Transcript Chapter 2

Chapter 5
Reactions in Aqueous
Solutions
Chapter goals
• Understand the nature of ionic substances
dissolved in water.
• Recognize common acids and bases and
understand their behavior in aqueous
solution.
• Recognize and write equations for the
common types of reactions in aqueous
solution.
• Recognize common oxidizing and reducing
agents and identify common oxidationreduction reactions.
• Define and use the molarity in solution
stoichiometry.
Solution
• homogeneous mixture
• can be gas, liquid, or solid
• solvent: component present in highest
proportion
exception - water
• solute: component(s) in solution other
than solvent
We will mostly study aqueous solutions:
human body is 2/3 water.
Examples
• mixture of 35% naphthalene
and 65% benzene
solvent - benzene
solute – naphthalene
• mixture of 10% ethanol, 40% methanol,
and 50% propanol
solvent - propanol
solute - ethanol and methanol
Examples
• mixture of 40% ethanol, 40% methanol,
and 20% butanol
solvent - ethanol/methanol
mixed solvent
solute – butanol
• mixture of 40% ethanol, 50% propanol, and
10% water
solvent - water
solute - ethanol and propanol
Next…
We will focus on compounds that
produce ions in aqueous solutions.
They are named electrolytes and may be
salts, acids, or bases.
Salts
Salts: ionic compounds made of cations other
than H+ and anions other than OH− or O2−, O22−
NaCl: Na+ & Cl−
K2SO4: K+ & SO42−
FeBr3: Fe3+ & Br−
Zn3(PO4)2: Zn2+ & PO43−
Ca(HCO3)2: Ca2+ & HCO3−
Electrolyte
• substance that dissolves to produce an
electrically conducting medium
• forms ions in solution (dissociates/ionizes)
• examples
soluble ionic compounds
H2O
KBr(s)  K+(aq) + Br–(aq)
H2O
Acids, HCl(g)  H+(aq) + Cl–(aq)
bases, NH3 + H2O
NH4+ + OH–
double arrow means equilibrium
Nonelectrolytes
• do not form ions in solution
• do not form electrically conducting media
upon dissolution
• Examples: molecular compounds
(alcohols, sugars & acetone)
H 2O
CH3OH(l)  CH3OH(aq) N.D.
Glucose C6H12O6(s)  C6H12O6(aq)
N.D.
Sucrose C12H22O11(s)  C12H22O11(aq) N.D.
N.D. = no dissociation/ionization
Types of Electrolytes
• Strong: dissociate ~100%
most ionic compounds (soluble salts), strong
acids, and strong bases
H2O
KBr(s)  K+(aq) + Br–(aq)
HCl(g)  H+(aq) + Cl–(aq)
NaOH(s)  Na+(aq) + OH−(aq)
Weak: insoluble salts, weak acids and bases,
water (H2O), and certain gases (e.g. CO2)
H2O
dissociate only slightly in water
HF(g)
H+(aq) + F–(aq)
Also acetic acid, CH3COOH
NH3 + H2O
NH4+(aq) + OH–(aq)
Solubility of Ionic compounds in Water:
Solubility Rules
Soluble Compounds
1. alkali metal salts (Li+, Na+, K+, Rb+…, )
except potassium perchlorate
2. ammonium (NH4+) salts
3. all nitrates(NO3−), chlorates (ClO3−),
perchlorates (ClO4−), and acetates (C2H3O2−),
except silver acetate and potassium
perchlorate
4. all Cl−, Br−, and l− are soluble except for
Ag+, Pb2+, and Hg22+ salts
5. all SO42− are soluble except for Pb2+, Sr2+,
and Ba2+ salts
Solubility of Ionic compounds in Water: Rules
Insoluble or slightly soluble Compounds
6. metal oxides (O2−) except those of the
alkali metals, Ca2+, Sr2+, and Ba2+
7. hydroxides (OH−) except those of the
alkali metals, Ba2+, Sr2+, and NH4+. Calcium
hydroxide is slightly soluble
8. carbonates, phosphates, sulfides, and
sulfites except those of the alkali metals and
the ammonium ion (NH4+)
9. for salts of Cr2O72−, P3−, CrO42−, C2O42−,
assume they are insoluble except for IA
metals and NH4+ salts
•
•
•
•
Precipitation Reactions:
A Driving Force in Chemical Reactions
formation of insoluble solid (precipitate,
ppt) is a common reaction in aqueous
solutions:
reactants are generally water-soluble
ionic compounds
once substances dissolve in water they
dissociate to give the appropriate cations
and anions
if the cation of one compound forms an
insoluble compound with the anion of
another, precipitation will occur
Precipitation Reaction:
A Double Replacement (Metathesis) Reaction
• Both ionic compounds trade partner ions
__________
|
|
AB(aq) + CD(aq)  AD(s) + CB(aq)
|_______|
AD is an insoluble or slightly soluble salt
A+, B−, C+, and D− are ions
AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
weak electrolyte
(unionized precipitate)
Precipitation Reaction:
A Double Replacement (Metathesis) Reaction
A (solid) precipitate is formed.
