Chapter 4
Aqueous solutions
Types of reactions
mixture
homogeneous
one phase
Tab water
solvent
more abundant
component of mixture
water in tab water
Water always solvent
even in 98% H2SO4
heterogeneous
two phases
oil/water
milk
Solute(s)
Less abundant or other
component(s) of
mixture
salts in tab water
Nonelectrolytes
Electrolytes
Solute undergoes
Dissociation
strong
weak
complete
dissociation
NaCl / H2O
HCl / H2O
NaCl( s ) 
 Na
H 2O

( aq)
Sugar / H2O
partial
dissociation
HAc / H2O
 Cl

( aq)
2O
HAc(l ) H

H  ( aq)  Ac ( aq)
Hydration of Solid Solute



At edges, fewer oppositely charged ions around
– H2O can come in; Ion-dipole forces; Remove ion
New ion at surface
– Process continues until all ions in solution
Hydration of ions
– Completely surrounded by solvent
4
Molecular Compounds In Water

When molecules dissolve in water
– Solute particles are surrounded by water
– Molecules are not dissociated
5
Electrical conductivity
of electrolyte solutions
Ex. Sugar, alcohol
Weak acids and bases
Ionic compounds
Ex. Acetic acid
(HC2H3O2),
ammonia (NH3)
Strong acids and bases
Ex. NaBr, KNO3,
HClO , HCl, KOH
Learning Check
Write the equations that illustrate the
dissociation of the following salts:
Na3PO4(aq)
→
Al2(SO4)3(aq)
CaCl2(aq)
3 Na+(aq) + PO43(aq)
→
Ca2+(aq) + 2 Cl(aq)
→
Ca(MnO4)2(aq)
→
7
Solubility

Maximum amount of a substance that can be
dissolved in a given amount of solvent at a
given temperature.

Usually g/100 mL.
g solute needed to make saturated solution
Solubility 
100 g solvent
Saturated solution: Solution in which no more solute can be
dissolved at a given temperature
Unsaturated solution: Solution containing less solute than max.
amount; Able to dissolve more solute
Solubilities of Some Common Substances
Solubility
(g/100 g water)
Substance
Formula
Sodium chloride
NaCl
35.7 at 0°C
39.1 at 100°C
Sodium hydroxide
NaOH
42 at 0°C
347 at 100°C
Calcium carbonate
CaCO3
0.0015 at 25°C
9
“Like dissolves Like”

Ethanol (C2H5OH) dissolves in water:
polar ↔ polar

Glucose (C6H12O6) and sucrose (C12H22O11)
dissolve in water: polar ↔ polar

Oil doesn’t dissolve in water: nonpolar ↔ polar

Oil dissolves in benzene: nonpolar ↔ nonpolar
Salts are polar.
soluble
NaCl
insoluble
AgCl
Water unable to
separate Ag+ from ClInteraction very strong
Relative Concentration
Solute-to-solvent ratio
Dilute solution
– Small solute to solvent ratio
Ex. Eyedrops
Concentrated
solution
– Large solute to solvent
ratio
Ex. Pickle brine

Dilute solution contains less solute per unit volume
than
more concentrated solution
12
quantitatively
nsolute
M
Vsolution / L


abbreviated M
1 M = 1 mol solute / 1 liter solution
nNaOH
M NaOH
mNaOH
11.5 g


 0.288 mol
M NaOH 40.0 g / mol
nNaOH
0.288 mol
mol


 0.192
 0.192 M
Vsolution / L
1.50 L
L
Preparing Solution of Known Molarity
a
a)
b)
c)
d)
e)
b
c
d
Weigh solid and transfer to volumetric flask
Add part of the water
Dissolve solute completely
Add water to reach etched line
Stopper flask and invert to mix thoroughly
e
14
Concentration of each type of ions in 0.50 M
Co(NO3)2(aq)?
Co( NO3 ) 2 ( aq)  Co
2
( aq )
 2 NO3

( aq )
1 mol
1 mol
2 mol
In 1.00 L
0.50 mol
0.50 mol
1.00 mol
Molarity
0.50 M
0.50 M
1.00 M
Concentration of each type of ions in 0.50 M
Fe(ClO4)3(aq)?
Fe(ClO4 )3( aq)  Fe
Molarity
0.50 M
3
( aq )
0.50 M
 3ClO4

