Transcript Ch. 16: Solutions

Ch. 7: Solutions
Chem. 20
El Camino College
1
Terminology





The solute is dissolved in the solvent.
The solute is usually in smaller amount, and the
solvent is usually in larger amount
A concentrated solution contains a relatively large
amount of solute
A dilute solution contains a relatively small amount
of solute
Solubility measures how much of a solute can
dissolve in a solvent at a given temperature.
2
3
4
Terminology





An unsaturated soln can hold more of a solute
A saturated soln cannot hold more of a solute. If more
solute is added (without the temp changing), it won’t
dissolve
A supersaturated soln is very unstable. It’s made by
creating a saturated soln at high temp, and then cooling
slowly. Adding a crystal will cause the excess solute to
crystallize out
Miscible refers to liquids that are soluble in each other
Immiscible liquids are not soluble, like oil and water.
5
Terminology

“Like dissolves like” means
Polar molecules dissolve polar (and many ionic)
substances
 Nonpolar molecules dissolve nonpolar
substances

Water is a very polar molecule.
 Which will dissolve in water:
NaCl, sugar, benzene (a nonpolar molecule)
 Is sugar polar or nonpolar?

6
7
Solution Formation
When water dissolves salt crystals, the d+
end of water is attracted to the Cl- ions, and
the d- end is attracted to the Na+ ions
 The ions become hydrated, or surrounded
by water molecules.

8
9
10
Electrolytes
Substances can be classified as strong-, weak-,
or non-electrolytes in water
 Strong electrolytes contain a lot of ions that
conduct electricity
 Weak electrolytes contain only a few ions, and
they conduct electricity weakly
 Nonelectrolytes do not contain ions. They don’t
conduct electricity

11
12
13
Strong Electrolytes
When a strong electrolyte is placed in water, it
dissolves and forms ions
Write the eqn for Ca(NO3)2(s) dissolving in H2O.
HO
Ca(NO3)2(s)  Ca2+(aq) + 2 NO3-(aq)
ions only
Write the eqn for NaCl(s) dissolving in H2O.
HO
NaCl(s)  Na+(aq) + Cl-(aq)
ions only

2
2
14
Weak Electrolytes
Some substances, such as weak acids, only
form a few ions when placed in water
 For weak acids, H+ pops off a small % of
molecules
 Since there are only a few ions in the soln, weak
electrolytes conduct electricity weakly

H2O
HF(g) 
 HF(aq) (minor: H+(aq) + F-(aq))
mostly molecules, few ions
15
Nonelectrolytes

Many molecules do not form ions when they
dissolve
H2O
C12H22O11(s)  C12H22O11(aq)
molecules only
16
Classify These
Based on the given information, decide whether
ions only, few ions, or molecules only form in
water. Are they strong, weak, or nonelectrolytes?
HO
LiCl(s) Li+(aq) + Cl-(aq) ions only,
strong electrolyte
note: LiCl is water soluble
2
molecules only,
CH3OH(l) CH3OH(aq) nonelectrolyte
H2O
note: CH3OH is a molecule that dissolves, but doesn’t form ions
H2O
HNO2(g) 
H+(aq) + NO2-(aq) few ions,
note: HNO2 is a weak acid
weak electrolyte
17
Factors Affecting Solubility

Pressure


More gas dissolves at higher pressure. Pressure has
no effect on solubility of solids or liquids
Temperature


More solid dissolves at higher temps
More gas dissolves at lower temps.
18
19
Fig. 16-5, p. 439
Solution Concentration
mass % =
g solute x 100
g solution
mass % =
g solute x 100
(g solute+g solvent)
Ex. What is the % concentration of 250.0 g solution which contains
8.75 g solid?
g solute x 100
8.75 g
% by mass =
=
x 100 = 3.50 %
g solution
250.0 g
Ex. What is the % conc of a soln which contains 5.33 g NaCl and
10.99 g water?
g solute x 100
mass % = g solute + g solvent
5.33 g x 100
mass % = 5.33 g + 10.99 g
= 32.7 %
20
Molarity



Molarity is (moles solute)/(liters solution)
M=mol/L
Ex. Calculate the molarity of 3.5 L of solution that
contains 0.50 mol hydrogen chloride.
0.50 mol
3.5 L
= .14 M
21
Using Molarity as a
Conversion Factor


Note, if molarity is given in the problem, use it as a
conversion factor with the units mol/L
A solution is listed as 5.60 M. Write a conversion
factor with mol on top. Write a conversion factor with
mol on the bottom.
5.60 mol
1L
1L
5.60 mol
22
Given Molarity, Find Moles

Ex. How many moles of sodium chloride are in 400.
mL of a 1.45 M solution?
400. mL

1L
1.45 mol
= 0.580 mol
1000 mL
1L
Ex. How many moles of HBr are in 755 mL of a 3.50
M solution?
755. mL
1L
3.50 mol
= 2.64 mol
1000 mL
1L
23
Given Molarity, Find Grams

Ex. How many g of sodium chloride are in 50. mL of a
4.0 M solution?
50. mL

1L
1000 mL
4.0 mol
1L
58.44 g
1 mol
= 12 g
Ex. How many g of I2 are in 2.00 L of a 0.75 M soln?
2.00 L 0.75 mol
1L
253.8 g
1 mol
= 381 g
24
Given Molarity, Find Volume

Ex. How many mL of 5.60 M sucrose soln contain
2.50 mol sucrose?
2.50 mol

1L
5.60 mol
1000 mL = 446 mL
1L
Ex. How many mL of 12.0 M HCl soln contain 0.330
mol HCl?
0.330 mol
1L
12.0 mol
1000 mL = 27.5 mL
1L
25
Colligative Properties
When a solute is dissolved in a pure liquid, the
freezing point (fp) and boiling point (bp) of the
soln will change.
 FP depression means that a solute will lower
the fp of the soln compared to the pure solvent
 BP elevation means that a solute will raise the
bp of a soln compared to the pure solvent

26
Colligative Properties



Colligative Properties are only determined by the
number of solute particles in a quantity of solvent
Molality
m=(mol solute)/(kg solvent)
Freezing Point Depression DTf = Kfm




DTf is the change in temperature
Subtract DTf from the freezing pt (for H2O, the freezing
pt is 0 oC)
Kf is the molal freezing pt depression constant
For H2O, Kf = 1.86 oC/m
27
Colligative Properties

Boiling Point Elevation




DTb = Kbm
DTb is the change in temperature
Add DTb to the boiling pt (for H2O, the bp is 100 oC)
Kb is the molal boiling pt elevation constant
For H2O, Kb = 0.52 oC/m
28

Ex. Determine the DTf of 3.0 kg of water when 5.0
mol antifreeze is added.
DTf=Kfm = 1.86 oC
m

5.0 mol solute
3.0 kg H2O
= 3.1 oC
Calculate the freezing pt of the above soln.
fp = 0 - DTf =
- 3.1 oC
29

Ex. Determine DTb of 3.0 kg of water when 5.0 mol
antifreeze is added.
DTb=Kbm = 0.52 oC
m

5.0 mol solute
3.0 kg H2O
= 0.87 oC
Determine the bp of the above soln.
bp = 100 + DTb = 100.87 oC
30

Ex. Determine the freezing pt of 1.53 kg of water
when 6.2 mol formic acid (HCO2H) is added.
DTf=Kfm = 1.86 oC 6.2 mol HCO2H = 7.5 oC
m
1.53 kg H2O
fp = -7.5 oC
31