Example: complete and balance the equation
(NH4)3PO4(aq) + MgSO4(aq)  MgPO4(s) + NH4SO4(aq)
we will write the right subscripts later
Using the solubility rules, predict if at least one
product is going to be insoluble in water.
According to rule 8, MgPO4 (subscripts not
right) is not soluble in water.
Ions are Mg2+, PO43−, NH4+, and SO42−;
subscripts
(NH4)3PO4(aq) + MgSO4(aq)  Mg3(PO4)2(s) + (NH4)2SO4(aq)
balancing
2(NH4)3PO4(aq) + 3MgSO4(aq)  Mg3(PO4)2(s) + 3(NH4)2SO4(aq)
Example: complete and balance the equation
• Na2SO4(aq) + BaBr2(aq) 
Na2SO4(aq) + BaBr2(aq) BaSO4(s) + NaBr
Na2SO4 + BaBr2 BaSO4(s) + 2NaBr(aq)
driving force = formation of insoluble
barium sulfate (precipitate)
• Os(NO3)5(aq) + Rb2S(aq) 
Os(NO3)5 + Rb2S  Os2S5(s) + RbNO3(aq)
2 Os(NO3)5 + 5 Rb2S  Os2S5(s) + 10 RbNO3
driving force = form. of insoluble Os5+ sulfide
(precipitate)
Net Ionic Equations: Spectator Ions
The equation
AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
is not quite correct, because three salts are
dissociated in ions while AgCl is a precipitate.
Ag+(aq) + NO3−(aq) + Na+(aq) + Cl−(aq)  AgCl(s) + Na+(aq) + NO3−(aq)
before reaction
after reaction
Na+ and NO3− are present on both sides of equation, i.e.,
before and after reaction. They are called spectator ions;
they do not participate in net reaction; they can be removed
from the equation, but they remain in the solution.
Ag+(aq) + Cl−(aq)  AgCl(s) is the net ionic equation
Net Ionic Equations: Spectator Ions
For two previous examples:
2(NH4)3PO4(aq) + 3MgSO4(aq)  Mg3(PO4)2(s) + 3(NH4)2SO4(aq)
6NH4+(aq) + 2PO43−(aq) + 3Mg2+(aq) + 3SO42−(aq)  Mg3(PO4)2(s) +
6NH4+(aq) + 3SO42−(aq)
after reaction
before reaction
3Mg2+(aq) + 2PO43−(aq)  Mg3(PO4)2(s)
is the net equation
spectator ions are eliminated from the equation
=======================
Na2SO4(aq) + BaBr2(aq) BaSO4(s) + 2NaBr(aq)
2Na+(aq) + SO42−(aq) + Ba2+(aq) + 2Br−(aq)  BaSO4(s) + 2Na+(aq) + 2Br−
Ba2+(aq) + SO42−(aq)  BaSO4(s)
net ionic equation
Net Ionic Equations: Spectator Ions
For the metathesis reaction
2 KF(aq) + Pb(NO3)2(aq)  PbF2(s) + 2 KNO3(aq)
formula unit equation
spectator ions are eliminated
2K+ + 2F– + Pb2+ + 2 NO3–  PbF2 + 2K+ + 2NO3–
ionic equation
PbF2 is the precipitate
2F–(aq) + Pb2+(aq)  PbF2(s)
net ionic equation
Net Ionic Equations: Spectator Ions
NH4Cl(aq) + KNO3(aq)  NH4NO3(aq) + KCl(aq)
NH4+ + Cl– + K+ + NO3–  NH4+ + NO3– +
K+ + Cl–
all ions are spectators; all can be cancelled
no net ionic equation
no driving force for reaction
N.R. (there is no reaction)
The two salts are just dissolved in water.