1.50 M
( aq )
Moles of Cl- 1.75 L of 1.0×10-3 M ZnCl2(aq)?
M ZnCl2 
nZnCl2
nZnCl2  M ZnCl2  Vsolution / L
Vsolution / L
nZnCl2  1.0 10 3 M 1.75 L
nZnCl2
mol
 1.0 10
1.75 L  1.75 10 3 mol
L
ZnCl2 ( aq)  Zn
3
2
( aq )
 2Cl
1
( aq )
2
1.75×10-3 mol
mCl   nCl   M Cl 

? n  2 1.75 103 mol
Cl

g
 2 1.75 10 mol  35.45
mol
3
nCl   3.5 10 3 mol
M NaCl 
nNaCl
nNaCl
nNaCl
Vsolution 
M NaCl
Vsolution / L
mNaCl
1.0 10 3 g
5


 1.7 10 mol
M NaCl 58.44 g / mol
5
nNaCl 1.7 10 mol
Vsolution 

 1.2 10 4 L  0.12 mL
M NaCl
0.14 mol / L
Practice

How many grams of HCl would be required to make
50.0 mL of a 2.7 M solution?

What would the concentration be if you used 27g of
CaCl2 to make 500. mL of solution? What is the
concentration of each ion?

Describe how to make 1.00 L of a 0.200 M K2CrO4
solution.

Describe how to make 250. mL of an 2.0 M copper
(II) sulfate dihydrate solution.

Calculate the concentration of a solution made by
dissolving 45.6 g of Fe2(SO4)3 to 475 mL. What is the
concentration of each ion?
Describe how to make 1.00 L of a 0.200 M K2CrO4 solution.
M K 2CrO4 
nK 2CrO4
Vsolution / L
nK 2CrO4  M K 2CrO4  Vsolution / L
nK 2CrO4  0.200 M 1.00 L  0.200 mol
mK 2CrO4  nK 2CrO4  M K 2CrO4  0.200 mol  294.20 g / mol  58.8 g
No solid K2CrO4 available in the lab .
But 2.00 M K2CrO4 solution is available .
M K 2CrO4 
nK 2CrO4
Vsolution / L
nK 2CrO4  M K 2CrO4  Vsolution / L
nK 2CrO4  0.200 M 1.00 L  0.200 mol
n

Dilution
K 2CrO4 before dilution

 nK 2CrO4

after dilution
n  M V
M
K 2CrO4
VK 2CrO4



M

V
K
CrO
K 2CrO4
2
4
before dilution
M 1  V1  M 2 V2
M 1  V1  M 2  V2
2.00 M  V1  0.200 M 1.00 L
0.200 M 1.00 L
V1 
 0.100 L
2.00 M

after dilution
Prepare 150 mL of 0.100 M H2SO4 from 16.0 M solution.
M 1  V1  M 2  V2
16.0 M  V1  0.100 M  150 mL
0.100 M  150 mL
V1 
 0.938 mL
16.0 M



What volume of a 1.7 M solution is needed to make 250 mL of a
0.50 M solution?
18.5 mL of 2.3 M HCl is added to 250 mL of water. What is the
concentration of the solution?
You have a 4.0 M stock solution. Describe how to make 1.0 L of a
0.75 M solution.
ReductionOxidation
Metathesis
Double Replacement
Electron transfer
AB + CD  AD + CB
precipitation
reaction
a solid is formed
from solution
precipitate
Acid-Base
Reaction
Formation of a
weak electrolyte
Formation of a gas
Precipitation reactions
Molecular equation
Ionic equation
2 K  ( aq)  CrO4
2
( aq )
 Ba 2 ( aq)  2 NO3

( aq )
 2 K  ( aq)  2 NO3

( aq )
 BaCrO4 ( s )
Net ionic equation: describes what really happens.
Ba
2
( aq )
 CrO4
Spectator ions:
K

2
( aq )
( aq)
 BaCrO4 ( s )
, 2 NO3

( aq)
A reaction takes place if it has a net ionic equation
Solubility Rules
 All nitrates and acetates are soluble
 Salts of alkali metals ions and NH4+ ions are soluble.
 Chlorides, bromides and iodides (salts of Cl-, Br- and I-) are
soluble except those of Ag+, Pb2+, and Hg22+.
 Most sulfates are soluble, except those of Pb2+, Ba2+, Hg2+,
and Ca2+.
 Most hydroxides are slightly soluble (insoluble) except those
of alkali metals (Ba(OH)2, Sr(OH)2 and Ca(OH)2 are
marginally soluble).
 Sulfides (S2-), carbonates (CO32-), chromates (CrO42-) and
phosphates (PO43-), are insoluble except those of alkali
metals and NH4+.
Does the following mixing process involve a
chemical reaction?
AgNO3( aq)  KCl( aq)  ??
AgNO3( aq)  KCl( aq)  AgCl  KNO3
AgNO3( aq)  KCl( aq)  AgCl( s )  KNO3( aq)
Ag

