Chapter 4: Aqueous Reactions
Solution: homogeneous mixture of solute and solvent
Solvent:
substance present in the larger amount
Solute:
substance(s) dissolved in solvent, generally present in
lesser amounts than solvent
Chapter 4: Aqueous Reactions
Ionic Compounds
When ionic compounds dissolve in water, they dissociate completely
-
-
+
-
+
-
+
-
+
-
+
+ H2O
+
-
NaCl (aq)
→
Na+ (aq)
+ Cl- (aq)
Chapter 4: Aqueous Reactions
Molecular Compounds
Most molecular compounds do not dissociate in water
+ H2O
methanol
Methanol dissolved in water
Chapter 4: Aqueous Reactions
Molecular Compounds
Some molecular compounds dissociate (ionize) in water (acids)
Strong acids, such as hydrochloric acid, dissociate completely:
HCl (aq)
→
H+ (aq)
+
Cl- (aq)
Weak acids, such as acetic acid, dissociate only partially:
CH3COOH (aq)
H+ (aq)
+
CH3COO- (aq)
Chapter 4: Aqueous Reactions
HW: 1,3,18,37
Aqueous solutions that contain ions, conduct electricity
Electrolytes: substances that generate ions when dissolved in water
For example:
Non-Electrolytes:
●
Ionic compounds
strong electrolyte
●
Strong acids
strong electrolyte
●
Weak acids
weak electrolyte
●
Strong bases
strong electrolyte
●
Weak bases
weak electrolyte
substances that do not generate ions when
dissolved in water
Chapter 4: Aqueous Reactions
Strong, Weak, and Non- Electrolytes
AgI
NaCl
HCl
Ag+
Cl-
Cl-
I-
Na+
H+
Ionic
Ionic
Molecular
(dissociated acid)
sugar
C12H22O11
Molecular
Chapter 4: Aqueous Reactions
Strong, Weak, and Non- Electrolytes
Electrolytes and Non-Electrolyte definition
only refers to the molecules/ions that are dissolved
Formic acid
HCOOH
HCOOH+
Molecular
Chapter 4: Aqueous Reactions
Some reactions involving ionic compounds:
Exchange ot Metathesis Reactions
+AX
+
+BY
→
+AY
+
+BX
Chapter 4: Aqueous Reactions
Some reactions involving ionic compounds:
Exchange ot Metathesis Reactions
If one of the products in insoluble, the reaction is a precipitation reaction:
AgNO3 (aq)
AgNO3 (aq)
+
+
NaCl (aq)
NaI (aq)
→
AgCl (s) +
white precipitate
NaNO3 (aq)
AgI (s)
+ NaNO3 (aq)
→
brownish precipitate
Chapter 4: Aqueous Reactions
Precipitation Reactions
AgNO3 (aq)
+
NaI (aq)
AgI (s)
+ NaNO3 (aq)
→
brownish precipitate
heterogeneous mixture!
Chapter 4: Aqueous Reactions
HW: 15
Precipitation Reactions
Ni(NO3)2 (aq) + 2 NaOH (aq)
I) Identify ions: A: Ni2+
X: NO3-
→
Ni(OH)2 (s) +
2 NaNO3 (aq)
II) Exchange X and Y: A: Ni2+
Y: OHB: Na+
X: NO3
B: Na+
Y: OH
III) Determine stoichiometry
of compounds formed:
Ni2+
+ OH-
Na+
+
=> Ni(OH)2
NO3- => NaNO
3
IV) Balance equation!
V) Is there an insoluble product?
Chapter 4: Aqueous Reactions
Precipitation Reactions
How do you know which ionic compounds are soluble?
Chapter 4: Aqueous Reactions
Precipitation Reactions
Digest of solubility rules:
Salts of the following ions are always soluble:
●
Group IA metals
●
Li+, Na+, K+ ...
