Chapter 11 Review (sections 1-8)

  1. Solutions
    1. Homogeneous mixtures ARE solutions.  You can't see the different parts (salt water looks just like water), they are usually transparent (clear) and they may or may not have color.  Remember heterogeneous mixtures are those where you can see the different ingredients like granite.
    2. A solution is a solute dissolved in a solvent.  Refer to Table 11.1  
      Solute Solvent Solution (example)
      gas gas air
      gas liquid soda pop, HCl(aq)
      liquid liquid martini
      solid liquid saline, Koolaid
      solid solid brass, steel, white gold
  2. Energy of Solutions
    1. IMF (intermolecular forces) are real important in solutions (except gas-gas solutions which have very, very little IMF)
    2. There are three main IMF involved for a solid to dissolve in water:
      1. IMF between solute molecules (if ionic solids - then ion-ion forces which are ionic bonds, if covalent solids - then H bonding forces or dipole dipole forces likely) These must be broken as the solute must be pulled apart to mix with the solvent. So heat is needed to break them apart and DH is positive = endothermic. Breaking an ionic solid apart is the same of overcoming the lattice energy LE. 
      2. IMF between solvent molecules (if water then H bonding forces) These must be broken also as we need room between solvent molecules to insert the solute. Breaking these apart requires heat so DH is positive = endothermic.
      3. IMF between solute and solvent are formed once the solution mixes (for ionic solid dissolved in water - ion-dipole forces are formed) Forming IMF gives off heat so DH is negative = exothermic
      4. The total heat of solution (DHsoln) is the sum of the three above. DHsoln= DHsolute + DHsolvent + DHsolute-solvent.  Because the third step is exothermic while the others are endothermic the total DHsoln may add up to be a positive or negative number. See Figure 11.4.  Try examples 11.1 and problems 11.1 and 11.2.
    3. Energy changes when dissolving salts in water.
      1. Some solids dissolving in water are exothermic (heat released or given off so DH is negative or <0) so they are used in heat packs.  Magnesium sulfate is one example.
      2. Some solids dissolving in water are endothermic (heat required or taken in so DH is positive or >0) so they are used in cold packs.  Ammonium nitrate is one example.
      3. Here you can dissolve many different salts in water and see if they are exothermic or endothermic.
    4. Remember "like dissolves like" which means polar compounds dissolve polar compounds and nonpolar dissolves nonpolar, etc...
    5. For NaCl(aq) the ion-ion forces (ionic bonds) of NaCl(s) are broken and the H bond forces of water are broken so that we make salt water NaCl(aq) and ion-dipole forces are formed.  Remember NaCl(aq) really means Na+(aq) and Cl-(aq).  Here is a great movie. So for salt water NaCl(aq) the ion-dipole forces formed must be greater than the ion ion and H bond forces broken. 
    6. Solvated ions (definition) = ions that are surrounded by solvent.  If water is the solvent we call them hydrated ions.  Here is a picture of hydrated ions.
    7. Skip the part about DG and DS of solution
  3. Concentration Units
    1. Molarity M = moles solute / liters solution  (mol/L)
    2. Mole Fraction c = moles solute / moles total (no units as mol cancel)
    3. mass% = grams solute / total grams x 100 (unit is %)
    4. ppm = grams solute / total grams x 106 (unit is ppm)
    5. ppb = grams solute / total grams x 109 (unit is ppb)
    6. molality m = moles solute / kg solvent (mol/kg) (this is the only one where the denominator is not a total solution number, it's just the solvent)
    7. See Table 11.3 and try examples 11.2 - 6 and problems 11.3 - 10.  Practice these!!!
  4. Solubility Factors
    1. Definitions
      1. unsaturated = solution where more solute can dissolve
      2. saturated = solution where no more solute can dissolve, if more solid solute is added, it will simply sink to the bottom and remain solid
      3. supersaturated = solution where more solute at the temperature is dissolved that should be.  What???  Well if you take really hot water and dissolve the maximum amount of salt possible, then COOL the solution down then more is dissolved than could be if you started with cold water.  The solute can solidify easily if you add just one tiny speck of solid to it. Then it will all crystallize quickly.
      4. solubility = the maximum grams of a solute than can dissolve in one liter of water
    2. Affecting solubility
      1. As temperature increases the solubility of solids usually increases also.  Why?  As KE increases the solid can more easily break apart and mix with the solvent.
      2. As temperature increases the solubility of gases decreases.  Why?  As KE increases the gas can more easily reach the surface and break away from the solution and float away. This is why you put pop in the refrigerator - the cooler temperature keeps the carbonation.  Another example is rivers warming - the oxygen gas is less soluble so there is less oxygen for the fish and they may die.
      3. As pressure increases the solubility of solids is unaffected.  Why?  Solids are not compressible.
      4. As pressure increases the solubility of gases increases also.  Click on activities then Henry's Law activity.  See the pressure increasing holds the gas molecules in the solvent. This is why you put a lid on soda pop - it holds the carbonation in.
  5. Colligative Properties
    1. They depend on the molality of the ions
      1. boiling point elevation
      2. freezing point depressions
      3. vapor pressure lowering
      4. osmosis
    2. Solutions are very different from their pure solvent.  Salt water is very different from plain water.
  6. Vapor Pressure lowering
    1. Consider solid in liquid solutions (don't worry about others or the van Hoff crap)
    2. In the first picture above the solvent is pure water and because it is a closed container a vapor pressure has set up.  Then in the second picture we have a solution.  Notice the vapor pressure is LOWER.  It is because the solid can't evaporate and the solution is no longer pure solvent.  So the vapor pressure depends on how much of the solution is solvent and not solute.  The more solute added, the lower the VP.  Psoln = XsolventPsolvent
    3. As solute goes up, the solvent fraction X goes down, sot the VP goes down. Click on activites then vapor pressure activity to see this in action. 
  7. Boiling point elevation and Freezing point (melting point, same thing) depression
    1. A solution of a solid solute in water is harder to boil since we have additional ion-dipole IMF and becasue the VP is lower as we just learned.  So it takes MORE energy to get the liquid to boil and turn into gas.
    2. A solution of a solid solute in water is also harder to freeze since we have solute particles interfering in the crystal lattice of the ice.  So is must be COLDER to get the water to freeze into ice.  See figure 11.12.
    3. How large is this bp and mp change in temperature?  It depends on the molality of ions in solution.  Equations:  DTb = Kbm and DTf = Kf m where the Kb or f  is a constant for boiling or freezing found in Table 11.4.
    4. Calculate the bp and mp of a 33.33 grams calcium chloride in 477 mg of water.  First we need the moles of ions.  33.33 g (1 mol / 110.98 g) (3 ions / mol) = 0.90097 mol ions. (CaCl2 ionizes into 3 ions:  one Ca2+ and two Cl- ions) Now we need molality.  m = 0.90097 mol ions / 0.477 kg water = 1.8888 mol/kg.  Now plug into the equations.  
      1. DTb = (0.51oC kg / mol)(1.8888 mol / kg) = 0.96 so the new bp = 100.96oC since water's original bp is 100
      2. DTf = (1.86oC kg / mol)(1.8888 mol / kg) = 3.5 so the new fp = -3.5oC since water's original fp is 0
    5. Try examples 11.11 and problems 11.18, 11.20 and 11.22
  8. Osmosis
    1. Consider two solutions are separated by a semipermeable (allows solvent not solute to pass) membrane with different concentrations. The goal when we have two solutions with different concentrations is to equalize the concentrations.  We do this by diluting the higher concentration solution by having solvent molecules pass through the membrane.  Thus solvent flows from lower to higher solute concentrations. The volume on the higher conc. side will increase as it is diluted, and the volume on the lower conc. side will decrease as it is concentrated.  Eventually the conc will be equal unless pressure build up stops the flow. 
    2. Example:  We have a solution on the left that is 1 M and a solution on the right that is 5 M.  They are connected by a semipermeable membrane.  Solvent will flow from the 1 M solution to the 5 M solution thus increasing the volume of the 5 M solution (diluting it) and the volume of the 1 M solution will decrease (thus concentrating it).  So the concentrations get closer to each other - 1 M increases and 5 M decreases.

