Solutions Test II

1
Unit XIV

• Solutions Test II

Objs8,9,12-18
2
Obj. 8Precipitates

• precipitate

an insoluble solid that will separate from
a solution during a reaction (double
displacement).
using solubility rules, we can predict
precipitates.
3
Obj. 8 cont

• Predict the precipitate

precipitate
1
-1
1
-1
LiNO3
AgI
LiI (aq) AgNO3 (aq)
(s)
(aq)
precipitate
1
-1
-2
3
NaOH (aq) Cr2(SO4)3 (aq)
Na2SO4
Cr(OH)3
(aq)
(s)
4
Obj. 9Net Ionic Equations

• a net ionic equation shows the chemical change
  in a

solution (formation of precipitate).
all soluble (aq) compounds are separated into
ions.
insoluble (s) compounds are written as
compounds.

• To write

write the complete ionic equation.

• separate (aq) compounds.

• leave (s) compounds intact.

cross out any spectator ions

• ions that appear on both sides of equation.

5
Obj. 9 cont

• Ex

LiI (aq) AgNO3 (aq)
LiNO3 (aq)
AgI (s)

• write complete ionic equation

Li1
I-1
Ag1
NO3-1
Li1
NO3-1
AgI (s)

• cross out spectator ions

• what is left is the net ionic equation!

I-1
Ag1
AgI (s)

• Ex

NaOH (aq) Cr2(SO4)3 (aq) Na2SO4
(aq)
Cr(OH)3 (s)
Na1
OH-1
Cr3
Cr(OH)3 (s)
SO4-2
Na1
SO4-2
OH-1
Cr3
Cr(OH)3 (s)
6
Obj. 12-13Solution Calculations

• we said that dilute and concentrated are very

general terms for solution concentration.

• Molarity indicates how many moles of solute are

dissolved in one liter of solution.
moles
molarity
units M
Liters

• Ex

What is the molarity of a salt water solution
containing
9.0 moles of salt dissolved in 3.0 liters of
solution?
9.0 moles

3.0 M
3.0 liters
7
Obj. 12-13 cont

• Ex

23
35
What is the molarity of 174 grams of NaCl
dissolved in
500.0 mL of solution?
moles
molarity
Liters
174 grams
1 mole
1000 mL

500.0 mL
1 liter
58 grams NaCl
6.00 M
8
Obj. 12-13 cont

• Molality indicates how many moles of solute are

dissolved in one kilogram of solvent.
moles
molality
units m
Kg

• Ex

59
160
What is the molality of 199 grams of NiBr2 in
500.0
grams of water?
199 grams NiBr2
1 mole
1000 grams

500.0 grams
219 grams NiBr2
1 Kg
1.82 m
9
Obj. 12-13 cont

• Other types of solution concentration include

part
mass of solute or solvent
mass

x
100
whole
mass of solution
part (solute)

• Ex

What is the by mass of 62.0 grams of KCl
dissolved in
475 grams of water?
solvent
part
mass
whole
whole solution
solute solvent!
62.0 475 537 grams of solution
62.0
mass

x
100
11.5
537
10
Obj. 12-13 cont
part (solute)
whole (solution)

• Ex

How many grams of KOH are required to prepare
450.0
grams of a 30.0 solution?

part
x
mass
whole
x
whole
part x whole
whole

KOH
0.300
x
450.0
135 g
pph
parts per hundred
ppt
parts per thousand
parts per million
ppm
11
Obj. 12-13 cont

• Dilution calculations

a dilution a solution is made less
concentrated
(weaker) by adding more solvent.

• changes concentration (molarity).

• changes volume of solution.