Acids and Bases
Acid
• Arrhenius definition
substance that ionizes in water to produce
H+, hydrogen ion, and hence increases the
concentration of this ion
HCl(aq)  H+(aq) + Cl–(aq)
• Brønsted-Lowry definition
substance capable of donating H+
HCl + H2O  H3O+ + Cl–(aq)
Acids and Bases
Base
• Arrhenius definition
substance that increases the concentration of
OH– in aqueous solution
KOH(aq)  K+(aq) + OH–(aq)
NH3 + H2O
NH4+ + OH–
• Brønsted/Lowry definition
substance capable of accepting H+
KOH(aq)  K+(aq) + OH–(aq)
OH– + H+  H2O (OH– from NaOH accepts H+)
NH3 + H2O
NH4+ + OH– (NH3 accepts H+)
Water can act as both an acid and a base: it
is an amphoteric substance
HClO4 + H2O  H3O+
acid
base
(accepts H+ from HClO4)
+ ClO4–
NH3 + H2O
NH4+ + OH–
base
acid
(donates H+ to NH3)
Strong Acids
dissociate ~100%
• HCl, HBr, HI (no HF)
hydro…ic acid
• HNO3
nitric acid
• HClO3
chloric acid (moderate)
• HClO4
perchloric acid
• H2SO4 (first proton) sulfuric acid
H2SO4(aq)  H+(aq) + HSO4−(aq)
2nd weak: HSO4−(aq)
H+(aq) + SO42−(aq)
Weak Acids
• dissociate <100%
• most other acids
HF
hydrofluoric acid
HCN
hydrocyanic acid
HNO2
nitrous acid
CH3CO2H acetic acid
H2CO3
carbonic acid (both protons)
H3PO4
phosphoric acid (all protons)
H2SO3
sulfurous acid (both protons)
oxalic acid H2C2O4(aq)
H+(aq) + HC2O4−(aq)
Strong Bases
dissociate ~100%
• alkali metal hydroxides
LiOH, NaOH, KOH, RbOH
name: lithium hydroxide
• hydroxide of
Ca
Ca(OH)2 calcium hydroxide
Ba
Ba(OH)2
Sr
Sr(OH)2
Ammonia, NH3, is a weak base
Neutralization Reactions
• acid + OH-ctg. base  salt + water
(a double replacement reaction)
HF(aq) + KOH(aq)  KF(aq) + H2O
HF(aq) + K+(aq) + OH–(aq)  K+(aq) + F–(aq) + H2O
HF(aq) + OH–(aq)  F–(aq) + H2O net ionic
spectator ions are eliminated from equation
HF is a weak acid and HCl is a strong acid
• acid + non-OH-ctg base  salt
HCl(aq) + NH3(aq)  NH4Cl(aq)
H+(aq) + Cl–(aq) + NH3(aq)  NH4+(aq) + Cl–(aq)
H+(aq) + NH3(aq)  NH4+(aq) net ionic equation
Neutralization Reactions
• Strong acid + strong base  salt + water
(a double replacement reaction)
HClO3(aq) + NaOH(aq)  NaClO3(aq) + H2O
chloric acid
H+(aq) + ClO3−(aq) + Na+(aq) + OH–(aq) 
Na+(aq) + ClO3–(aq) + H2O
H+(aq) + OH–(aq)  H2O
net ionic equation
spectator ions, ClO3− + Na+, are eliminated from
equation
Formation of a Weak Acid or Base as a Driving Force
(another double replacement reaction)
• HNO3(aq) + KCN(aq)  HCN(aq) + KNO3(aq)
H+(aq) + NO3–(aq) + K+(aq) + CN–(aq)  HCN (aq) +
K+(aq) + NO3–(aq)
H+(aq) + CN–(aq)  HCN(aq) (a weak acid)
• NH4Cl + NaOH(aq)  NH4OH + NaCl(aq)
NH4+(aq) + Cl–(aq) + Na+(aq) + OH–(aq)  NH4OH
Na+(aq) + Cl–(aq)
NH4+(aq) + OH–(aq)  NH4OH (a weak base)
NH4OH is NH3 in water, i.e., NH3 + H2O
When no Weak Electrolytes are Formed
• HNO3(aq) + KCl(aq)  HCl(aq) + KNO3(aq)
H+(aq) + NO3–(aq) + K+(aq) + Cl–(aq) 
H+(aq) + Cl–(aq) + K+(aq) + NO3–(aq)
There is no net reaction: N.R. No driving force
All ions are spectators.
• BaCl2(aq) + 2NaOH(aq)  Ba(OH)2(aq) + 2NaCl(aq)
Ba2+(aq) + 2Cl–(aq) + 2Na+(aq) + 2OH–(aq) 
Ba2+(aq) + 2OH–(aq) + 2Na+(aq) + 2Cl–(aq)
There is no net reaction: N.R. No driving force
Gas Forming Reactions (a Driving Force)
Some of the weak acids and bases that are formed at
double replacement reactions decompose to form a
gas and water
CO2
Na2CO3(aq) + 2HCl(aq)  H2CO3(aq) + 2NaCl(aq)
H2CO3(aq)  H2O + CO2(g)
Na2CO3(aq) + 2HCl(aq)  H2O + CO2(g) + 2NaCl(aq)
SO2
Na2SO3(aq) + 2HCl(aq)  H2SO3(aq) + 2NaCl(aq)
H2SO3(aq)  H2O + SO2(g)
Na2SO3(aq) + 2HCl(aq)  H2O + SO2(g) + 2NaCl(aq)
Redox (Oxidation-Reduction) Reactions
•
•
•
•
involve transfer of electron(s)
oxidation: loss of electron(s)
reduction: gain of electron(s)
some can be identified when an
uncombined element is a reactant or a product
•
•
•
•
•
eg. Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
Zn  Zn2+
Zn(s)  Zn2+(aq) + 2 e–, oxidation
Cu2+  Cu
Cu2+(aq) + 2e–  Cu(s), reduction
Single Displacement Reactions
•
•
•
•
Zn(s) + CuCl2(aq)  Cu(s) + ZnCl2(aq)
Zn oxidized to Zn2+
Cu2+ reduced to Cu
occurs because zinc is more active than
copper
•
•
•
•
Cl2(g) + CuBr2(aq)  Br2(l) + CuCl2(aq)
Br oxidized from Br– to Br2
Cl reduced from Cl2 to Cl–
Cl is more active than Br
Oxidation Numbers
• also an accounting tool
• very useful
• oxidation numbers of all atoms in
substance add up to charge on substance
e.g. (charge of species)
zero for Al2(SO4)3 and H3PO4
+1 for NH4+
–2 for Cr2O72–
Assigning Oxidation Numbers, ON
• ON = 0 for all atoms in any substance in
most elemental form, Na(s), Zn(s), Hg(l)
H2(g), Cl2(g), I2(s), O2(g), C(s), P4(s), S8(s)
• ON = charge for monatomic ions (Na+, S2−)
• ON = –1 for F in all compounds
• ON = –2 for O in compounds, usually
– exceptions: peroxide, O22–, ON = –1
–
superoxide, O2–, ON = –1/2
• ON = +1 for H in compounds, usually
– exception: ON = –1 in metallic hydrides
Assigning Oxidation Numbers, ON
• ON = +1 for alkali metals in compounds
• ON = +2 for alkaline earth metals in
compounds
• ON = +3 for Al in compounds
• ON = −1 for Cl, Br, and I (iodine) in binary
compounds except for those with oxygen
(in these cases they variable positive)
• F is always −1
Assign ON to Each Atom in the
Following Substances
?
