hg2+

all these cations form compounds with HS− and S2− that are insoluble in water at low pH

precipitated by the addition of H2S in HCl

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Group 3Group 3

group three cations are Fe2+, Co2+, Zn2+, Mn2+, Ni2+ precipitated as sulfides; as well as Cr3+, Fe3+, and Al3+

precipitated as hydroxides

all these cations form compounds with S2− that are insoluble in water at high pH

precipitated by the addition of H2S in NaOH

group three cations are Fe2+, Co2+, Zn2+, Mn2+, Ni2+ precipitated as sulfides; as well as Cr3+, Fe3+, and Al3+

precipitated as hydroxides

all these cations form compounds with S2− that are insoluble in water at high pH

precipitated by the addition of H2S in NaOH

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Group 4Group 4

group four cations are Mg2+, Ca2+, Ba2+

all these cations form compounds with PO43− that are

insoluble in water at high pH

precipitated by the addition of (NH4)2HPO4

group four cations are Mg2+, Ca2+, Ba2+

all these cations form compounds with PO43− that are

insoluble in water at high pH

precipitated by the addition of (NH4)2HPO4

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Group 5Group 5

group five cations are Na+, K+, NH4+

all these cations form compounds that are soluble in water – they do not precipitate

identified by the color of their flame

group five cations are Na+, K+, NH4+

all these cations form compounds that are soluble in water – they do not precipitate

identified by the color of their flame

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Complex Ion FormationComplex Ion Formation

transition metals tend to be good Lewis acids

they often bond to one or more H2O molecules to form a hydrated ion H2O is the Lewis base, donating electron pairs to form

coordinate covalent bondsAg+(aq) + 2 H2O(l) Ag(H2O)2

+(aq)

ions that form by combining a cation with several anions or neutral molecules are called complex ions e.g., Ag(H2O)2

+

the attached ions or molecules are called ligands e.g., H2O

transition metals tend to be good Lewis acids

they often bond to one or more H2O molecules to form a hydrated ion H2O is the Lewis base, donating electron pairs to form

coordinate covalent bondsAg+(aq) + 2 H2O(l) Ag(H2O)2

+(aq)

ions that form by combining a cation with several anions or neutral molecules are called complex ions e.g., Ag(H2O)2

+

the attached ions or molecules are called ligands e.g., H2O

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Complex Ion EquilibriaComplex Ion Equilibria

if a ligand is added to a solution that forms a stronger bond than the current ligand, it will replace the current ligand

Ag(H2O)2+

(aq) + 2 NH3(aq) Ag(NH3)2+

(aq) + 2 H2O(l)

generally H2O is not included, since its complex ion is always present in aqueous solution

Ag+(aq) + 2 NH3(aq) Ag(NH3)2

+(aq)

if a ligand is added to a solution that forms a stronger bond than the current ligand, it will replace the current ligand

Ag(H2O)2+

(aq) + 2 NH3(aq) Ag(NH3)2+

(aq) + 2 H2O(l)

generally H2O is not included, since its complex ion is always present in aqueous solution

Ag+(aq) + 2 NH3(aq) Ag(NH3)2

+(aq)

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Formation ConstantFormation Constant

the reaction between an ion and ligands to form a complex ion is called a complex ion formation reaction

Ag+(aq) + 2 NH3(aq) Ag(NH3)2

+(aq)

the equilibrium constant for the formation reaction is called the formation constant, Kf

the reaction between an ion and ligands to form a complex ion is called a complex ion formation reaction

Ag+(aq) + 2 NH3(aq) Ag(NH3)2

+(aq)

the equilibrium constant for the formation reaction is called the formation constant, Kf

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Formation ConstantsFormation Constants

42

200.0 mL of 1.5 x 10-3 M Cu(NO3)2 is mixed with 250.0 mL of

0.20 M NH3. What is the [Cu2+] at equilibrium? 200.0 mL of 1.5 x 10-3 M Cu(NO3)2 is mixed with 250.0 mL of

0.20 M NH3. What is the [Cu2+] at equilibrium?

Write the formation reaction and Kf expression.Look up Kf value

Determine the concentration of ions in the diluted solutions

Cu2+(aq) + 4 NH3(aq) Cu(NH3)42+(aq)

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200.0 mL of 1.5 x 10-3 M Cu(NO3)2 is mixed with 250.0 mL of

