Principles of Solubility Nature of the solute and solvent
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Transcript of Principles of Solubility Nature of the solute and solvent
CH 10 Solutions
Principles of SolubilityNature of the solute and solvent
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Electrolytes will dissolve in a polar solvent (like water)
Non-electrolytes will dissolve more easily in non polar or slightly polar solvents (like benzene, or toluene)
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Non electrolytes (molecular substances) dissolving in water.
This ‘unusual’ case will involve Hydrogen bonding or otherwise polar molecules.
Ex: alcohols (-OH
group)Methanol,
Ethanol Hydrogen peroxide
H2O2
Solubilities of Alcohols in Water
Formula Name Solubility in Water (g/100 g)
CH3OH methanol infinitely soluble
CH3CH2OH ethanol infinitely soluble
CH3(CH2)2OH propanol infinitely soluble
CH3(CH2)3OH butanol 9
CH3(CH2)4OH pentanol 2.7
CH3(CH2)5OH hexanol 0.6
CH3(CH2)6OH heptanol 0.18
CH3(CH2)7OH octanol 0.054
CH3(CH2)9OH decanol insoluble in water
“greasy end” is larger proportion in large alcohols.
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DDT no more mosquitoes!
Pesticide used widely in the 60’s
Soluble in non polar solvents.
Concentrates in tissues of fish, birds, and other wildlife. (biological ½ life of 8 years)
Thins eggshells.CH 10 Solutions
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Principles of SolubilityEffect of Pressure/Henry’s Law
For gases only Solubility is directly proportional
to pressure.
Soda is bottled under a pressure of 4atm.
This drives more of the CO2 into solution,
when the cap is released, the pressure drops to 1 atm, and CO2
bubbles and rapidly escapes solution.
Cg =k Pg Concentration in solution = constant x Pressure of gas
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Principles of Solubility Effect of Temperature
Solid + Water Solution Usually + ΔH heat in
Increased T favors an endothermic process
So: Solubility of a solid usually increases with increased temperature
Gas + Water Soln. Usually –ΔH heat out
Increased T is detrimental for an exothermic process
• So: Solubility of a gas usually decreases with higher temperature.
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Concentration Unitsproblem type 1. Molarity (M)
Molarity = moles of solute/
liters of solution
Units are moles/liter
Tricks:
change mass to moles
change mL to Liters
Typical Problems Making a volume of
solution from the solid. Creation
Preparing a volume of a solution from another (stock) solution. Dilution
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ProblemsMolarity (M)
Prepare 35.0 mL of 0.200 M Aluminum nitrate from a solid sample.
1. How many moles of Al-nitrate would be in the solution?
2. Measure out that many grams and then add water to make 35.0 mL
Prepare 35.0 mL of 0.20 M Aluminum Nitrate solution from a 0.50M stock solution.
How many moles do you need? Remember
M1V1=M2V2
McVc=MdVd
Meausure out the volume of stock you need, then dilute
Molar mass (gfm) of aluminum nitrate is 213.03 g/n
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Problem type 2.
Mole Fraction (X) Similar to percent
composition…you are interested in part of the whole
Moles of substance/ total moles
Typical problems will have you determine the mole fraction
from % mass. determine the mole fraction
of solute in solution
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ProblemsMole Fraction (X)
If you dissolve 12.0 g of methanol in 100.0 g of water.
What is the mole fraction that is methanol?
1. Determine the moles of water
2. Determine the moles of methanol
3. Add together for the total moles, then take part/whole
Methane is CH4
Methanol is CH3OH
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Problem type 3. Mass % Solute
Simply determine the percent composition by mass.
Trick is to change the decimal to a percent
(x100 + “%”)
Solute is the solid dissolved in
solution
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Problem Type 4. Molality (m)
This is similar to molarity, but uses mass (kilograms) of the solvent, rather than liters of solvent.
Typical problems will have you determine the molality from other concentration units.
Molarity (M) = # moles solute/ Liter Solvent
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Problems : Molality (m)
What is the molality of methanol if 12g of methanol are dissolved into 100g of water?
1. Find Moles of solute (methanol)
2. Find kilograms of solvent
3. Divide
if the solvent is water, M and m are numerically equal
12g/32.0 g/n = 0.375 n
100g/ 1000g/kg = 0.100kg
= 3.75m
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Conversion between units-decide on a fixed amount of solution
When the original concentration [ ] is:
Start with:
Mass Percent 100 g solution
Molarity (M) 1.00L of solution
Molality (m) 1000g of solvent (1Kg)
Mole Fraction (X) 1 mol (solute+solvent)
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Part III: Colligative PropertiesVapor Pressure Lowering
VPL = X2Pº1
VPL= (vapor pressure lowering)
X2= mole fraction of the solute
Pº1= vapor pressure of the PURE solvent
Each liquid has a vapor pressure.
The pressure to escape the surface into the gas phase.
Dissolved particles get in the way and block the escape, thereby reducing the vapor pressure.
The higher the concentration (X2), the more the VP is lowered.
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Raoult’s Law
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Part III: Colligative Properties*Boiling Point Elevation ΔTb
*Freezing Point Depression ΔTf
ΔTb = kbm
ΔTf = kfmThe change in temperature
from the normal boiling/freezing points is a constant (k) times the molality(m).
The higher the concentration of ‘contaminant’, the greater the change in freezing/boiling points
Boiling Elevation:
The particles lower the vapor pressure.
Kb for water is 0.52 °C/m
Freezing Depression:
The particles obstruct the formation of the solid crystal. There must be a lower temperature to form solid structure
Kf for water is 1.86 °C/m
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Problems: Colligative Propertiesfreezing point depression
ΔTf = kfm Change in freezing
temp is (Kf) x (molality)
Find the freezing point of a solution containing 20.0 g of ethanol in 50.0g of water.
What is the molality of the solution?
Grams to moles Moles / Kg of
Solvent Molality x Kf
Kf for water is 1.86 °C/m
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Problems: Colligative Propertiesboiling point elevation
ΔTf = kbm Change in boiling temp is (Kb) x (molality)
Find the boiling point of a solution containing 20.0 g of ethylene glycol in 50.0g of water.
What is the molality of the solution?
Grams to moles Moles / Kg of
Solvent Molality x Kb
Kb for water is 0.52 °C/m
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Colligative PropertiesOsmotic Pressure (π)
π = M R T
M = molarityR= .0821 L * atm/n *K
T = Temperature
Water moves through a semi-permeable membrane from an area of high vapor pressure to low vapor pressure.
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Problems:Using freezing point depression to find molar mass
A solution is made using .0100g of a substance in 1.00g of water.
The freezing point depresses 1.00 °C
What is the molar mass?
Solve for molarity