( aq)
 Cl

( aq)
 AgCl( s )
Precipitation reactions

NaOH(aq) + FeCl3(aq) 

NaOH(aq) + FeCl3(aq) NaCl + Fe(OH)3

NaOH(aq) + FeCl3(aq)  NaCl(aq) + Fe(OH)3(s)

Na+(aq)+OH-(aq) + Fe3+ (aq) + Cl-(aq) Na+ (aq) +
Cl- (aq) + Fe(OH)3(s)

OH-(aq) + Fe3+ (aq)  Fe(OH)3(s)
Precipitation reactions

BaCl2(aq) + KNO3(aq) 

BaCl2(aq) + KNO3(aq) KCl + Ba(NO3)2

BaCl2(aq) + KNO3(aq) KCl(aq) + Ba(NO3)2(aq)

Ba2+(aq)+2 Cl-(aq) + K+ (aq) + NO3-(aq) K+ (aq) +
Cl- (aq) + Ba2+(aq)+ 2 NO3-(aq)

No net ionic equation

No reaction
Practice



iron (III) sulfate and potassium sulfide
Lead (II) nitrate and sulfuric acid.
solutions of NaOH and NiCl2 are mixed.
AgNO3( aq)  NaCl( aq)  AgCl( s )  NaNO3( aq)
1
1
0.15 mol
?
nNaCl  0.150 mol
mNaCl  nNaCl  M NaCl
g
 0.150 mol  58.44
 8.77 g
mol
n AgNO3  M AgNO3 VAgNO3  0.100 M 1.50 L  0.150 mol
1.25 L of 0.0500 M Pb(NO3)2 mixed with 2.0 L of
0.0250 M Na2SO4. Calculate the mass of precipitate.
Pb( NO3 ) 2 ( aq)  Na2 SO4( aq)  ??
Pb( NO3 ) 2( aq)  Na2 SO4( aq)  PbSO4( s )  2 NaNO3( aq)
1
1
0.0625 mol
?
1
1
0.0500 mol
?
nPbSO4  0.0625 mol
nPbSO4  0.0500 mol
m  m  M  15.2 g
nPb( NO3 ) 2  M Pb( NO3 ) 2  VPb( NO3 ) 2  0.0500 M 1.25 L  0.0625 mol
nNa2 SO4  M Na 2 SO4 VNa2 SO4  0.0250 M  2.00 L  0.0500 mol
Stoichiometry of Precipitation

What mass of solid is formed when 100.00 mL
of 0.100 M Barium chloride is mixed with
100.00 mL of 0.100 M sodium hydroxide?

What volume of 0.204 M HCl is needed to
precipitate the silver from 50.0 ml of 0.0500 M
silver nitrate solution ?

25 mL 0.67 M of H2SO4 is added to 35 mL of
0.40 M CaCl2 . What mass CaSO4 Is formed?
Arrhenius Acid

Substance that reacts with water to produce the
hydronium ion, H3O+
HCl(g) + H2O

Cl–(aq) + H3O+(aq)
Acid + H2O  Anion + H3O+
HA + H2O
 A– + H3O+
HC2H3O2(aq) + H2O  H3O+(aq) + C2H3O2−(aq)
Bronsted-Lowry Acid: H+ donor
Arrhenius Bases

