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Unit 2: Solubility and K sp Actually the most important unit
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K sp The equilibrium reaction quotient Constant for any reaction Can be calculated using concentrations of all reactants and products at equilibrium
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Calculating K sp The quotient of the product of the concentrations of products in a reaction divided by the product of the concentrations of the reactants at equilibrium AB A + B K sp = [A][B]/[AB] Rules Any solids are not counted in the calculation and are given a value of 1 In the calculation, the concentration of the reactants and products is raised to their respective coefficient in the balanced reaction equation AB 2A + B K sp = [A] 2 [B]/[AB]
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Solubility Rules
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LeChâtlier Principle When a stress (pressure, temperature, concentration, is put on a system, the system shifts to reach an equilibrium Effects of Changes If concentration of product is put into a system, more reactant is produced If concentration of reactant is put into a system, more product is produced If pressure is increased, the equilibrium shifts to the side with the fewest moles of gas particles If pressure is decreased, the equilibrium shifts to the side with the most moles of gas particles ExothermicEndothermic Increase in TemperatureShift towards productsShifts towards reactants Decrease in TemperatureShifts towards reactantsShifts towards products
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ICE Charts I Initial concentration C Change in concentration E Equilibrium concentration Can be used to find equilibrium concentrations from a given set of concentrations
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Solving an Ice Chart 2 NH 3 (g) ‹—› N 2 (g) + 3 H 2 (g) K c = 0.0076 @ 900 K Find reaction quotient If Q > K sp the equilibrium shifts left If Q < K sp the equilibrium shifts right Q=[.4][1] 3 /[.600] 2 =1.11 (Shifts left) Solve for X NH 3 N2N2 H2H2 Initial0.6000.4001.00 Change+2x-x-3x Equilibrium1.032.184.352
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Saturation Saturation curves relate maximum concentration of a solute to the temperature On the line is saturated, under the line is unsaturated, over the line is supersaturated To supersaturate, heat the solution and saturate it, then let cool
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Henry’s Law S g =K H P gas Henry’s Law relates gas solubility (M) to the partial pressure (atm) of the gas K H is unique to each gas
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Molality and the van ‘t Hoff Factor Molality is (mol of solute)/(kg of solvent)=m Easily calculated in the lab The van ‘t Hoff factor is the (concentration of particles when dissolved)/(concentration based on mass), roughly how many particles it will split into when dissolved Used in colligative properties, properties dependent on concentrations and not the identity of solutes Examples: C 6 H 12 O 6 gives 1 NaCl gives 2 CaCl 2 gives 3 (NH 4 ) 3 PO 4 gives 4
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Boiling/Freezing Point Elevation/Depression Freezing point depression: ΔT F =iK F C molal ΔT F = change in freezing point (C⁰) i = van ‘t Hoff factor K F = a constant dependent on the solvent (K kg/mol) C molal = molal concentration Analogous boiling point elevation: ΔT B =iK B C molal
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Osmotic Pressure Π=C molar RT Osmotic pressure is the force from osmosis, the movement of solvent across a membrane
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Vapor Pressure P vap =χP⁰ vap Vapor pressure is the pressure of the evaporating or subliming gas from a substance It is related to the Clausius-Clapeyron equation
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Crystal Structure NameCellParticles/CellRadius/Side Simple Cubic8*1/8=11/2 Face-Centered Cubic 6*1/2+8*1/8=4(√2)/4 Body-Centered Subic 1*1+8*1/8=2(√3)/4
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