Henderson-Hasselbalch Equation HA H + + A -. Titration Curve for a Weak Acid or Base.
1 Titration Curve of a Weak Base with a Strong Acid.
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Transcript of 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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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
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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?
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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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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