Substance that reacts with water to give OH–.
a. Metal Hydroxides
NaOH(s)  Na+(aq) + OH–(aq)
Mg(OH)2(s)  Mg2+(aq) + 2OH–(aq)
b.
Basic Anhydrides
CaO(s) + H2O  Ca(OH)2(aq)
Ca(OH)2(aq)  Ca2+(aq) + 2OH–(aq)
c. Molecular bases:
NH3(aq)+H2O  NH4+(aq)+ OH-(aq)
Bronsted Base:
H+ acceptor
Acid-Base Reactions
HCl( aq)  NaOH ( aq)  ??
HCl( aq)  NaOH ( aq)  NaCl( aq)  H 2O(l )
Ionic equation
H  ( aq)  Cl  ( aq)  Na ( aq)  OH  ( aq)  Na ( aq)  Cl  ( aq)  H 2O(l )
Weak electrolyte:
Net ionic equation:
Any strong acid + strong base
H2O + H2O  H3O+(aq)+ OH-(aq)
H  ( aq)  OH  ( aq)  H 2O(l )
HC2 H3O2( aq)  NaOH ( aq)  ??
HC2 H3O2( aq)  NaOH ( aq)  NaC2 H3O2( aq)  H 2O(l )
Formation of Weak electrolyte:
HAc + H2O  H3O+(aq)+ Ac-(aq)
Acid - Base Reactions are often called
neutralization reaction Because the acid
neutralizes the base.
Volume of 0.100 M HCl needed to neutralize
25.0 mL of 0.350 M NaOH ?
HCl( aq)  NaOH ( aq)  NaCl( aq)  H 2O(l )
1
1
?
nHCl  8.75 10
8.75×10-3
3
n  M V
mol
n 8.75 10 3 mol
V

 8.75 10 2 L
mol
M
0.100
L
nNaOH  M NaOH VNaOH  0.350 M  0.0250 L  8.75 10 3 mol
28.0 mL of 0.250 M HNO3 mixed with 53.0 mL of 0.320 M KOH;
1) Amount of water formed
2) Concentrations of H+ and OH- at the end of rct
HNO3( aq)  KOH ( aq)  KNO3( aq)  H 2O(l )
1
1
7.0 mmol
? nH O  7.0 mmol

2
1
1
17.0 mmol
? nH O  17 mmol
nHNO3  M HNO3  VHNO3
2
mol
 0.250
 28.0 mL  7.0 mmol
L
nKOH  M KOH  VKOH  0.320
mol
 53.0 mL  17.0 mmol
L
HNO3 is Limiting reactant: reacts completely
1) No HNO3 left
nH
0
M


0M
H
2) HNO3 → H+ + NO3Vsolution 28  53 mL
3) No H+ at the end of reaction


How much remains from KOH?
nKOH remaining  nKOH initially  nKOH reacted
nKOH remaining  nKOH initially  nHNO reacted  17.0  7  10 mmol
3
1) KOH → K+ + OH10 →
10 mmol
M OH  
nOH 
Vsolution

10 mmol
10 mmol

 0.123 M
28  53 mL 81 mL
Volumetric analysis: Titration
Controlled addition of 1 reactant to another until rxn is complete.
Acid-Base Titration: Very common type of titration
Ex. Analysis of citric acid in orange juice by neutralization with NaOH

An indicator is needed: organic substance that
changes color according to solution acidity
Phenolphthalein
Acidic
Basic

Where the indicator changes color is the endpoint.

Endpoint must be very close to the equivalence point.
Acid (Base) added equivalent to base (acid) present
nacid
equivalent to nbase
Standardization of NaOH solution
•Know the exact concentration!
•Its weight is inaccurate .
•NaOH is hygroscopic and it absorbs CO2.
•Cannot be used to prepare solutions with exactly known M.
•Not a primary standard.
•KHP is a primary standard: high purity, no weighing problems,
Potassium hydrogen phthalate: KHC8H4O4.
Monoprotic!
41.2 mL of NaOH solution is needed to react
exactly with 1.300 g of KHP (M=204.22 g/mol).
MNaOH=?
nNaOH  nKHP
M NaOH VNaOH
M NaOH
mKHP

Mwt KHP
mKHP
1.300 g


 0.1546 M
Mwt KHP  VNaOH 204.22 g  0.04120 L
mol
nNaOH  nacid
M NaOH VNaOH
macid

Mwt acid
macid  M NaOH  VNaOH  Mwtacid
macid
mol
g
 0.1546
 0.01059 L 122.12
 0.1999 g
L
mol
macid
0.1999 g
%acid 
100% 
100%  56.82%
msample
0.3518 g
practice

75 mL of 0.25M HCl is mixed with 225 mL of
0.055 M Ba(OH)2 . What is the
concentration of the excess H+ or OH- ?

A 50.00 mL sample of aqueous Ca(OH)2
requires 34.66 mL of 0.0980 M Nitric acid for
neutralization. What is [Ca(OH)2 ]?
Equivalence when
nCa (OH ) 2
1
 nHNO3
2