 

Practice Problems on Concentrations

  1. How many moles are in 234 mL of a 1.29 M solution?
  2. What is the molarity if 9.25 moles of sodium sulfide is dissolved to make a 787 mL solution?
  3. How many grams of magnesium fluoride do you need to prepare 838 mL of a 2.42 M solution?
  4. The salt is blood serum is 0.14 M.  How many grams of sodium chloride are in 5.00 mL of blood serum?
  5. I used 54.7 grams of sodium phosphate to make a 0.250 M solution.  What is the volume in mL?
  6. What is the ppm if I have 45.0 mg of lead in 2500 mL of water?
  7. What is the ppm of dioxin if there is 0.0223 mg in 1550 mL of water?
  8. I had 25.0 mL of a 5.00 M solution.  How much water did I add if I diluted the solution to 2.30 M?
  9. Concentrated HCl is 6.00 M.  In lab we used 1.00 M HCl.  If we need 500.0 mL in lab, how much of the concentrated do I need to dilute?

Answers

  1. 1.29 mol / L = x moles / 0.234 L, x = (1.29 mol/L)(0.234 L) = 0.302 moles
  2. M = (9.25 moles / 0.787 L) = 11.8 M
  3. (0.838 L)(2.42 mol/L) = 2.03 moles MgF2 needed, 2.03 mol MgF2 (62.3 g/mol) = 126 grams
  4. (0.14 mol/L)(0.00500 L) = 7.00 x 10-4 moles, 7.00 x 10-4 moles NaCl (58.5 g/mol) = 0.0410 g
  5. 54.7 g Na3PO4 (mol / 164 g) = 0.334 moles,   0.334 mol ( 1 L / 0.250 mol) = 1.34 L = 1340 mL
  6. Remember for water 1 g = 1 mL.  So we have 2500 grams of water.  ppm = (0.0450 g / 2500 g)(106) = 18.0 ppm
  7. ppm = (2.23 x 10-5 g / 1550 g)(106) = 0.0144 ppm
  8. (5.00 M)(25.0 mL) = (2.30 M) V2,  V2 = 54.3 mL so I added 29.3 mL
  9. (6.00 M) V1 = (1.00 M)(500.0 mL),   V1 = 83.3 mL