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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
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
Strong electrolyte
Ionic
Strong electrolyte
Molecular
(dissociated acid)
Strong electrolyte
sugar
C12H22O11
Molecular
Non-electrolyte
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
If the solution contains any type of
ion, it is an electrolyte
If all the dissolved particles are
ions, the solution is a strong electrolyte
HCOOH
HCOOH+
Molecular
Weak electrolyte
If only some of the dissolved particles are
ions, the solution is a weak electrolyte
Chapter 4: Aqueous Reactions
Aqueous solutions that contain ions, conduct electricity
Substance
Ionic
Molecular
soluble
insoluble
highly soluble
in H2O
very little soluble
in H2O
strong
electrolyte
strong
electrolyte
Acid
weak
Base
Neither acid
nor base
strong
non-electrolyte
weak
strong
electrolyte electrolyte
weak
weak
electrolyte
strong bases are
usually ionic
compunds
Chapter 4: Aqueous Reactions
Sample 1
Sample 2
Sample 3
Chapter 4: Aqueous Reactions
(1)
(2)
soluble in water
(1) AgNO3 (aq)
(2) Pb(NO3)2 (aq)
+
NaCl (aq)
+
2 NaI (aq)
insoluble in water (precipitate)
→
AgCl (s) +
NaNO3 (aq)
→
PbI2 (s) +
2 NaNO3 (aq)
Chapter 4: Aqueous Reactions
Some reactions involving ionic compounds:
Exchange or Metathesis Reactions
+AX
+
+BY
→
+AY
+
+BX
Chapter 4: Aqueous Reactions
Exchange or 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
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+
Na+
+
2 OH- => Ni(OH)2
+
NO3-
=> NaNO3
IV) Balance equation!
V) Is there an insoluble product?
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 [CH3COO-, AcO-]
Chapter 4: Aqueous Reactions
Precipitation Reactions
How do you know which ionic compounds are soluble?
Chapter 4: Aqueous Reactions
Precipitation Reactions
(NH4)2SO4
soluble (NH4+ salt)
AgCl
insoluble
CuSO4
soluble
FeNO3
soluble (NO3- salt)
Cu(OH)2
insoluble
CaCO3
insoluble
LiCl
soluble (Li+ salt)
Ca(C2H3O2)2
soluble (C2H3O2- [acetate]
salt)
Chapter 4: Aqueous Reactions
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) → CaCO3 (s)
+ 2 NaNO3 (aq)
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(aq) + 2 NaNO3 (aq)
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
Acids
produce CO2 when
reacting with carbonates
●
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
Which of the following is insoluble?
a. (NH4)CO3
b. ZnS
c. K2CO3
d. Zn(NO3)2
Chapter 4: Aqueous Reactions
Properties of Acids and Bases
Strong Acids
when dissolved, 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
Weak Acids
when dissolved, do not dissociate completely in water
= weak electrolytes
for example:
HF
Hydrofluoric acid
CH3COOH
Acetic acid
HCOOH
Formic 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 ...
Chapter 4: Aqueous Reactions
Properties of Acids and Bases
Weak Bases
generate few ions when dissolved
= weak electrolytes
for example:
NH3
Ammonia
NH3 (aq) + H2O (l) → NH4+ (aq) + OH- (aq)
Chapter 4: Aqueous Reactions
Neutralization Reactions
acid
base
HNO3 (aq) + NaOH (aq) →
WATER
salt (ionic compound)
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 (ionic compound)
Chapter 4: Aqueous Reactions
What is the net ionic reaction for the reaction between insoluble
lead(II) hydroxide and nitric acid?
base
acid
Pb(OH)2 (s) + 2 HNO3 (aq) →
salt
water
Pb(NO3)2 (aq) + 2 H2O(l)
When writing an ionic equation, which compounds are written in ionic form?
a. All compounds in the equation
b. Reactants only
c. Compounds that generate ions
d. Hydroxide ions and protons
(aq)
Chapter 4: Aqueous Reactions
What is the net ionic reaction for the reaction between insoluble
lead(II) hydroxide and nitric acid?
base
acid
Pb(OH)2 (s) + 2 HNO3 (aq) →
→ Pb2+ (aq) + 2 NO3- (aq) + 2 H2O (l)
Pb(OH)2 (s) + 2 H+(aq) + 2 NO3- (aq)
Pb(OH)2 (s) + 2 H+(aq)
salt
water
Pb(NO3)2 (aq) + 2 H2O(l)
→
Pb2+ (aq) + 2 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
Someone is preparing to cook pasta by adding 5.0 g of table salt
(NaCl, 49.5 g/mol) to 400 mL of boiling water. What is the molarity of
the resulting NaCl solution?
5.0 g
M
MM
=
moles
concentration (molarity = mol/L)
Volume
moles of NaCl
L of solution
(I) convert g NaCl into mol NaCl:
1mol NaCl
5.0 g NaCl 
 0.10 mol NaCl
49.5 g NaCl
(II) calculate molarity:
0.10 mol NaCl
M
400 mL
×
1000 mL
1L
=
mol NaCl
0.25
= 0.25 M NaCl
L
Chapter 4: Aqueous Reactions
How many moles of HF are needed to make 0.15 L of a 0.13M HF solution?
0.13 M (= mol/L) HF
0.13mol HF
L
x
Volume
0.15 L =
mol HF
0.02 mol HF
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+]
-pH = log [H+]
10-pH = [H+]
[H+] = 10-4.3
= 5.0 x 10-5 M
[H+] x [OH-] = 10-14
[OH-] = 10-14
[H+]
= 10-14 =
5.0 x 10-5
2 x 10-10 M
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?
Dilution: adding solvent without changing number of moles
higher concentration
M conc × V conc
=
M dil × V dil
Vconc = 15mL = 0.015L
M dil 
lower concentration
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
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 (l) +
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
Chapter 4: Aqueous Reactions
What is the molarity of Na+ ions in a 0.2 M solution of Na2SO4 ?
molarity Na+ ions
molarity Na2SO4
empirical formula
Na2SO4 (aq)
→
2 Na+ (aq) + SO42- (aq)
Na+
Na+
SO42Na+
Na+
+
0.2 M Na2 SO 4 ×
2 Na
1 Na2 SO 4
= 0.4 M Na+
SO42-
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