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CHAPTER 12: SOLUTIONS

Part One: Introduction A. Terminology.

1. Solution - 2. Consists of a solvent and one or more solutes. 3. Solute = is a substance being dissolved, and usually present in small amounts. 4. Solvent = Solvent is usually the most abundant species in the mixture. 5. Solubility = the amount of a given solute that dissolves in a given quantity of solvent.

NaCl in water, e.g., solubility of NaCl in water = 3.6 g/mL. 6. Two fluids that mix in all proportions to form a solution are said to be .

(alcohol and water). Oil and water would be said to be immiscible.

Figure 12.1

B. Importance.

1. Extremely important in all life – 2. A solid, liquid, or gas can act as either solvent or solute. Usually the solvent is a

liquid. 3. Sea water is an aqueous solution of many salts & some gases. 4. Tap water is an aqueous with trace amounts of various ions, molecular

compounds, and gases. (Distilled water is pure).

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5. Air is a solution of gases. 6. Solids solutions are called . One example is steel, which is mostly Fe with

small amounts of additives like carbon, chromium, titanium, etc. Another example is dental fillings are solid solutions called by the special name amalgams = .

Part Two: The Solution Process

A. Solubility and Saturated Solutions 1. Solution formation is a process of . Example, dissolving table salt in

water: NaCl(s) Na+(aq) + Cl-(aq)

Figure 12.2

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Ions are continually leaving the solid crystal into the solution and returning to the crystal, until eventually:

a) all the solid is dissolved, OR b) the solution reaches its saturation point, at which some solid remains.

Which case pertains depends on the of NaCl in water.

Figure 12.3

B. Factors that influence solubility.

1. General rule = e.g., oil dissolves in gasoline because both are hydrocarbons, which are non-polar

molecules water, however, is a polar molecule, such that oil does not dissolve in it, but alcohols

do (which are also polar).

Therefore, intermolecular forces are involved. 2. There is some natural tendency of all substances to mix, based on tendency toward a

state of increased disorder (ENTROPY CHANGE) 3. This is balanced by the tendency of a system to have the lowest energy possible

(ENERGY CHANGE). In other words, is heat absorbed or released in forming a solution? Release of energy (exothermic) favors solution formation.

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The Entropy effect and the Energy effect combine to determine the solubility. A process is favored by: a. b.

4. Energy change is called the heat of solution, ΔHsolution. Depends on how strongly solute

and solvent particles interact.

5. Relative strengths of these interactions determine the extent of solubility (refer to

figure below) a. Strong solvent-solute attractions favor solubility. b. Weak solvent-solvent attractions favor solubility. c. Weak solute-solute attractions favor solubility.

6. Solution formation is always accompanied by an increase in the disorder of both

solvent and solute, always favorable to solubility. C. Dissolving Solids in Liquids, viewed as a stepwise process (Section 12.2)

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1. Take an ionic solid. Must overcome crystal lattice energy (solute-solute interactions):

2. Must overcome solvent-solvent interactions. If solvent is water, must break Hydrogen bonds:

3. Must “solvate” the solute particles:

Figure 12.8

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4. Combined Steps 2 and 3 is called “solvation,” or in case of water, “hydration.” 5. Overall energy involved: ΔHsolution =

Figure 12.10

6. Energy is not the only factor. Entropy increase upon solution is a favorable factor

which can sometimes overcome unfavorable energy factor if latter is not too large. 7. Example: NaCl(s) + H2O(aq) → Na+(aq) + Cl-(aq)

ΔHsolution is (+), endothermic (unfavorable). Still occurs due to slightly larger favorable entropy consideration.

8. Ionic compounds which have very large crystal lattice energies will be insoluble in

H2O because solvation and entropy can’t overcome energy holding lattice together.

9. Some ionic solids dissolve with release of heat (exothermic) - both energy and entropy

favor solution process.

CaCl2, Na2SO4, ...

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10. Nonpolar solids (e.g., paraffin) do not dissolve in H2O because interactions with water (hydration) is very weak compared to water-water interactions.

Weak interaction Strong Hydrogen Weak interaction btwn. in nonpolar solid bonding solvent- solvent & nonpolar (should be easy to solvent interactions. solute. Nothing gained break up). (difficult to break up) here except entropy of HOWEVER... mixing.

D. Dissolving Liquids in Liquids – Molecular solutions (Miscibility). (Section 12.2)

1. Same factors involved. Here, though, a solid does not need to be broken up, so no

lattice energy to overcome. 2. Case 1: Polar or Hydrogen-bonding solute with polar or Hydrogen bonding solvent.

Methanol dissolves in water.

H-bonding. Moderately H-bonding. Moderately H-bonding. Nothing strong solute-solute strong solvent-solvent lost, entropy of interactions. interactions. mixing favors.

3. Case 2: Nonpolar solute with polar or Hydrogen-bonding solvent.

CCl4 does not dissolve in H2O.

Weak solute-solute H-bonding. Moderately Weak solute-solvent interactions. (only strong solvent-solvent. interactions. (loss of London forces) solvent-solvent int.)

(hydrophobic effect)

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4. Case 3: Nonpolar solute and nonpolar solvent.

CCl4 does dissolve in benzene.

Weak solute-solute Weak solvent-solvent Weak solute-solvent interactions. (only interactions. (only (London). Entropy London forces) London forces) of mixing favors this.