●
Ammonium
NH4+
●
Nitrate
NO3Acetate
C 2 H 3 O2 -
HW: 10,19,22
Chapter 4: Aqueous Reactions
Precipitation Reactions
(NH4)2SO4
AgCl
CuSO4
FeNO3
Cu(OH)2
CaCO3
LiCl
Ca(C2H3O2)
2
HW: 90
Chapter 4: Aqueous Reactions
HW: 23
Net Ionic Equations
Molecular Equation (shows undissociated compounds):
Ni(NO3)2 (aq)
+
2 NaOH (aq) → Ni(OH)2 (s) +
Ionic Equation:
2 NaNO3 (aq)
spectator ions
Ni2+ (aq) + 2 NO3- (aq) + 2 Na+ (aq) + 2 OH- (aq) →
Ni(OH)2 (s) + 2 Na+ (aq) + 2 NO3- (aq)
Net Ionic Equation:
Ni2+ (aq) + 2 OH- (aq) → Ni(OH)2 (s)
The spectator ions do not participate in the reaction!
Chapter 4: Aqueous Reactions
Ca(NO3)2 (aq) +
Na2CO3 (aq) → CaCO (s)
+ 2 NaNO3 (aq)
3
Spectator ions: 2 Na+, 2 NO3-
Net ionic equation:
Ca2+ (aq) + CO32- (aq) →
CaCO3 (s)
Chapter 4: Aqueous Reactions
Ca(NO3)2 (aq) + 2 NaC2H3O2 (aq) → Ca(C2H3O2)2
+ 2 NaNO3
Spectator ions: ALL !
If all salts are soluble, no precipitation reaction will take place
Chapter 4: Aqueous Reactions
Properties of Acids and Bases
Acids
Bases
●
taste sour
●
taste bitter
●
turn blue litmus red
●
turn red litmus blue
produce CO2 when
reacting with carbonates
●
produce H2 when
reacting with metals
●
generate protons, H+, when
dissolved in water
●
generate hydroxide ions, OH-,
when dissolved in water
●
Chapter 4: Aqueous Reactions
Properties of Acids and Bases
Strong Acids
= dissociate completely in water = strong electrolytes
HCl
Hydrochloric acid
HBr
Hydrobromic acid
HI
Hydroiodic acid
HNO3
Nitric acid
H2SO4
Sulfuric acid
HClO3
Chloric acid
HClO4
Perchloric acid
Chapter 4: Aqueous Reactions
Properties of Acids and Bases
Strong Bases
strong electrolytes
Metal Hydroxides of Group IA metals:
NaOH, KOH ...
Metal Hydroxides of Group IIA metals:
Ca(OH)2 , Mg(OH)2 ...
HW: 36
Chapter 4: Aqueous Reactions
The reaction of acids with carbonate salts:
MgCO3 (s) + 2 HCl (aq) →
MgCl2 (aq) +
H2CO3 (aq)
unstable
H2CO3 (aq)
→
CO2 (g) + H2O (l)
Overall:
MgCO3 (s) + 2 HCl (aq) →
MgCl2 (aq) +
CO2 (g) + H2O (l)
Chapter 4: Aqueous Reactions
HW: 39
Neutralization Reactions
acid
base
HNO3 (aq) + NaOH (aq) →
WATER
salt
H2O (l) + NaNO3 (aq)
net ionic equation:
H+(aq) + NO3- (aq) + Na+ (aq) + OH- (aq) → H2O (l) + Na+ (aq) + NO3- (aq)
H+(aq) + OH- (aq) →
H2O (l)
In a neutralization reaction, an acid and a base react to form
water and a salt
Chapter 4: Aqueous Reactions
Complete and balance the following neutralization reaction:
Mg(OH)2 (aq) + 2 HBr (aq) → MgBr2 (aq)
+ 2 H2O (l)
net ionic equation:
Mg2+ (aq) + 2 OH- (aq) + 2 H+ (aq) + 2 Br- (aq)
→
Mg2+ (aq) + 2 Br- (aq) + 2 H2O (l)
net: 2OH- (aq) + 2H+(aq) →
OH- (aq) + H+(aq) →
2H2O (l)
H2O (l)
Chapter 4: Aqueous Reactions
Concentrations of Solutions
... are measured in Molarity
Molarity
=
moles of solute
Volume of solution in L
The concentration of 0.4 L of solution containing 0.25 moles of sugar is
0.25 mol
mol
= 0.6
0.4 L
L
= 0.6 M
Chapter 4: Aqueous Reactions
HW: 62,73
Someone is preparing to cook pasta by adding 5.0 g of table salt
(NaCl, formula mass = 49.5 g/mol) to 400 mL of boiling water. What is
the molarity of the resulting NaCl solution?