–1
WCl6
x + 6(–1) = 0
x –6 = 0
x = +6
+1
?
–2
Na2S2O3
+2 + 2x – 6 = 0
2x = 6 – 2
x = +2
ON of S
–2
+1
?
Na2S4O8
+2 + 4x –16 = 0
4x = 16 – 2
+14
+7
x = —— = ——
4
2
ON of S
?
–2
Cr2O72–
2x –14 = –2
2x = 14 – 2
+12
x = —— = +6
2
ON of Cr
+1
–2
H2C2O4
+2 + 2x – 8 = 0
2x = 8 – 2
+6
x = —— = +3
2
ON of C
?
–1
MoBr5+
x – 5 = +1
x=5+1
x=6
ON of Mo = +6
Oxidizing and Reducing Agents
In every redox reaction there is (at least) a
Reducing agent (the one that is oxidized) and
an oxidizing agent (the one that is reduced)
ON increases
The species is
oxidized
+7
+6
+5
+4
+3
+2
+1
0
−1
−2
−3
−4
−5
−6
−7
ON decreases
The species is
reduced
• Activity (Electromotive) Series
for metals
Activity
increases
Li
K
Ba
Ca
Na
Mg
Al
Mn
Zn
Cr
Fe
Co
Ni
Sn
Pb
(H2)
Cu
Ag
Hg
Pt
Au
Activity
decreases
Examples
• Complete and balance each of the
following chemical equations
Li
A free and chemically active metal displacing K
Ba
a less active metal from a compound
Ca
• Mg + FeCl3 
Na
Mg
Mg + FeCl3  Fe + MgCl2
Al
3
2
Mn
Mg  Mg2+, oxidized; Mg reducing agent Zn
Cr
Fe3+  Fe, reduced; Fe3+ oxidizing agent Fe
Co
Ni
3
Sn
Sn + CrF3  No Reaction
Pb
H2
Cu
3 3
Hg
Pb(s) + Au(ClO3)3(aq)  Au(s) + Pb(ClO3)2(aq) Ag
Pt
3Pb(s) + 2Au(ClO3)3(aq)  2Au(s) + 3Pb(ClO3)2(aq) Au
+3
3Mg(s) + 2FeCl (aq)  2Fe(s) + 3MgCl (aq)
• Sn + CrF  Sn is less reactive than Cr
• Pb(s) + Au(ClO ) (aq) 
Pb is oxidized
Au
is reduced
A free and chemically active metal displacing
a less active metal from a compound
• Zn + CrBr3 
Zn + CrBr3  Cr + ZnBr2
3Zn(s) + 2CrBr3(aq)  2Cr(s) + 3 ZnBr2(aq)
• Zn oxidized to Zn2+; Zn reducing agent
• Cr3+ reduced to Cr; Cr3+ oxidizing agent
• Ag(s) + Hg(NO3)2  No Reaction
Ag is less reactive than Hg
Li
K
Ba
Ca
Na
Mg
Al
Mn
Zn
Cr
Fe
Co
Ni
Sn
Pb
H2
Cu
Hg
Ag
Pt
Au
A free and chemically active metal displacing
Li
K
Hydrogen from acids or water
Ba
Ca
• Fe + HBr 
Na
Mg
Fe + HBr  H2 + FeBr3
0
+1
0
+3
2Fe + 6HBr  3H2 + 2 FeBr3
Fe oxidized to Fe3+; Fe reducing agent
H+ reduced to H2;
H+ oxidizing agent
• Cu + HBr  Cu less active than H2
Cu + HBr  No Reaction
Al
Mn
Zn
Cr
Fe
Co
Ni
Sn
Pb
H2
Cu
Hg
Ag
Pt
Au
A free and chemically active metal displacing
Li
K
Hydrogen from acids or water
Ba
• K(s) + H2O(l) 
Ca
0
+1
+1
0
2K(s) + 2H2O(l)  2 KOH(aq) + H2(g)
K oxidized to K+; K reducing agent
H+ reduced to H2;
H+ oxidizing agent
• Ag(s) + H2O(l)  Ag less active than H2
Ag(s) + H2O(l)  No Reaction
• Ni(s) + H2SO4(aq) 
0
+1
+2
0
Ni(s) + H2SO4(aq)  NiSO4(aq) + H2(g)
Ni oxidized to Ni2+; Ni reducing agent
H+ reduced to H2;
H+ oxidizing agent
Na
Mg
Al
Mn
Zn
Cr
Fe
Co
Ni
Sn
Pb
H2
Cu
Hg
Ag
Pt
Au
• Which one of the following metals could
be used safely for lining a tank intended
for storage of sulfuric acid?
