Chapter 12 Solutions

Minimum energy (exothermic)Maximum entropy (disorder) dissolving increases entropy

Solids can dissolve if the energy is + but small in value because of increasing in entropySolids do not dissolve If the energy is + and large

4EnergySeparate solute and solvent particles endothermic (+ E) Solvation -attraction between solvent and solute particles exothermic (- E) Hydration (when water is solvent) Hydration energy (supplies energy to break bonds)2Will a solid dissolve?E to separate solute + E E to separate solvent + EEnergy of solvation (hydration) - EUsually if overall energy is negative the solute will dissolve Usually if overall energy is positive the solute will not dissolve

3Like dissolves LikePolar sub dissolve Polar subNon polar sub dissolve Non polar subPolar and non polar usually do not dissolve. Why?5Soaps and detergentsSoaps 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 tailThe hydrophilic end dissolves in waterSee page 450 and 451

6

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

7Solubility of a gas in liquid (temp)Gases will decrease in solubility as the temperature increases

8Solubility of a gas in liquid (pressure)Gases will increase in solubility as the pressure increases

9Concentrations 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

10 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

11 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

12Dilution 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) 13Molarity of ions in solutionWhat is the concentration of each ion in a .25 M sodium phosphate solution?14Stoichiometry and MolarityWhat 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

15Heating/cooling curve of pure substance

16Boiling/Freezing points of solutionsPure 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)

17What does the solute do to the V.P?Adding a solute lowers the vapor pressure of the liquid which raises the B.P

18Heating/cooling curve of pure substance

19Colligative propertyProperty that depends on the number of particlesEx. Boiling point elevation and freezing point depressionTb = Kb x moles of solute particles/kilograms of solventTf = Kf x moles of solute particles/kilograms of solvent

20Molecular vs ionic solutesC6H12O6(s) -------> C6H12O6 (aq)

NaCl(s) ------->Na+(aq) + Cl- (aq)

21If 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?Tb = Kb x moles of solute particles/kilograms of solvent

22Heating/cooling curve of pure substance

23Line A-B heating (or cooling) of solid- temperature change q = c m T K.E changing q= heat energy c= specific heat of solid m= mass of solid T= change in temperature

Line B-C melting ( or freezing) - heat of fusion- amount of energy added to melt (or removed to freeze) at the substances 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 changeLine C-D heating (or cooling) of liquid- temperature change q = c m T K.E changing q= heat energy c= specific heat of liquid m= mass of liquid T= change in temperature Line D-E boiling( or condensing) - heat of vaporization amount of energy added to boil(or removed to condense) at the substances 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 T K.E increasing q= heat energy c= specific heat of gas m=mass of liquid T= change in temperature

specific heat (c)

Solid = 2.13 J/goC

Liquid=4.18 J/goC Gas=2.01 J/goCheat 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 oCEx. 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 H2OSketch a heating /cooling graph and determine how many energy changes the substance will undergo