 # The Solution Process Chapter 13 Brown-LeMay. I. Solution Forces Solution = Solvent + Solute Attractions exist between A. solvent and solute B. solute.

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The Solution Process Chapter 13 Brown-LeMay

I. Solution Forces Solution = Solvent + Solute Attractions exist between A. solvent and solute B. solute and solute C. solvent an solvent If solute by solute attraction is greater that solute by solvent (water) then a precipitate forms.

I. Solution Forces Lattice Energy – describes the attractive forces between solute molecules or ions Solvation – solvent – solute attraction is strong enough so the solute dissolves. Hydration – when water is the solvent.

II. Enthalpy of Solvation H (sol,total)= H 1 + H 2 + H 3 H 1 = Energy of dispersing solute particles (+) endothermic H 2 = Energy of dispersing solvents particles (+) endothermic H 3 = Energy of solvated solute molecules (-) exothermic

II. Enthalpy of Solvation If the energy released from solvation is greater than that required for dispersion (delta 1 + delta 2) then the process is exothermic If delta H of solvation is negative or exothermic it is energetically favorable or spontaneous

Although some delta H solvations are endothermic they may proceed spontaneously 1) the increase in disorder is the driving force that can overcome slightly positive enthalpy changes 2) remember disorder is measured in degrees of entropy delta S a) high entropy = disorder b) low entropy = ordered c) systems tend to be disordered move from low to high

III. Ways to Express Conc. A. Dilute vs Concentration 1) Dilute – weak– few solute particles vs solvent particles 2) Concentrated– strong- greater solute vs solvent particles B. Mass % - mass of solute X 100 mass of solution Ex: A 100g NaCl solution evaporated to dryness weighed 5g. What was the mass Percent of the solution. 5g/100g X 100 = 5%

III. Ways to Express Conc. C. Parts Per Million ppm = mass g of solute x 10 6 mass g of solution 1) a mass % of 1 would equal 10,000 ppm 1g/100g x 10 6 2) 1 ppm would be equal to 0.0001% 1% = xg x 10 6 100g 1 = xg  100 = 10 6 x = 0.0001% 10 6 100g

A 100ml sample of water is evaporated to dryness and as 75ug Pb 2+. of How many parts per million of lead II does it have? (d water is 100g per milliliter, 1ug=1x10 6 g) 75ug (1g/1x10 6 ug) = 7.5 x 10 -5 g Pb 2+ 100g of water x 1 x 10 6 = 7.5 x 10 -1 ppm 7.5 x 10 -5 g Pb 2+

D. Mole Fraction X = moles of solute moles of solution X Pb 2+ = sum of moles fractions in a solution (including the solution itself) must equal one. E. Molarity = Moles of solute (M) liters of solution F. Molality = Moles of solute (m) kg of solvent

Ex. 1 a certain beverage contains 7% ethanol C 2 H 5 OH by mass. Calculate the mole fraction, molarity, and molality. (mm ethanol = 46.1 g/mol) 7g ethanol 100g(C 2 H 5 OH+ H 2 O) (93g H2O+7g C 2 H 5 OH) X eth = 0.152 mol eth = 0.0286 (0.152 mol eth + 5.17 mol H 2 0)

Molarity = moles of solute (M) liters of solution 7g ethanol 100g solution(1g/ml) assume ethanol has no effect on density 7g(1mole/46.1g) 0.152 moles = 1.52 M 100ml = 0.1 liter 0.1 liter Molality= moles of solute 0.152 Kg of solvent 93g(1kg/1000g) = 1.63 m

Saturated solutions and solubility A. Two opposing forces 1. dissolving – hydration of individual ions 2. collision of ions – ions unite and increase crystal mass (crystallization) Solute + solvent dissolve  solution crystallize  solution

IV saturated solutions and solubility If dissolution>crystallization then the crystals in the solvent get smaller If crystallization>dissolution crystals in the solvent get larger If crystallization = dissolution the system is in dynamic equilibrium (saturated) no more solute will be dissolved

IV saturated solutions and solubility The concentration of solute present at saturated is known as the solubility of the solute At higher temperatures usually more solute can be dissolved and solubility is higher

b. Supersaturated Solutions Solute is dissolved at high temperature to saturation Solution is carefully cooled to a lower temperature At the lower temperature the concentration of solute is higher than at the equilibrium concentration at that temperature The introduction of a “seed” crystal will stimulate rapid crystal formation

V. Factors Affecting Solubility A. Gases in Water 1. solute-solvent interactions gases have weak I.F. primarily London dispersion L.D. forces L.D. forces increase with increased molecular weight – solubility of a gas typically decreases with increasing mass of a gas molecule

Gases in Water If a gas molecule appears to be more soluble than its mass would indicate a chemical reaction may have taken place B. Polar solutes in polar solvents 1. interactions between polar solutes are typically dipole-dipole (or hydrogen bonding) 2. interactions between molecules of polar solvents are the same as the solutes

B. Polar solutes in polar solvents 3. thus the energy associated with disrupting solute-solute and solvent-solvent interactions are aprox. Equal 4. entropic forces drive the dissolution process 5. polar liquids tend to dissolve in polar solvents Note: liquids that mix are miscible- don’t mix are immiscible

c. Alcohol’s Short chained alcohol’s are miscible in water - polar - D.D. and hydrogen bonding Long chained alcohol’s tend not to be miscible The hydrocarbon chain is held together by C-H bonds that are primarily L.D. forces. Single OH groups at the end are not enough to make the liquids miscible.

c. Alcohol’s In general “like dissolves like” substances with the same or similar intermolecular forces tend to be soluble in one another D. Effect of pressure on gases and solubility – increasing the pressure at constant temp. results in more collisions of gas molecules, per unit time with the surface of the solvent resulting in greater solubility

E. Temperature effects Solvated gas molecules with enough K.E. can escape from the surface of a liquid K.E. increases with temperature Increased temperature reduces the solubility of gas molecules in a solvent

F. Temp. effects on Solid solutes and solubility Insolubility – inability of solvent to overcome the solute – solute attraction Increase temp increases K.E. of solvent and solute molecules Both separate more readily and the effect of the solvent is increased Increasing temperature increases the solubility of solid solutes

Colligative Properties Vapor Pressure lowering Depend upon the collective number of particles in a solution The physical bases for the behavior is the effect of the solute upon the vapor pressure of the water Vapor pressure - The pressure of the gas that collects above a liquid in a closed container

Colligative Properties Vapor Pressure lowering  Raoult's Law - The vapor pressure of the solvent above a solution is equal to the product of the mole fraction of the solvent and the vapor pressure of the pure solvent: P a = vp solution (cont. solute) P a o = pure solvent –

Colligative Properties Vapor Pressure lowering Ideal Solutions – solutions that obey Raoult’s law – generally for dilute solutions where the mole fraction is closer to one, also where I.F. are similar among solute and solvent Deviations – hydrogen bonding or when solute solvent interactions are extremely strong

Vapor pressure lowering expression Given Raoult’s law holds true P =P a o Xb(mole fraction solute)

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