H2SO4
• aluminum
• iron
• chromium
• mercury
• copper
• tin
Li
K
Ba
Ca
Na
Mg
Al
Mn
Zn
Cr
Fe
Co
Ni
Sn
Pb
H2
Cu
Hg
Ag
Pt
Au
Nonmetal Activity Series
F
Cl
Br
I
same order as in periodic table
F > Cl > Br > I
Complete and Balance
Cl2 + FeBr3 
Cl2(g) + FeBr3(aq)  Br2(l) + FeCl3(aq)
•
0
−1
0
−1
3Cl2(g) + 2FeBr3(aq)  3 Br2(l) + 2 FeCl3(aq)
Cl2 oxidizing agent; Br– reducing agent
•
I2(s) + NaF(aq)  I2 is less active than F2
I2(s) + NaF(aq)  No Reaction
•
F2(g) + NaCl(aq) 
0
−1
o
−1
F2(g) + 2 NaCl(aq)  Cl2(g) + 2 NaF(aq)
F2 oxidizing agent; Cl– reducing agent
•
Br2 + FeCl3  Br less active than Cl
Br2 + FeCl3  No Reaction
Identifying Oxidizing and Reducing Agents
0
+3
0
+2
3Zn(s) + 2CrBr3(aq)  2Cr(s) + 3 ZnBr2(aq)
• Zn oxidized to Zn2+; Zn reducing agent
• Cr3+ reduced to Cr; Cr3+ oxidizing agent
0
+1
+2
0
Ni(s) + H2SO4(aq)  NiSO4(aq)) + H2(g)
Ni oxidized to Ni2+; Ni reducing agent
H+ reduced to H2;
H+ oxidizing agent
0
-1
0
-1
F2(g) + 2 NaCl(aq)  Cl2(g) + 2 NaF(aq)
F2 reduced; Cl– oxidized
F2 oxidizing agent; Cl– reducing agent
Identifying Oxidizing and Reducing Agents
• The device for testing breath for the presence
of alcohol is based on the following reaction.
Identify the oxidizing and reducing agents
ON: –1
+6
3CH3CH2OH(aq) + 2Cr2O72–(aq) + 16H+ 
ethanol
orange-red
+3
3CH3CO2H(aq) + 4 Cr3+(aq) + 11H2O
acetic acid
green
ethanol is oxidized (reducing agent)
Cr2O72– (dichromate ion) is reduced (oxidizing agt.)
Balancing redox equations
in acidic media (presence of H+)
Cr2O72–(aq) + Fe2+(aq) 
Cr3+(aq) + Fe3+(aq)
ON: +6
Cr2O72– + 6e−  2Cr3+
Fe2+ 
Fe3+ + e−
(this the reduction)
(this is the oxidation)
Cr2O72– + 6e− + 14H+  2Cr3+
(to have +6 = +6, charge)
Cr2O72– + 6e− + 14H+  2Cr3+ + 7H2O (to balance H & O)
Cr2O72– + 6e− + 14H+  2Cr3+ + 7H2O
6(Fe2+ 
Fe3+ + e−)
to equal # of
electrons, 6
Cr2O72– + 14H+ + 6Fe2+  2Cr3+ + 7H2O + 6Fe3+
Balancing redox equations
in basic media (OH−) +4
+6
MnO4–(aq) + SO32−(aq)

MnO2(s) +
SO42−(aq)
ON:
+7
+4
MnO4– + 3e−  MnO2
SO32−  SO42− + 2e−
(this the reduction)
(this is the oxidation)
MnO4– + 3e−  MnO2 + 4OH−
SO32− + 2OH−  SO42− + 2e−
(to equal charges, −4)
(to equal charges, −4)
2(MnO4– + 3e− +2H2O  MnO2 + 4OH−)
3(SO32− + 2OH−  SO42− + 2e− + H2O)
to equal H, O,
and electrons
Measuring Concentrations of Compounds in
Solutions
Concentration Terms
Parts Per
• Hundred (percent, %)
weight/weight, %(w/w)
(most common)
mass solute (g)
———————— x 100
mass solution (g)
volume/volume, %(v/v)
V solute (mL)
———————— x 100
V solution (mL)
liquid solute in
liquid solvent
Parts Per
• Hundred (percent, %)
weight/volume, %(w/v)
mass solute (g)
———————— x 100
V solution (mL)
Learning Check
A solution is prepared by mixing 15.0 g of Na2CO3
and 235 g of H2O. The final V of solution is 242
mL. Calculate the %w/w and %w/v concentration
of the solution.
g solution = 15.0 g Na2CO3 + 235 g H2O = 250. g
15.0 g solute
%w/w = ——————— x 100 = 6.00% Na2CO3
250. g solution
15.0 g solute
%w/v = ——————— x 100 = 6.20% Na2CO3
242 mL solution
Molarity, M
The Molarity, M, usually known as the molar
concentration of a solute in a solution, is the
number of moles of solute per liter (1000 mL)
of solution / or mmoles per mL of solution.
