chem unit8
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SolutionsSolutions
Mixtures: Homogeneous and Heterogeneous
Uniform Non-uniform
Mixtures: Homogeneous and Heterogeneous
Uniform Non-uniform
SolutionsSolutions
• Concentrated or Dilute
• Units of concentration:
• Ppm/ppb vs. Molarity
• Concentrated or Dilute
• Units of concentration:
• Ppm/ppb vs. Molarity
ConcentrationConcentration
• Parts per million (ppm)
mass of soluteppm = mass of solution x 100%
• Parts per million (ppm)
mass of soluteppm = mass of solution x 100%
ConcentrationConcentration
• Molarity = moles/Liter• Molarity = moles/Liter
MolarityMolarity
• Vinegar
• What is the molarity if 3.78 g of CH3CO2H are dissolved in 100 mL?
• Vinegar
• What is the molarity if 3.78 g of CH3CO2H are dissolved in 100 mL?
MolarityMolarity
• Vinegar
• What is the molarity if 3.78 g of CH3CO2H are dissolved in 100 mL?
1 mole3.78 g x 60 g = 0.063 moles
Molarity = 0.063 moles/0.1 L = 0.63 M
• Vinegar
• What is the molarity if 3.78 g of CH3CO2H are dissolved in 100 mL?
1 mole3.78 g x 60 g = 0.063 moles
Molarity = 0.063 moles/0.1 L = 0.63 M
Effects of SolutesEffects of Solutes
• Adding a dissolved substance to a solvent:
• Adding a dissolved substance to a solvent:
Effects of SolutesEffects of Solutes• Adding a dissolved substance to a
solvent:
• Less solvent molecules evaporate at the regular boiling point:
Higher boiling point
• Adding a dissolved substance to a solvent:
• Less solvent molecules evaporate at the regular boiling point:
Higher boiling point
Boiling Point ElevationBoiling Point Elevation
• Adding salt to cooking pasta
• Makes the pasta cook faster
• Adding salt to cooking pasta
• Makes the pasta cook faster
Effect of SolutesEffect of Solutes
Added solute makes it harder for the solvent to freeze
So a lower freezing point occurs
Added solute makes it harder for the solvent to freeze
So a lower freezing point occurs
Ice Cream!Ice Cream!
• Milkfat +
+dissolved sugar
+ Air
+ flavor
• Milkfat +
+dissolved sugar
+ Air
+ flavor
SolubilitySolubility
• Like dissolves LikeMiscible
• Oil + Water= immiscible
• Like dissolves LikeMiscible
• Oil + Water= immiscible
SurfactantsSurfactants
Surface – active Agent
Polar/ non-polar go-between
Surface – active Agent
Polar/ non-polar go-between
Solubility & TemperatureSolubility & Temperature
• Solubility increases with temperature
• Solubility increases with temperature
SaturationSaturation
• When the maximum amount of solute is dissolved in a solution
• When the maximum amount of solute is dissolved in a solution
Henry’s LawHenry’s Law
• The solubility of a gas in a solution increases as the partial pressure of the gas on the surface of the solution increases.
• The solubility of a gas in a solution increases as the partial pressure of the gas on the surface of the solution increases.
Henry’s LawHenry’s Law
• Carbonated Drinks:What happens when you open the
bottle?
• Carbonated Drinks:What happens when you open the
bottle?
Molarity & StoichiometryMolarity & Stoichiometry
• Recall Displacement Reactions:
Cu + 2 AgNO3 ---> 2 Ag + Cu(NO3)2
How much 0.5 M Silver Nitrate solution is needed to react with 0.317 g Cu metal?
• Recall Displacement Reactions:
Cu + 2 AgNO3 ---> 2 Ag + Cu(NO3)2
How much 0.5 M Silver Nitrate solution is needed to react with 0.317 g Cu metal?