0.20 M NH3. What is the [Cu2+] at equilibrium? 200.0 mL of 1.5 x 10-3 M Cu(NO3)2 is mixed with 250.0 mL of

0.20 M NH3. What is the [Cu2+] at equilibrium?

Create an ICE table. Since Kf is large, assume all the Cu2+ is converted into complex ion, then the system returns to equilibrium

[Cu2+] [NH3] [Cu(NH3)22+]

Initial 6.7E-4 0.11 0

Change -≈6.7E-4 -4(6.7E-4) + 6.7E-4

Equilibrium x 0.11 6.7E-4

Cu2+(aq) + 4 NH3(aq) Cu(NH3)42+(aq)

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200.0 mL of 1.5 x 10-3 M Cu(NO3)2 is mixed with 250.0 mL of

0.20 M NH3. What is the [Cu2+] at equilibrium? 200.0 mL of 1.5 x 10-3 M Cu(NO3)2 is mixed with 250.0 mL of

0.20 M NH3. What is the [Cu2+] at equilibrium?

Cu2+(aq) + 4 NH3(aq) Cu(NH3)22+(aq)Substitute in and

solve for x

confirm the “x is small” approximation

[Cu2+] [NH3] [Cu(NH3)22+]

Initial 6.7E-4 0.11 0

Change -≈6.7E-4 -4(6.7E-4) + 6.7E-4

Equilibrium x 0.11 6.7E-4

since 2.7 x 10-13 << 6.7 x 10-4, the approximation is valid

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The Effect of Complex Ion Formation on Solubility

The Effect of Complex Ion Formation on Solubility

the solubility of an ionic compound that contains a metal cation that forms a complex ion increases in the presence of aqueous ligands

AgCl(s) Ag+(aq) + Cl−(aq) Ksp = 1.77 x 10-10

Ag+(aq) + 2 NH3(aq) Ag(NH3)2

+(aq) Kf = 1.7 x 107

adding NH3 to a solution in equilibrium with AgCl(s) increases the solubility of Ag+

the solubility of an ionic compound that contains a metal cation that forms a complex ion increases in the presence of aqueous ligands

AgCl(s) Ag+(aq) + Cl−(aq) Ksp = 1.77 x 10-10

Ag+(aq) + 2 NH3(aq) Ag(NH3)2

+(aq) Kf = 1.7 x 107

adding NH3 to a solution in equilibrium with AgCl(s) increases the solubility of Ag+

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47

Solubility of Amphoteric Metal Hydroxides

Solubility of Amphoteric Metal Hydroxides

many metal hydroxides are insoluble

all metal hydroxides become more soluble in acidic solution shifting the equilibrium to the right by removing OH−

some metal hydroxides also become more soluble in basic solution acting as a Lewis base forming a complex ion

substances that behave as both an acid and base are said to be amphoteric

some cations that form amphoteric hydroxides include Al3+, Cr3+, Zn2+, Pb2+, and Sb2+

many metal hydroxides are insoluble

all metal hydroxides become more soluble in acidic solution shifting the equilibrium to the right by removing OH−

some metal hydroxides also become more soluble in basic solution acting as a Lewis base forming a complex ion

substances that behave as both an acid and base are said to be amphoteric

some cations that form amphoteric hydroxides include Al3+, Cr3+, Zn2+, Pb2+, and Sb2+

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Al3+Al3+

Al3+ is hydrated in water to form an acidic solution

Al(H2O)63+

(aq) + H2O(l) Al(H2O)5(OH)2+(aq) + H3O+

(aq)

addition of OH− drives the equilibrium to the right and continues

to remove H from the molecules

Al(H2O)5(OH)2+(aq) + OH−

(aq) Al(H2O)4(OH)2+

(aq) + H2O (l)

Al(H2O)4(OH)2+

(aq) + OH−(aq) Al(H2O)3(OH)3(s) + H2O

(l)

Al3+ is hydrated in water to form an acidic solution

Al(H2O)63+

(aq) + H2O(l) Al(H2O)5(OH)2+(aq) + H3O+

(aq)

addition of OH− drives the equilibrium to the right and continues

to remove H from the molecules

Al(H2O)5(OH)2+(aq) + OH−

(aq) Al(H2O)4(OH)2+

(aq) + H2O (l)

Al(H2O)4(OH)2+

(aq) + OH−(aq) Al(H2O)3(OH)3(s) + H2O

(l)

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