To calculate it we need moles of solute
and V(in liters) of solution (or mmol and mL)
and to divide
mol solute
mmol solute
M = —————— = ———————
V(L) solution
V(mL) solution
Calculation of Molarity, M
What is the molarity of 500. mL NaOH solution if
it contains 6.00 g NaOH?
500. mL  1000 = 0.500 L (volume in liters)
FW (NaOH) = 40.0 g/mol (from periodic table)
How many moles of NaOH?
1 mol NaOH
6.00 g x —————— = 0.150 mol NaOH
40.0 g NaOH
mol solute
0.150 mol
M = —————— = ————— = 0.300 M NaOH
V(L) solution
0.500 L
(0.300 mol in 1 L or 0.300 mmol in 1 mL)
Formality, F
is the same as molarity, but referred to ionic
compounds in aqueous solution
FW (formula weights) of solute per L of solution
1 FW
# FW = g solute x ————— = # of FW = # of moles
g solute
# FW
F = ——————
V(L) solution
Formality = Molarity
(# of formula weights)
(volume in liters)
Molality, m
is the amount (moles) of solute per kg of
solvent (usually but not necessarily water).
What is the m of a solution prepared by
dissolving 25.3 g Na2CO3 in 458 g water?
458 g = 0.458 kg (after dividing by 1000)
1 mol Na2CO3
25.3 g Na2CO3 x —————— = 0.239 mol Na2CO3
106.0 g Na2CO3
mol solute
0.239 mol
m = —————— = ————— = 0.522 m Na2CO3
kg H2O
0.458 kg
(0.522 mol in 1 kg H2O)
Mole Fraction, X
is the amount (mol) of a given component of a
solution per mol of solution.
Here we need # moles of every component
(solute(s) and solvent) and the total
e.g. for a solution with n1, n2, n3, … mol
ni
ni
mol fraction Xi = —————— = —— (no units)
n1+ n2+ n3 + …
nt
ni: moles of component i (1, 2, 3, …)
nt: total number of moles
How many g of NaCl are contained in
250.0 mL of 0.2193 M NaCl solution?
250.0 mL  1000 = 0.2500 L
Now, M as a conversion factor
0.2193 mol
0.2500 L x ————— = 0.0548 mol NaCl
1 L sol’n
grams out of moles and formula weight (58.44)
58.44 g NaCl
0.0548 mol x —————— = 3.204 g NaCl
1 mol NaCl
Making Solutions
• Consider the making of 1.00 L of 1.00 M NaCl
solution.
• need 1.00 mol NaCl or 58.5 g NaCl
• dissolve 58.5 g NaCl in 1.00 L water?
NO!!
• dissolve 58.5 g NaCl in less than 1.000 L water
and dilute to a total volume of 1.000 L
Example: Describe the preparation of 300.0
mL of 0.4281 M silver nitrate solution.
300.0 mL  1000 = 0.3000 L
AgNO3 FW = 169.97 g/mol
0.4281 mol
0.3000 L x ————— = 0.1284 mol AgNO3
1 L sol’n
169.97 g
0.1284 mol x ————— = 21.82 g AgNO3
1 mol AgNO3
• Dissolve 21.82 g AgNO3 in 300.0 mL water?
NO!!
• Dissolve 21.82 g AgNO3 in water and dilute to
300.0 mL.
How many milliliters of 2.00 M HNO3 contain
24.0 g HNO3?
HNO3 FW = 63.0 g/mol
How many moles?
1 mol HNO3
24.0 g HNO3 x —————— = 0.381 mol HNO3
63.0 g HNO3
Now the M with the volume on top to get mL
1 L sol’n
1000 mL
0.381 mol x —————— x ———— = 191 mL
2.00 mol HNO3 1 L sl’n
How many grams of AlCl3 are needed to
prepare 25 mL of a 0.150 M solution?
25 mL  1000 = 0.025 L
FW (AlCl3) = 133.5 g/mol
All at once:
V(L) of sol’n and
M to calculate
mol of AlCl3
mol and FW
to calculate
g of AlCl3
0.025 L x 0.150 mole x 133.5 g = 0.500 g AlCl3
1L
1 mole
Dilution
the process of decreasing the concentration
of solutes in a solution by addition of solvent
or another solution that does not contain the
same solutes
• volume increases and concentration
decreases.
concentrated
diluted solution
volumetric flask: calibrated to contain, tc
pipet: calibrated to deliver, td
Example: Calculate the concentration of a
solution made by diluting 25.0 mL of 0.200 M
methanol, CH3OH, solution to 100.0 mL.
