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solutions.notebook March 06, 2013

Solutions Review Voc. and the following concepts for quiz.

crystall lattice of hydration hydration energy miscibility immiscible saturated unsaturated supersaturated

Dissolution depends on 2 factors. Name them.

Ionic solids dissolve by endo or exo process?

Which way does equilibrium shift when you raise the temperature of an endothermic reaction?

What is thermal pollution? What effect does it have on the environment?

The magnitude of the crystal lattice energy increases, with an increase or decrease charge and increase or decrease size of atoms.

Describe how you dilute an acid? Why do you do it this way?

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The colligative properties of a solution depend on the relative numbers () of solute and particles, they do not depend on the nature of the particles. Colligative properties change in proportion to the concentration of the solute particles. We distinguish between four colligative properties: • vapor pressure lowering • freezing point depression • boiling point elevation • osmotic pressure.

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Enthalpy of Solution ΔHsoln = ΔH1 + ΔH2 + ΔH3

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Factors Affecting Structure Effects Hydrophobic (water­fearing) non­polar molecules are not likely to be watersoluble Hydrophilic (water­loving) – molecules with many H­bonding sites

Pressure Effects Henry’s Law ­ gas solubility is directly proportional to the pressure of the gas over the solution. C = kP If the gas reacts with the solution (i.e., NH3, CO2, O2 in blood) c will be much higher

Temperature Effects Saturated Solution ­ maximum amount of substance is dissolved at a given T Unsaturated Solution

Supersaturated solution ­ unstable solubility vs T ­ solubility doesn’t change as ΔHsoln

Fractional Crystallization ­ separation o f a into pure components on basis of varying solubility. Works best if substance has steep sol vs. T curve or if amt of impurity is small

Gas solubility and T ­ solubility always decreases with increasing T Thermal pollution ­ causes deoxygenization of water

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Hydrophobic (water­fearing) non­polar molecules are not likely to be watersoluble Hydrophilic (water­loving) – molecules with many H­bonding sites

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Compounds With That Decrease with Increasing Temperature

CaSO4 Ca(OH)2 Na2SO4∙10H2O Ca(C2H3O2)2 Ce2(SO4)3 Li2SO4

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http://www.youtube.com/watch?v=8yU5y­cFXoo

***Watch the units some chemist use the equation S=KP (solubility= Henrys constant x pressure) Henrys constant will have the units of M/atm for this equation

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P=KC P=pressure of the gas (atm) K=Henry's Law Constant (atm/M) C=Concentration of the Gas (M)

Examples­The Henry's law constant for He gas in water at 30°C is 2.70 x 103 atm/M.

3 The constant for N2 at the same temperature is 1.67 x 10 atm/M. If the two gases are each present at 1.43 atm pressure, calculate the solubility, in M, of each gas. C=P/K

He

1.43 atm / 2.70x103 atm/M = 5.30 x10­4 Molar

N2 1.43 atm/ 1.67 x 103 atm/M = 8.56 x10­4 Molar

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Vapor­pressure lowering ­ The equilibrium vapor pressure of a liquid is achieved when the rates of evaporation and condensation are equal. When particles of solute are added, the rate of evaporation of solvent is diminished, simply because the number of molecules close to the surface is reduced.

EXAMPLE: Determine the vapor pressure of a solution of 25.00 g sugar

(C12 H22 O11 ) dissolved in 100.00 g water. The vapor pressure of pure water at room temperature is 23.76 mm Hg.

SOLUTION: Sugar is a nonvolatile non­, so the vapor pressure of the solution is the partial pressure of water by Raoult's Law. Determine the of water:

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Negative deviation occurs when unlike molecules are more strongly attracted to each other than are like molecules, thus, it is harder for molecules to escape the liquid phase. ΔHsoln < 0 Example, nitric acid and water.

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Positive deviation occurs when unlike molecules are less strongly attracted to each other than are like molecules, thus, it is easier for molecules to escape the liquid phase. ΔHsoln > 0 Example, ethanol and water.

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Freezing­point depression is more sensitive to the number of moles of solute than the boiling­point elevation because for the same solvent Kf is larger then Kb. The most sensitive method for the determination of the molar mass of a solute is to measure the osmotic pressure. The equation for osmotic pressure then becomes useful in the following form:

These methods are particularly useful for determining the molar masses of large molecules such as proteins and polymers.

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