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Unit 2: Solubility and K sp Actually the most important unit. By David Lichko and Keith HArris. K sp. The equilibrium reaction quotient Constant for any reaction Can be calculated using concentrations of all reactants and products at equilibrium. Calculating K sp.
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Unit 2: Solubility and Ksp Actually the most important unit By David Lichko and Keith HArris
Ksp • The equilibrium reaction quotient • Constant for any reaction • Can be calculated using concentrations of all reactants and products at equilibrium
Calculating Ksp • 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 Ksp= [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 Ksp= [A]2[B]/[AB]
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
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
Solving an Ice Chart • 2 NH3(g) ‹—› N2(g) + 3 H2(g) Kc = 0.0076 @ 900 K • Find reaction quotient • If Q > Ksp the equilibrium shifts left • If Q < Ksp the equilibrium shifts right • Q=[.4][1]3/[.600]2=1.11 (Shifts left) • Solve for X
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
Henry’s Law • Sg=KHPgas • Henry’s Law relates gas solubility (M) to the partial pressure (atm) of the gas • KH is unique to each gas
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: • C6H12O6 gives 1 • NaCl gives 2 • CaCl2 gives 3 • (NH4)3PO4 gives 4
Boiling/Freezing Point Elevation/Depression • Freezing point depression: ΔTF=iKFCmolal ΔTF = change in freezing point (C⁰) i = van ‘t Hoff factor KF = a constant dependent on the solvent (K kg/mol) Cmolal = molal concentration • Analogous boiling point elevation: ΔTB=iKBCmolal
Osmotic Pressure • Π=CmolarRT • Osmotic pressure is the force from osmosis, the movement of solvent across a membrane
Vapor Pressure • Pvap=χP⁰vap • Vapor pressure is the pressure of the evaporating or subliming gas from a substance • It is related to the Clausius-Clapeyron equation