Key for calculations: moles of solute taken
from the concentrated solution are the same
in the diluted solution (only solvent is added.)
moles = concentration x V = M x V
Cc x Vc = Cd x Vd
Mc x Vc = Md x Vd
one of the four is unknown
c: concentrated d: diluted
L or mL can be used for volume
Example: Calculate the concentration of
a solution made by diluting 25.0 mL of
0.200 M methanol, CH3OH, to 100.0 mL.
Mc x V c = Md x V d
Md is the unknown
25.0 mL x 0.200 M = 100.0 mL x Md
25.0 mL x 0.200 M
Md = ————————— = 0.0500 M
100.0 mL
0.0500 mole/L
How many mL of a 0.515 M NaBr solution
must be diluted to produce 500.0 mL of a
0.103 M NaBr solution?
Mc x V c = Md x V d
Vc is the unknown
Vc x 0.515 M = 500.0 mL x 0.103 M
500.0 mL x 0.103 M
Vc = ————————— = 100.0 mL
0.515 M
Serial Dilutions
Example: Calculate the M of NaOH in a solution
made by diluting 25.0 mL of 0.928 M NaOH to
200.0 mL and then diluting 50.0 mL of the
second solution to 100.0 mL.
Mc x Vc = Md x Vd
First dilution:
25.0 mL x 0.928 M
Md = ————————— = 0.116 M
200.0 mL
Second dilution:
50.0 mL x 0.116 M
Md = ————————— = 0.0580 M
100.0 mL
Solution Stoichiometry
Use of M, V, and coefficients in equations to
calculate any amount of reagent or product.
• Example: What volume of 0.273 M
potassium chloride solution is required to
react with exactly 0.836 mmol of silver
nitrate?
• KCl(aq) + AgNO3(aq) 
• KCl(aq) + AgNO3(aq)  KNO3(aq) + AgCl(s)
• driving force, formation of precipitate AgCl
• Key: work with mmol of KCl and AgNO3
Solution Stoichiometry
What volume of 0.273 M KCl solution is required
to react with exactly 0.836 mmol of AgNO3 ?
KCl(aq) + AgNO3(aq)  KNO3(aq) + AgCl(s)
First, calculate mmol of KCl:
1 mmol KCl
0.836 mmol AgNO3 x—————— = 0.836 mmol
1 mmol AgNO3
KCl
Second, calculate volume of KCl solution:
1 mL
0.836 mmol KCl x ———— = 3.06 mL of solution
0.273 mmol
(KCl)
Titration
Buret
Titrant
Erlenmeyer (conical) Flask
Titrate
Titration
• buret: calibrated td, fine tip to deliver small
volumes accurately, stopcock for flow
control
• Erlenmeyer flask: sloped walls allow
swirling of solutions without spilling or
splashing
• Titrant: solution containing reagent that
will react with sample in well known
manner
• Titrate: solution containing the sample
•
•
•
•
•
Titration
equivalence point: point in a titration at
which the exact stoichiometric amount of
titrant has been added to react with the
titrate
end point: the point in a titration at which
the indicator changes, titration stopped
here and volume of titrant read.
ideally, end point = equivalence point
reality, not so, error introduced, hopefully
small error
Four parameters: V, M of titrant and V, M of
titrate. Usually one is unknown.
Titration
Example: Calculate the concentration of
hydrochloric acid in a solution if 35.0 mL of
it required 28.9 mL of 0.178 M potassium
hydroxide solution for titration.
HCl:
titrated, V known, M unknown
(in this problem)
KOH: titrant, V and M known
Titration
Buret
0.178 M KOH
HCl + KOH  KCl + H2O
Erlenmeyer (conical) Flask
35.0 mL sample of HCl Sol’n,
unknown M + drops indicator
Titration
Strategy: mmol of KOH are calculated first.
Second, by using stoichiometry coefficients
mol of HCl are calculated.
Finally, M of HCl is calculated with mmol and V(mL)
0.178 mmol KOH 1 mmol HCl
28.9 mLx———————— x————— = 5.14 mmol
1 mL
1 mmol KOH
HCl
Based on the end point concept (mol ratio)
5.14 mmol
MHCl = ————— = 0.147 M (mol/L)
35.0 mL
(mmol/mL)
Example: Calculate the concentration of
arsenic acid in a solution given that a 25.0
mL sample of that solution required 42.2 mL
of 0.274 M potassium hydroxide for titration.
• H3AsO4 + KOH  K3AsO4 + H2O
• H3AsO4 + 3 KOH  K3AsO4 + 3 H2O
M of H3AsO4 is the unknown.
1 mol H3AsO4 reacts with 3 mol KOH. That is
the mole ratio.
We can use mL and mmol instead of L and mol
Example: Calculate the concentration of
arsenic acid in a solution given that a 25.0
mL sample of that solution required 42.2 mL
of 0.274 M potassium hydroxide for titration.
