Title: Solution Composition Energy of Solution Factors Vapor Pressure of Solutions BPE
1Unit 11
- Solution Composition Energy of Solution
Factors Vapor Pressure of Solutions BPE FPE
Osmotic Pressure Colligative Properties
2Equations Table
- Draw yourself a table to help you study
Make at least 6 rows maybe more
3Solution Composition How do we
describe the parts of a solution quantitatively?
- 1. Mass Percent of a solutes mass in a total
solution - mass (g of solute/g of solution) x 100
- Example A solution is prepared by adding 5.84g
of formaldehyde, H2CO to 100.0 g of water. The
final volume of solution was 104.0 ml . Was is
mass percent? - 5.84g/105.84 g x 100 5.52 H2CO
4Another way to describe solution composition
Mole Fraction
- 2. Mole Fraction na/ntotal
- Example A solution is prepared by adding 5.84g
of formaldehyde, H2CO to 100.0 g of water. The
final volume of solution was 104.0 ml . Mole
fraction of formaldehyde. - Calculate moles of formaldehyde
- mass of formaldehyde/ formula mass
- (5.84g H2CO)/(30.03 g/ H2CO) .194 m H2CO
- Calculate mols of water 100 g H2O / 18.02 g g
H2O 5.55 mol H2O - Calculate total mols N H2O N H2CO .194 mol
5.55 mol 5.75mol - Calculate mol fraction na/ntotal 1.94 m H2CO/
5.75 mol 0.0338
5Molality and Molarity Example A solution is
prepared by adding 5.84g of formaldehyde, H2CO to
100.0 g of water. The final volume of solution
was 104.0 ml .
- Molality mol of solute/kg of solvent
- .194 mol of H2CO/ .100 Kg H2O 1. 94 m H2CO
- Molarity mol of solute/L solvent
- .194 mol of H2CO/ .104 L Solution 1.87 M H2CO
- Please note the difference between m and M!!!
They are not interchangeable and they DO matter!
6Compare Mass Mole Fraction Molality Molarity
- Which way of describing solution composition is
dependant upon temperature? - Hint Think of your gas laws.
7Molarity
- Molarity is the only one that calculates using
volume and when temperature changes, volume
changes. - Remember Charles Law V1/T1 V2/T2
- What will happen to Molarity if temperature
rises? - decreases
8Normality
- N grams equivalent of solute/ liter of solution
- An equivalent is the mass of the solute that can
accept one mol of protons H (in an acid/base) - This is a VERY awkward term and will become
unused in the future. And it is not tested. - For now an example each mol of phosphoric acid,
H3PO4 supplies 3 mol of H so that means - 3 equivalents per 97.99 g/mol 32.66 g/eq
9Energy of Solution
- There are three steps that must occur so that a
solution will form - 1. Expanding a solute endothermic ?H
- 2. Expanding a solvent endothermic ?H
- 3. Combining often exothermic ?H
- The sum of these three steps is often - or
exothermic but this doesn - Remember that the overall assumption that must be
made is that Like Will Dissolve Like covalent
dissolves covalent, etc. - SO polarity will need to be determined of each,
the more polar, the more soluble in water.
Remember to consider dipoles.
10Remember you are looking for a general ratio of
polar to non-polar bonds are there any polar
bonds?
- The higher the ratio of polar to nonpolar bonds,
the more likely the covalent substance is to be
water soluble - Dont forget to consider hydrogen bonding as
polar - Unless there is no dipole moment and then it is
insoluble
11Example Determine whether each of the following
are likely to be water soluble
- a. CH3-CHNH2 d. CH3(CH4)4CH2NH2
- OH
- H
- b. H2C N C6H5 e. H2C - CH2
- NC6H5 OH OH
- c. C4H9CHCH2
12- Solution
- Water soluble small molecule with 3 polar bonds
- Insoluble large molecule with one polar bond (N,
1 lone pair of electrons) - Insoluble nonpolar no dipole moment
- Insoluble large molecule only 1 polar bond (N)
- Water Soluble small molecule with 2 polar bonds
13Henrys Law PkC
- P partial pressure in atm
- C concentration of dissolved gas (mol/L or M)
- k constant for a solution (Latm/mol)
- Relationships
- When the temperature of in two solutions is the
same , then the constant is the same and the
constant can be used to link two P/C - In general, the solubility of most solutes
increases as temperature increases and
dissolution of the solid occcurs more quickly at
higher temperatures
14An Example The solubility of Nitrogen in blood
at 37 C and .80 atm is 5.6 x 10-4 M. If a deep
sea diver breathes compressed air from a tank at
a partial pressure of 3 atm what would the
solubility of Nitrogen be in the divers blood at
37 C?
- P kC k is constant at the same temperature
P/C P/C - (.8atm/5.6 x 10-4M) (3/C)
- C 3/ (.8atm/5.6 x 10-4M) .0021 M
15Vapor Pressure of Solutes Non- Volatile Solutes
and Volatile Solutes
- Raoults Law (Non-Volatile) Psoln xsolvent
Psolvent - The addition of a nonvolatile solute will cause
the vapor pressure of the solution to fall in
direct proportion to the mole fraction of the
solute - Psoln Vapor Pressure of Solution( atm or
Torr) - xsolvent mole fraction of solvent
- Psolvent vapor presssure of pure solvent
- Raoults Law (Volatile Solutes)
- PtotalPsolute Psolvent
- Psolute Xsolute Psolute
16Example The vapor pressure of pure hexane
(C6H12)at 60.0C is 573 torr. That of pure
benzene(C6H6) at 60.0C is 391 torr. What is the
expected vapor pressure of a solution prepared by
mixing 58.9g hexane with 44.0 g benzene?
- Calculate mols nhexane 58.9g/86.0g 0.685 mol
hexane - nbenzene 44.0g/ 78.0 g 0.564
molbenzene - Calculate mole fraction
- xhexane0.685/(.685. 564) 0.548
- xbenzene 0.564/(.685.564) 0.452
- PtotalPsolute Psolvent
-
- xsolvent Psolvent xsolute Psolute
- 0.548(573) 0.452(391) 491 torr
17Boiling Point Elevation (the increase in the
boiling point of a solution due to the addition
of a non-volatile solute)
- ?T kbmsoluten
- ?T change in temperature
- kb molal boiling point constant
- msolute molality of solute (mol solute/kg
solvent) - n number of particles dissociated
- Example a solution is prepared by adding
31.65g of sodium chloride to 220.0 ml of water at
34.0C (density .994g/ml) kb for water 0.51
kg/mol Calculate boiling point elevation of the
solution. - molssolute31.65g/(58.44g/mol)0.542 mol
- kg solvent 220.0 ml(.994 g/ml)(1kg/1000g)
.219kg
18continued
- msolute 0.542 mol/ 0.219kg 2.47 m
- n 2 because NaCl dissociates into two types of
particles, Na and Cl- - ?T kbmsoluten
- ?T(.051)(2.47) (2)
- 2.5 C
- The boiling point will therefore be waters
normal boiling point, 100 C ?T - 102.5 C
19Freezing point depression works much the same way
- ?T kf msolute n
- ?T change in temperature
- kf molal freezing point constant
- msolute molality of solute (mol solute/kg
solvent) - n number of particles dissociated
- The only difference really is decrease the
temperature so normal freezing point will be
lowered by the ?T