5. Generalization -

E. Gases in Liquids. (Section 14.4)

1. Same factors as before. 2. Polar gases dissolve in polar liquids, nonpolar gases dissolve in nonpolar liquids. 3. CO2 and O2 are nonpolar, so only slightly dissolve in H2O. 4. CO2 more so because of ionization: 5. HCl and NH3 are polar covalent gases, dissolve strongly in H2O.

Part Three: Effects of Environmental Variables on Solubility (Section 12.3) A. Temperature Effect. (Section 12.3)

1. Explained by LeChatelier’s Principle = 2. When heat of solution is exothermic:

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3. When heat of solution is endothermic:

4. NaCl solubility↑ as T↑; Na2SO4 solubility↓ as T↑.

Figure 12.12

5. O2 dissolves in H2O exothermically: T↑ O2 solubility↓ . Basis of “thermal pollution.”

B. Pressure Effect.

1. Henry’s Law = 2. Example: Carbonated beverage stored with pressurized CO2. 3. Henry’s Law:

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Part Four: Expressions of Concentration (Section 12.4) A. Molarity.

1. Molarity =

B. Mass Percentage of solute.

1. Mass % solute =

2. Example problem: A certain solution of an alcohol and water is 2.0 M alcohol and has a density of 0.95 g/mL. The alcohol has a molar mass of 126 g/mol. What is the mass percent alcohol in this solution?

C. Molality.

1. m = 2. Example: What is the molality of a solution in which 1.0 gram glucose (C6H12O6) has

been added to 100 mL of H2O? 3. Note that:

Why? Because under these conditions:

D. Mole Fraction.

1. XA =

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2. Unitless. 3. Example: What is Xglucose and Xwater in previous problem?

Part Five: Colligative Properties of Solution (Sections 12.5 through 12.8)

A. Definition.

1. Colligative property =

2. Examples: a. b. c. d.

3. Depends on: B. Vapor Pressure Lowering - Raoult’s Law. (Section 12.5)

1. A solution containing a nonvolatile solute has a vapor pressure than the pure solvent.

(“nonvolatile solute” means solute has no vapor pressure of its own)

2. Physical Picture:

3. V.P. lowering depends on fraction of surface blocked by solute particles, and thus, mole fraction of solute and solvent, hence:

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4. Raoult’s Law:

5. Also could write:

6. Solutions obeying this perfectly are called ideal solutions. 7. Slight deviations occur when: -- attractions differ from -- attractions and -- attractions.

C. Boiling Point Elevation. (Section 12.6)

1. Closely related to vapor pressure lowering, because boiling occurs when

2. 3. 4. Each solvent has its own characteristic Kb, independent of what the solute is. See

Table 12.3. Kb = 0.512°C/molal for H2O 5. Problem: What is the normal boiling point of a glucose solution in which 270 g

glucose are added to 1.0 L of H2O? D. Freezing Point Depression. (Section 12.6)

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1. 2. Kf = 1.86°C/molal for H2O. 3. What is freezing point of the preceding glucose solution? 4. This is how antifreeze works, as well as salt on icy roads.

E. Determination of Molar Mass by Colligative Properties. (Section 12.6) 1. Add known mass of solute to known mass of solvent. Measure f.p. lowering ΔTf.

Calculate Molar Mass (M).

2. Problem: When 0.154 g of sulfur is finely ground and melted with 4.38 g of camphor, the

freezing point of the camphor is lowered by 5.47°C. What is the molecular weight of sulfur? What is its molecular formula? Given Kf = 40°C/m

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F. Colligative Properties of Ionic Solutions. (Section 12.8) 1. When solute is an electrolyte (breaks into ions in solution), colligative properties are

because of more particles present. Collig. Prop. depend on total concentration of ions produced.

2. Deviations from ideal behavior can be large since ions interact with each other in

solution and thus are not independent.

e.g. You don’t get quite a doubling of b.p. elevation when you dissolve 1 mole of NaCl vs. 1 mole of glucose because the Na+ and Cl- ions interact.

3. Effect of ionization is expressed by the van’t Hoff factor i:

€

i =ΔTactual

ΔTnon−elec

i = 2 if NaCl had no ion-ion interactions.

i = 1.83 in 1.0 m NaCl

ΔTb = i Kbm modified F.P. equation for ionic solutes

G. Osmotic pressure. (Section 12.7)

1. Osmosis =

2. Semipermeable membrane = allows solvent to flow but not solutes. (e.g., cell

membranes are semipermeable) 3. osmotic pressure π

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4. Working equation:

Part Six: Colloids (Sections 12.9) A. Definition.

1. Remember our classification of mixtures as: a. homogeneous b. heterogeneous

2. Colloids are somewhere between these two extremes. 3. Collodial dispersion = dispersed particles are small enough to remain suspended in the

solvent, so no settling takes place, but yet are large enough to make the sample cloudy.

4. Cloudiness is due to = scattering of light by suspended particles when

they become large enough that they are comparable to the wavelength of the light. B. Types of colloids - study Table 12.4.

Soap molecules solubilize oils in water by forming Colloids with the oil.

oil-soluble end (Hydrophobic) water-soluble end (Hydrophilic)

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Figure 12.30 - a micelle particle

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Figure 12.31 – how soap cleans fabrics

Figure 12.34 – cell membranes

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NOTES: