Transcript Solutions: Concentration

Solutions:
Concentration
Chapter 14
Solution
 Homogenous mixture of 2 or more
substances in single phase
 = 1 layer
 Component present in largest amt =
solvent
 Other component(s) = solute
 Alloys, air are all solutions (solns)
Colligative properties
 = change of mp, bp, osmotic pressure of
soln, change in vapor pressure
 Irrespective of solute identity
 Dependent on concentration
Concentration
 Molarity (M) = moles solute/L of solution
= mol/L
 We’re all familiar with this
 Drawback  won’t give you proper
amount of solvent used to make soln
 Molality (m) better
Problem
 Give the concentration (in M) of 0.0012
grams of NaCl in 545 mL of water
 MW of NaCl = 58.442 g/mol
Molality
 Molality (m) = moles
solute/kg of solvent
 Let’s look to the right
  molality  molarity
Problem
 Give the concentration (in m) of 0.0012
grams of NaCl in 545 mL of water
 Density of water @ 25°C = 0.9970 g/mL
 MW of NaCl = 58.442 g/mol
Mole fraction
 Mole fraction of A
(XA) = nA/ntot
 Amt of component
A/total components
 Soln contains 1.00
mol ethanol and 9.00
mol water
1.00
= 0.100
(1.00+9.00)
9.00
X water =
= 0.900
(1.00+9.00)
X ethanol + X water = 1.000
X ethanol =
Weight percent
solute mass
100%
solution mass
 Ex: 46.1 g ethanol &
162 g water
Weight % ethanol =
46.1g
 100% = 22.2%
(46.1g+162g)
 Commonly used in
household products
like vinegar & bleach
Problem
• Concentrated sulfuric acid has a density
of 1.84 g/cm3 and is 95.0% (w/w) H2SO4.
MW H2SO4 = 98.079 g/mol.
 Calculate the molarity and the molality of this
solution.
Solution: molarity
95.0gacid
mol
1.84g soln 1cm3 1000mL




 1.78M
3
100.0g soln 98.079gacid
cm
1mL
1L
Solution: molality
95.0g acid
mol
1000g


 190m
5.0 g water 98.079g acid 1kg water
Problem
• A 10.7 molal solution of NaOH has a
density of 1.33 g/cm3 at 20°C. MW NaOH
= 39.996 g/mol & MW H2O = 18.0153
g/mol.
 Calculate the mole fraction of NaOH, the
weight percentage of NaOH and the molarity
of the solution.
Solution
1000g
mol
1kg water

 55.5mol water
1kg 18.0153g
10.7mol
X NaOH  (
)  0.162
10.7mol 55.5mol
39.996g
10.7mol NaOH 
 428g NaOH
mol
428g NaOH
100%  30.0% NaOH (w/w)
428g NaOH  1000gwater
10.7molNaOH 1.33g soln cm3 1000mL



 9.97M
3
1428gsoln
cm
mL
L
More practice
 An aqueous soln of NaCl is created using
133 g of NaCl diluted to a total soln
volume of 1.00 L.
 Calculate the molarity, molality, and
mass percent of the soln, given a
density of 1.08 g/mL and MW of NaCl =
58.442 g/mol.
Solution
mol
1
133g

 2.28M
58.442g 1.00L
1.08gsoln 1000mL 1.08 103 g soln


mL
1L
L
1.08 103 g soln - 133g NaCl  947gsolvent
947gsolvent 0.947kg solvent
mol
1
133g NaCl 

 2.40m
58.442g 0.947kg solvent
133g NaCl
 100%  12.3% by mass
133g NaCl  947gsolvent
Part per million
 = PPM (in grams)
 Ex: 1.0 ppm = 1.0 g of substance in
system w/ 1.0 million g total mass
 @ STP water density  1.0 g/mL
 So, mg/L and ppm are 
 Used predominately by environmental
and analytical chemists
Solution process
 One can add only so much solute to
solvent
 Since no more dissolves  soln said to
be saturated
 NaCl = 35.9 g/100 mL water (25°C)
 Albeit, nothing changes visually, soln is
constantly dissolving and re-solidifying
ions
Solution process
 Essentially, solubility = solute concentration in
equilibrium w/undissolved solute in saturated
soln
 Unsaturated soln = soln w/less than saturated
amt of solute
 NaCl < 35.9 g/100 mL water (25°C)
 Supersaturated soln = soln w/more solute
than sat. soln
 NaCl > 35.9 g/100 mL water (25°C)
Making supersaturated
solutions
 Pour in excessive amount of solute
 Heat up the soln
 Stir until all solute dissolves
 Cool it slowly
 No shaking, no jarring of soln
 Gives lower freezing point
 Once disturbed (energy in), causes crystallization to occur 
excess crystallized out of soln
 Exothermic
 Heat packs of sodium acetate (can reach 50°C!)
 http://www.npr.org/programs/wesun/features/2001/dec/heatpack/0
11229.heatpack.html
 Your second lab deals with this
Liquids as solutes
 Miscible = mixable
 Immiscible = unmixable
 Used in language too:
 Mestizo, mischling
 Ability to dissolve based
on similar polarities (or
lack thereof) of
solute/solvent
 Like dissolves in like
Let’s try these
 Considering intermolecular forces, give
reasons for the following observations:
 a) Octane, C8H18, is very miscible with CCl4.
 b) Methanol, CH3OH, mixes in all proportions
with water.
 c) Sodium bromide is not very soluble in diethyl
ether (CH3CH2—O—CH2CH3).
 d) Octanol, C8H17OH, is not very soluble in
water.