Chapter 12 Solutions• Solute (smaller amount) • + Solvent (larger amount) • Solution
• Solute Solvent• S L (salt and water)• L L (alcohol and water)• G L (carbonated drink –CO2)• S S (any alloy – sterling silver- copper and silver )• G G ( air)• Water is universal solvent• (aq) dissolved in water
Energy
• Separate solute and solvent particles –endothermic (+ ΔE)
• Solvation -attraction between solvent and solute particles exothermic (- ΔE)
• Hydration (when water is solvent) • Hydration energy (supplies energy to break bonds)
Will a solid dissolve?
• E to separate solute + Δ E • E to separate solvent + ΔE• Energy of solvation (hydration) - ΔE
• Usually if overall Δ energy is negative the solute will dissolve
• Usually if overall Δ energy is positive the solute will not dissolve
• Minimum energy (exothermic)• Maximum entropy (disorder) dissolving increases entropy
• Solids can dissolve if the Δ energy is + but small in value because of increasing in entropy
• Solids do not dissolve If the Δ energy is + and large
Like dissolves Like
• Polar sub dissolve Polar sub• Non polar sub dissolve Non polar sub• Polar and non polar usually do not dissolve.
Why?
Soaps and detergents
• Soaps have a long nonpolar tail (hydrophopbic –does not like water)
• and a polar end (hydrophilic –does like water)
• Nonpolar substances like grease dissolve in the hydrophobic tail
• The hydrophilic end dissolves in water• See page 450 and 451
• Solubility- the maximum amount of solute that can be dissolved in agiven amount of solvent
Most solids will increase in solubility as the temperature increases
When the maximum amount of solute is dissolved the solution is SATURATED
Solubility of a gas in liquid (temp)
• Gases will decrease in solubility as the temperature increases
Solubility of a gas in liquid (pressure)
• Gases will increase in solubility as the pressure increases
Concentrations of Solutions
• Molarity (M) = moles of solute/liter of solution• % by mass = (grams of solute/grams of
solution)x100 %• % by volume =( volume of solute/volume of
solution) x100 %• molality (m) = moles solute/kg of solvent
• Explain how to prepare 400. ml of a .150 M solution of
NaCl. (from solid NaCl)• Moles?
• Grams?
• To make 400 mL of a .150 M solution of NaCl add ___________ grams of NaCl to a flask and add enough water until the total volume of solution is __________ mL
• Explain how to prepare 400. ml of a .150 M solution of
NaCl. (from a stock solution of. 500M) dilution NaCl)• Moles?• Volume?
• To make 400 mL of a .150 M solution of NaCl from a .500 M solution of NaCl measure _____________mL of
the stock solution and add enough water until the total volume of solution is __________ mL
Dilution
M st x V st = M dil x V of dil• Moles( in stock) = Moles (in dilute)• Explain how to prepare 400. ml of a .150 M solution of NaCl.
(from a stock solution of. 500M) dilution NaCl)
•
Molarity of ions in solution
• What is the concentration of each ion in a .25 M sodium phosphate solution?
Stoichiometry and Molarity
• What volume of .80M Al(NO3)3 is needed to completely react 2.0 grams of calcium?
• Balanced equation
• Find moles of Ca
• Find moles of Al(NO3)3
• Convert to volume
Heating/cooling curve of pure substance
Boiling/Freezing points of solutions
• Pure substances –definite B.P. and F.P solutions do not (B.P is higher, F.P is lower)
• What happens when a solute is added to a solvent? (salt to water)
What does the solute do to the V.P?
• Adding a solute lowers the vapor pressure of the liquid which raises the B.P
Heating/cooling curve of pure substance
Colligative property
• Property that depends on the number of particles
• Ex. Boiling point elevation and freezing point depression
• ΔTb = Kb x moles of solute particles/kilograms of solvent
• ΔTf = Kf x moles of solute particles/kilograms of solvent
Molecular vs ionic solutes
• C6H12O6(s) -------> C6H12O6 (aq)
• NaCl(s) ------->Na+(aq) + Cl- (aq)
• If 5.85 g NaCl are added to 100 g of water what temperature will the water begin boiling?
• How many moles of NaCl?• How many moles of particles?• ΔTb = Kb x moles of solute particles/kilograms
of solvent
Heating/cooling curve of pure substance
Line A-B heating (or cooling) of solid- temperature change q = c m ΔT K.E changing q= heat energy c= specific heat of solid m= mass of solid ΔT= change in temperature
Line B-C melting ( or freezing) - heat of fusion- amount of energy added to melt (or removed to freeze) at the substance’s M.P. (F.P) energy may be given per gram or mole- ex. kJ/mole or kJ/g P.E change No K.E change -no temperature change
Line C-D heating (or cooling) of liquid- temperature change q = c m ΔT K.E changing q= heat energy c= specific heat of liquid m= mass of liquid ΔT= change in temperature
Line D-E boiling( or condensing) - heat of vaporization –amount of energy added to boil(or removed to condense) at the substance’s B.P. energy may be given per gram or mole- ex. kJ/mole or kJ/g P.E change No K.E change-no temperature change
Line E-F heating ( or cooling) of gas- temperature change q = c m ΔT K.E increasing q= heat energy c= specific heat of gas m=mass of liquid ΔT= change in temperature
specific heat (c)
Solid = 2.13 J/goC
Liquid=4.18 J/goC Gas=2.01 J/goC
heat of fusion = 334 J/gram or 6.01 kJ/mol
heat of vaporization =2228 J/gram 40.1 kJ/mole
M.P. (F.P.) = 0 oCB.P. = 100 oC
Ex. If you have 100.0 grams of water at -5.0 C, how much energy will it take to change it to 25o C?
For H2O
Sketch a heating /cooling graph and determine how many energy changes the substance will undergo
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