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Unit 2 Liquids, solids, solubility, and equilibrium.

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Presentation on theme: "Unit 2 Liquids, solids, solubility, and equilibrium."— Presentation transcript:

1 Unit 2 Liquids, solids, solubility, and equilibrium

2 Solubility Rules Given reactants, can anticipate what products will form By looking at solubility rules, can predict whether or not a precipitate will form

3 Sample Problem KOH (aq) + AgNO 3 (aq)  KNO 3 (aq) + AgOH (s) AgOH precipitates because hydroxides are generally insoluble

4 Electrolytic Properties Electrolyte=solution containing ions –The more a substance dissociates, the stronger the electrolyte Non-electrolyte=solution containing no ions

5 Solids Retain shape & volume Virtually incompressible Diffusion occurs very slowly Strong intermolecular forces Particles close together

6 Liquids Assume shape of container Definite volume Does not expand to fill container Virtually incompressible Flows readily Diffusion occurs slowly Particles fairly close together

7 Gases Assume both volume & shape of container Compressible Flow easily Diffusion occurs rapidly Widely separated molecules Disorder Particles free to move

8 Phase Diagram

9 Phase Changes

10 Heating Curve (vaporization)

11 Intermolecular Forces of Attraction Ion-dipole: exist between an ion and the partial charge on end of a polar molecule Dipole-dipole: exist between neutral polar molecules, attraction between unlike charges on ends of molecules, significant only when molecules very close together London dispersion: exist only when molecules are very close together, nonpolar atoms/molecules experience temporary dipoles, momentary dipoles attract Hydrogen bonding: exist between a hydrogen atom in a polar bond and an electronegative element, (H-F, H-O, H-N)

12 Bonding in Solids Molecular solids: atoms/molecules held together by intermolecular forces (London dispersion, dipole-dipole, hydrogen bonds), soft, relatively low boiling points, poor thermal and electrical conduction, i.e. methane, sucrose, and dry ice Covalent-network solids: atoms held together by covalent bonds, very hard, high melting points, poor thermal and electrical conductors, i.e. diamonds Ionic solids: ions held together by ionic bonds, hard and brittle, high melting points, poor thermal and electrical conduction, i.e. salts Metallic solids: metal atoms held together by metallic bonds, vary in strength of bonding, wide range of physical properties (hardness, melting points), malleable and ductile, excellent thermal and electrical conductors, i.e. copper, iron, aluminum

13 Crystalline Structure Simple cubic: 1 atom, V=8r 3, e=2r Body-centered: 2 atoms, V=(4r/√3) 3, e=4r/√3 Face-centered: 4 atoms, V= (32r 3 /√2), e=4r/√2

14 Equilibrium aA + bB ↔ dD + eE K c =[D] d [E] e K p =[P D ] d [P E ] e [A] a [B] b [P A ] a [P B ] b Δn K P =K C (RT)

15 Le Châtlier’s Principle Any disturbance to a system in equilibrium will produce a shift in equilibrium that offsets the disturbance as much as possible –Add or subtract product/reactant, change temperature, change pressure, etc. Reaction Quotient: Q is the number obtained by substituting pressures or concentrations into an equilibrium-constant expression (K c or K p ) Q C = K C system at equilibrium Q C < K C needs to shift right to reach equilibrium Q C > K C needs to shift left to reach equilibrium

16 Sample Problem 4Liza + 2Mr. Hinton ↔ 7Juan Increase amount of Liza… Decrease amount of Mr. Hinton… Increase pressure… Increase volume…

17 Solutions K sp is the product of the concentration of the ions involved in equilibrium, each raised to the power of its coefficient in the equilibrium equation A (s) ↔ bB (aq) + dD(aq) K sp =[B] b [D] d Given the value of K sp, you can find the concentrations of ions in solution Remember that solids are not included in the K sp equation

18 Sample Problem PbSO 4 (s) ↔ Pb 2+ (aq) + SO 4 2- (aq) Given: K sp = 1.8 x 10 -8 K sp = [Pb 2+ ][SO 4 2- ] = x 2  x = √(1.8 x 10 -8 ) X = 1.3 x 10 -4 = [Pb 2+ ] = [SO 4 2- ]

19 Molarity, Molality, Mass Fraction Mole fraction= (moles of component)/(total moles of all components) Molarity= (moles solute)/(liters solution) Molality=(moles solute)/(kg of solvent) Mass fraction= (mass solute)/(mass solution)

20 Raoult’s Law P vapor =XP° vapor –X is the mole fraction of a solvent in solution –P° vapor is the vapor pressure of the pure solvent –P vapor is the partial pressure of a solvent over a solution

21 Henry’s Law S g =kP g S g is the solubility of the gas in the solution phase P g is the partial pressure of the gas over the solution k is a proportionality constant

22 Boiling Point Elevation and Freezing Point Depression Boiling Point Elevation: ΔT b =iK b m –i is the number of particles –K b is a constant –m is molality Freezing Point Depression: ΔT f =iK f m –i is the number of particles –K f is a constant –m is molality


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