Molarity & StoichiometryMolarity & Stoichiometry
Cu + 2 AgNO3 ---> 2 Ag + Cu(NO3)2
0.317 g Cu x 1 mole/63.45 g = 0.05 moles Cu
2 moles AgNO3
0.05 moles Cu x 1 mole Cu = 0.1 mole AgNO3
0.1 moles AgNO3 x 1 Liter/0.5 moles = 0.05 L
= 50 mL
Cu + 2 AgNO3 ---> 2 Ag + Cu(NO3)2
0.317 g Cu x 1 mole/63.45 g = 0.05 moles Cu
2 moles AgNO3
0.05 moles Cu x 1 mole Cu = 0.1 mole AgNO3
0.1 moles AgNO3 x 1 Liter/0.5 moles = 0.05 L
= 50 mL
Chemical EquilibriumChemical Equilibrium
• Balance point when the forward and reverse reaction rates are equal– Photochromatic lenses
• Balance point when the forward and reverse reaction rates are equal– Photochromatic lenses
Reversible ReactionsReversible Reactions
• Forward Reaction2 NO2 -----> N2O4
• Reverse ReactionN2O4 -----> 2 NO2
• Forward Reaction2 NO2 -----> N2O4
• Reverse ReactionN2O4 -----> 2 NO2
Reversible ReactionsReversible Reactions
• Forward Reaction2 NO2 -----> N2O4
• Reverse ReactionN2O4 -----> 2 NO2
2 NO2 N2O4
• Forward Reaction2 NO2 -----> N2O4
• Reverse ReactionN2O4 -----> 2 NO2
2 NO2 N2O4
Chemical EquilibriumChemical Equilibrium
• Balance point where the rate of the forward reaction equals the rate of the reverse reaction
• Concentrations of products and reactants remain unchanged
• Balance point where the rate of the forward reaction equals the rate of the reverse reaction
• Concentrations of products and reactants remain unchanged
Equilibrium ConstantEquilibrium Constant
• A + B C + D [C] [D]
• Keq = [A] [B]
if Keq > 1 mostly products (C & D)
if Keq < 1 mostly reactants (A & B)
if Keq = 1 Equal amounts of reactants, products
• A + B C + D [C] [D]
• Keq = [A] [B]
if Keq > 1 mostly products (C & D)
if Keq < 1 mostly reactants (A & B)
if Keq = 1 Equal amounts of reactants, products
Equilibrium ConstantsEquilibrium Constants
• Balance the equation
• Leave out any solids or liquids (s), (l)
• Raise the substance’s concentration to the power equal to its coefficient
• Balance the equation
• Leave out any solids or liquids (s), (l)
• Raise the substance’s concentration to the power equal to its coefficient
Equilibrium ConstantEquilibrium Constant
• COCl2(g) CO(g) + Cl2(g)
• Keq = ?
• COCl2(g) CO(g) + Cl2(g)
• Keq = ?
Equilibrium ConstantEquilibrium Constant
• COCl2(g) CO(g) + Cl2(g)
[CO][Cl2]
• Keq = [COCl2]
If [CO] = [Cl2] = 0.0178 and [COCl2] = 0.007, Keq = ???
• COCl2(g) CO(g) + Cl2(g)
[CO][Cl2]
• Keq = [COCl2]
If [CO] = [Cl2] = 0.0178 and [COCl2] = 0.007, Keq = ???
Equilibrium ConstantEquilibrium Constant
• COCl2(g) CO(g) + Cl2(g)
[CO][Cl2]
• Keq = [COCl2]
If [CO] = [Cl2] = 0.0178 and [COCl2] = 0.007, Keq = 0.045
• COCl2(g) CO(g) + Cl2(g)
[CO][Cl2]
• Keq = [COCl2]
If [CO] = [Cl2] = 0.0178 and [COCl2] = 0.007, Keq = 0.045
Equilibrium ConstantsEquilibrium Constants
• N2(g) + O2(g) 2 NO(g)
[NO]2
• Keq = [N2][O2]
• N2(g) + O2(g) 2 NO(g)
[NO]2
• Keq = [N2][O2]
Equilibrium ConstantsEquilibrium Constants
H2 + I2 2 HI
[HI]2
Keq = [H2][I2]
At 25°C, [H2] = [I2] = 0.05 M, [HI] = 0.389M [HI]2 (0.389)2
Keq = [H2][I2] = (0.05)(0.05) = 60.3
H2 + I2 2 HI
[HI]2
Keq = [H2][I2]
At 25°C, [H2] = [I2] = 0.05 M, [HI] = 0.389M [HI]2 (0.389)2
Keq = [H2][I2] = (0.05)(0.05) = 60.3
Equilibrium ConstantsEquilibrium Constants
H2 + I2 2 HI
[HI]2
Keq = [H2][I2]
At 0°C, [H2] = [I2] = 0.0057 M, [HI] = 0.0127M
[HI]2 (0.0127)2
Keq = [H2][I2] = (0.0057)(0.0057) = 4.96
H2 + I2 2 HI
[HI]2
Keq = [H2][I2]
At 0°C, [H2] = [I2] = 0.0057 M, [HI] = 0.0127M
[HI]2 (0.0127)2
Keq = [H2][I2] = (0.0057)(0.0057) = 4.96
Solubility ProductSolubility Product
• Ksp
• AgCl(s) Ag+(aq) + Cl–(aq)
• Ksp = [Ag+][Cl-]
• Ksp
• AgCl(s) Ag+(aq) + Cl–(aq)
• Ksp = [Ag+][Cl-]
Tooth EnamelTooth Enamel
• Hydroxyapatite Ca5(PO4)3OH
• Fluorine treatment– H2SiF6
– Ca5(PO4)3F
• Hydroxyapatite Ca5(PO4)3OH
• Fluorine treatment– H2SiF6
– Ca5(PO4)3F
Solubility ProductSolubility Product
• How much Hg2+ ion would be found in a pond where cinnabar was located?
• HgS Hg2+ + S2-
Ksp = 1 x 10-53
• How much Hg2+ ion would be found in a pond where cinnabar was located?
• HgS Hg2+ + S2-
Ksp = 1 x 10-53
Solubility ProductSolubility Product
• How much Hg2+ ion would be found in a pond where cinnabar was located?
• HgS Hg2+ + S2-
Ksp = 1 x 10-53
= [Hg2+][S2- ] ([Hg2+] = [S2-])
• How much Hg2+ ion would be found in a pond where cinnabar was located?
• HgS Hg2+ + S2-
Ksp = 1 x 10-53
= [Hg2+][S2- ] ([Hg2+] = [S2-])
Solubility ProductSolubility Product
• How much Hg2+ ion would be found in a pond where cinnabar was located?
• HgS Hg2+ + S2-
Ksp = 1 x 10-53
= [Hg2+][S2- ] ([Hg2+] = [S2-])
[Hg2+] = 3.16 x 10-27
• How much Hg2+ ion would be found in a pond where cinnabar was located?
• HgS Hg2+ + S2-
Ksp = 1 x 10-53
= [Hg2+][S2- ] ([Hg2+] = [S2-])
[Hg2+] = 3.16 x 10-27
Solubility Product KspSolubility Product Ksp
Equilibrium StressesEquilibrium Stresses
• LeChatelier’s Principle:
If a change occurs to a system at equilibrium, the system will react to reduce the stress
ConcentrationPressure
Temperature
• LeChatelier’s Principle:
If a change occurs to a system at equilibrium, the system will react to reduce the stress
ConcentrationPressure
Temperature
LeChatelier’s PrincipleLeChatelier’s Principle
• Changes in Concentration– (adding or removing reactant or
product)
– CO + H2 CH3OH
– To increase CH3OH: Add CO or H2
– OR remove CH3OH
• Changes in Concentration– (adding or removing reactant or
product)
– CO + H2 CH3OH
– To increase CH3OH: Add CO or H2
– OR remove CH3OH
LeChatelier’s PrincipleLeChatelier’s Principle
• Changes in Pressure
• 2 NO2(g) N2O4(g)
• Changes in Pressure
• 2 NO2(g) N2O4(g)
LeChatelier’s PrincipleLeChatelier’s Principle
• Changes in Pressure
• 2 NO2(g) N2O4(g)
2 moles of gas 1 mole of gas
Increasing Pressure: Shifts equiibrium to N2O4 side
Decreasing Pressure: Shifts equilibrium to NO2 side
(Remember if volume decreases, pressure increases)
• Changes in Pressure
• 2 NO2(g) N2O4(g)
2 moles of gas 1 mole of gas
Increasing Pressure: Shifts equiibrium to N2O4 side
Decreasing Pressure: Shifts equilibrium to NO2 side
(Remember if volume decreases, pressure increases)
LeChatelier’s PrincipleLeChatelier’s Principle
• Changes in Temperature
Exothermic reactionsHeat is a product
Endothermic ReactionsHeat is a reactant
• Changes in Temperature
Exothermic reactionsHeat is a product
Endothermic ReactionsHeat is a reactant
LeChatelier’s PrincipleLeChatelier’s Principle
• Changes in TemperatureExothermic reactionC2H4 + H2 C2H6 + heat
Adding heat is like adding product so raising the temperature shifts the equilibrium toward C2H4 and H2
• Changes in TemperatureExothermic reactionC2H4 + H2 C2H6 + heat
Adding heat is like adding product so raising the temperature shifts the equilibrium toward C2H4 and H2
LeChatelier’s PrincipleLeChatelier’s Principle
• Endothermic Reactions
H2 + F2 + heat 2 HF
Raising temperature = adding heat = adding
reactant Equilibrium shifts to form more HF
• Endothermic Reactions
H2 + F2 + heat 2 HF
Raising temperature = adding heat = adding
reactant Equilibrium shifts to form more HF