M1V1 M2V2
initial molarity
final molarity
initial volume
final volume
12
Obj. 12-13 cont
M1

• Ex

V1

• How many liters of a 12M solution are needed to
  create

2.0 liters of a 4.0M solution?
M1V1 M2V2
M2
V2
8.0
12
X

12
X
2.0
(4.0)

0.67 liters
X
12
12
V1

• Ex

M1

• What is the molarity of 1.5 liters of solution
  made from

600.0 mL of 10.0M NaOH?
M2
V2
6.0
1.5
X

4.0M
1.5
X
0.6
(10.0)

X
1.5
1.5
13
Obj. 12-13 cont

• Some dilution problems may contain
  concentrations

instead of molarity
you do NOT have to put in decimal formjust
be consistent!
1

• Ex

mass1

• How many grams of a 25.0 solution of a KCl are
  needed to

prepare 85.0 grams of a 40.0 solution?
2
mass2
1mass1 2mass2
25
X
3400
(40)
25
X
85

136 grams

X
25
25
14
Obj. 14Colligative Properties

• Colligative properites are properties of
  solutions that are

affected only by the of particles in the
solution.

• NOT affected by the type of particle!!!

• Ex

vapor pressure (VP)
freezing point (FP)
boiling point (BP)
15
Obj. 15Effect of Solutes on Vapor Pressure

• Vapor pressure (VP) is the P exerted at the
  surface of a

liquid by particles trying to escape the liquid.
16
Obj. 15 cont

• adding a nonvolatile solute to a solvent will
  cause the VP of

the solvent to decrease.
solute particles replace some solvent
particles at the
surface of the solution.
less solvent particles on surface less
evaporation
lower VP!
17
Obj. 16 and 18How Solutes Affect BP and FP

• Boiling pt. (BP) is temp. at which the VP of the
  liquid

atmospheric pressure.
adding solute lowers VP of solvent
must add more KE (heat) to equalize the
pressures
solutes RAISE the BP of solutions!
(i.e. we add salt before we boil water)
pure water
salt water
18
Obj. 16 and 18 cont

• Freezing pt. (FP) is temp. at which liquid turns
  into a solid.

enough KE is lost (removal of heat) that
molecules stop
moving around and lock into place.
adding solute lowers VP of solvent
even more KE (heat) must be lost to lock
molecules into
place.
solutes LOWER the FP of solutions!
(i.e. we add salt to icy roadssalt is used in
making ice cream )
19
Obj. 16 and 18 cont

• Ionic solutes lower the VP of solvents more than
  molecular

solutes!
ionic solutes dissociate (break up into ions)
in solvents.

• AlCl3 dissociates into 4 separate ions (1 Al3
  and 3 Cl-1)

molecular solutes stay intact in solvents.

• glucose (C6H12O6) breaks into separate units of
  C6H12O6

NOT Cs, Hs and Os!
20
Obj. 17BP and FP Calculations

• solutes raise (elevate) the BP of solvents.

• to calculate how high the BP is elevated

BP elevation ?TBP
m(KBP) normal BP
given except
BP constant
molality
water 100C
given

• solutes lower (depress) the FP of solvents.

• to calculate how low the FP is depressed

FP depression ?TFP
normal FP - m(KFP)
given except
FP constant
molality
water 0C
given
ionic solutes affect molality!!!
CaCl2 dissociates into 3 ions, so multiply m by 3!
Round all temps. to the hundredths place!
21
Obj. 17 cont

• Ex

If 52.34 grams of NiBr2 dissolve in 392.0 grams
of water, what is the BP of the resulting
solution?
BP m(KBP) normal BP
52.34 g NiBr2
1 mole
1000 g
x 3
1.829 m
0.6097 m

219 g NiBr2
392.0 g water
1 Kg
(ions)
0.936
(0.512)
100C
100.94C
BP
1.829
22
Obj. 17 cont

• Ex

If 95.67 grams of CCl4 dissolve in 981.0 grams of
benzene, what is the FP of the resulting
solution? (FP of Benzene 5.53C)
FP normal FP - m(KFP)
molecular, so dont multiply by anything!
95.67 g CCl4
1 mole
1000 g
0.6416 m

152 g CCl4
981 g benzene
1 Kg
3.285
2.25C
FP
5.53C
(5.12)
- 0.6416