moles of NaCl
M=
L of solution
(I) convert g NaCl into mol NaCl:
5.0 g NaCl 
1mol NaCl
 0.10 mol NaCl
49.5 g NaCl
(II) calculate molarity:
M
0.10 mol NaCl
400 mL
×
1000 mL
1L
=
0.25
mol NaCl
= 0.25 M NaCl
L
Chapter 4: Aqueous Reactions
What is the molarity of Na+ ions in a 0.2 M solution of Na2SO4 ?
Na2SO4 (aq)
→
2 Na+ + SO42-
Na+
SO4
Na+
2-
SO42-
Na+
Na+
Each formula unit of Na2SO4 that dissolves gives rise to
1 SO42- ion and 2 Na+ ions
+
2 Na
0.2 M Na2 SO 4 ×
1 Na2 SO 4
= 0.4 M Na+
Chapter 4: Aqueous Reactions
Proton Concentration in Aqueous Solutions
[H+] ≡ proton concentration
[H+] = 0.001 M
pH = -log [H+]
[H+] = 0.001 M = 1 x 10-3 M
and [H+] = 10-pH
pH = -log(1 x 10-3) = 3
[H+] x [OH-] = 10-14 = constant
Neutral solution: [H+] = [OH-] = 10-7 M
The pH scale
1
acidic
2
3
4
5
6
7
8
neutral
9
10
12
13
14
basic
Chapter 4: Aqueous Reactions
What are the proton and hydroxide concentrations in a solution that has a
pH of 4.3 ?
pH = -log [H+]
[H+] = 10-pH
[H+] = 10-4.3
[H+] x [OH-] = 10-14
[OH-] = 10-14
[H+]
= 10-14 =
5.0 x 10-5
Chapter 4: Aqueous Reactions
How many moles of HF are needed to make 0.15 L of a 0.13M solution?
(how many moles HF are in 0.15 L of a 0.13 M solution?)
1 L of a 0.13 M solution:
0.15 L of a 0.13 M solution:
0.13 mol HF
L
0.13mol HF
L
x
0.15 L = 0.02 mol HF
Chapter 4: Aqueous Reactions
Dilutions
add solvent
number of solute
molecules before dilution
number of moles of
solute before dilution
M conc =
=
number of solute
molecules after dilution
=
number of moles of
solute after dilution
mol solute
V conc
M conc × V conc
M dil =
=
mol solute
V dil
M dil × V dil
Chapter 4: Aqueous Reactions
What is the concentration of a solution that is made by adding 0.3L
of water to 15mL of a 0.65M solution?
M conc × V conc
=
M dil × V dil
Vconc = 15mL = 0.015L
M dil 
M dil
M conc  Vconc

Vdil
Mconc= 0.65M
0.65M  0.015 L
0.315 L
Vdil = 0.3L + 15mL
= 0.3L + 0.015L = 0.315L
 0.031 M
Chapter 4: Aqueous Reactions
HW: 79a,d
Solution Stoichiometry and Chemical Analysis
What volume of a 0.30 M HCl solution is needed to completely
react 3.5 g of Ca(OH)2 ?
2 HCl (aq)
+
Ca(OH)2 (aq) →
2 H2O (aq) + CaCl2 (aq)
Strategy:
3.5 g Ca(OH)2 → moles Ca(OH)2
molar mass = 74g/mol
→ moles HCl
stoichiometric factor
from equation
→
Liters HCl
molarity of
solution
Chapter 4: Aqueous Reactions
Solution Stoichiometry and Chemical Analysis
What volume of a 0.30 M HCl solution is needed to completely
react 3.5 g of Ca(OH)2 ?
2 HCl (aq)
+
Ca(OH)2 (aq) →
2 H2O (aq) + CaCl2 (aq)
1mol Ca (OH ) 2
2mol HCl
1L HCl
 0.32 L HCl
3.5 g Ca (OH ) 2 


74 g Ca (OH ) 2 1mol Ca (OH ) 2 0.30 mol HCl