• H3AsO4 + 3 KOH  K3AsO4 + 3 H2O
0.274 mmol KOH 1 mmol H3AsO4
42.2 mLx———————— x————— = 3.85 mmol
1 mL
3 mmol KOH
H3AsO4
3.85 mmol
M H3AsO4 = ————— = 0.154 M (mol/L)
25.0 mL
(mmol/mL)
Purity Analysis: A 1.034 g-sample of impure
oxalic acid is dissolved in water and an acidbase indicator added. The sample requires 34.47
mL of 0.485 M NaOH solution to reach the
equivalence point. What is the mass of H2C2O4
and what is its mass percent in the sample?
• H2C2O4 + 2 NaOH  Na2C2O4 + 2 H2O
Strategy: mol of H2C2O4 out of mol of NaOH (by
using V and M). Then, mol of H2C2O4 will be
used to calculate g of H2C2O4 and, hence, % of it
in the sample.
Coefficients are 1 for H2C2O4 and 2 for NaOH.
Oxalic acid: Oxac
M W = 90.04 g/mol
H2C2O4 + 2 NaOH  Na2C2O4 + 2 H2O
0.485 mol 1 mol Oxac
34.47 mLx————— x—————— = 0.00836 mol
1000 mL 2 mol NaOH
Oxac
Based on the end point concept (mol ratio)
90.04 g
0.00836 mol x ———— = 0.753 g oxalic acid
1 mol
0.753 g oxac
%Oxac = ——————— x 100 = 72.8%
1.034 g of sample
Molar Mass of an Acid: A 1.056 g of a pure acid,
HA, is dissolved in water and an acid-base
indicator added. The solution requires 33.78 mL
of 0.256 M NaOH solution to reach the
equivalence point. What is the molar mass of
the acid?
We don’t know what A is
• HA + NaOH  NaA + H2O
Strategy: mol of HA out of mol of NaOH (by
using V and M). Then, mol of HA and g of HA
will be used to calculate the molar mass.
Coefficients are 1 for HA and 1 for NaOH.
Molar Mass of HA
33.78 mL NaOH solution = 0.03378 L
HA + NaOH  NaA + H2O
0.256 mol NaOH 1 mol HA
0.03378 Lx———————— x————— = 8.65x10-3
1 L sol’n
1 mol NaOH
mol HA
Molar mass,
1.056 g HA
M W = ——————— = 122 g/mol
8.65x10-3 mol HA
122 grams per mol
Vinegar: A 25.0-mL sample of vinegar (which
contains the weak acetic acid, CH3CO2H)
requires 28.33 mL of a 0.953 M solution of NaOH
for titration to the equivalence point. What mass
of the acid, in grams, is in the vinegar sample
and what is the M of acetic acid in the vinegar?
• CH3CO2H + NaOH  NaCH3CO2 + H2O
Strategy: mol of CH3CO2H (HOAc) out of mol of
NaOH (by using V and M). Then, g of HOAc will
be calculated with the molar mass. The M will
be calculated by using the volume of vinegar.
Coefficients are 1 for HOAc and 1 for NaOH.
CH3CO2H: HOAc
M W = 60.05 g/mol
0.953 mol 1 mol HOAc
28.33 mLx————— x—————— = 0.0270 mol
1000 mL 1 mol NaOH
HOAc
60.05 g
0.0270 mol x ——— = 1.62 g acetic acid in vinegar
1 mol
For the molarity, Vvinegar = 25.0 mL = 0.0250 L (sol’n)
0.0270 mol HOAc
M HOAc = ————————— = 1.08 M
0.0250 L sol’n
Problem: To analyze an iron-containing
compound, you convert all the iron into Fe2+ in
aqueous solution and then titrate the solution
with standardized KMnO4. The balanced-net ionic
equation is
MnO4− + 5Fe2+ + 8H+  Mn2+ + 5Fe3+ + 4H2O
A 0.598-g sample of the iron compound requires
22.25 mL of 0.0123 M KMnO4 solution for titration
to the equivalence point. What is the mass % of
iron in the sample?
Strategy:
mL and M of MnO4− to  mol of MnO4− and Fe2+
Coefficients are 1 and 5 for MnO4− and Fe2+ respectively
Fe2+ and MnO4− (mole ratio is 5 to 1):
0.0123 mol MnO4− 5 mol Fe2+
22.25 mLx ———————— x————— = 1.37x10−3
1000 mL
1 mol MnO4− mol Fe2+
55.85 g Fe
g of iron: 1.37 x 10−3 mol Fe2+ x ————— = 0.0765
1 mol Fe2+
g Fe
Now, for the mass % of iron:
0.0765 g Fe
———————— x 100 = 12.8% Fe
0.598 g sample
Normality
• equivalents (eq) solute per liter solution
• milliequivalents (meq) solute per mL solution
eq solute
meq solute
N = —————— = —————— (calculated by dividing)
V(L) sol’n
V(mL) sol’n
• equivalent = 1 equivalent weight
• Equivalent Weight (EW); given in g/eq
acid/base: the mass of a substance required to
furnish or react with exactly 1 mol H+
redox reactions: the mass of substance able to gain
or lose 1 mol e−
Equivalent weight
HCl
• HCl  H+ + Cl–
• 1 mol HCl  1 mol H+,  EW = MW
H2SO4
• H2SO4  2H+ + SO42–
• 1 mol H2SO4  2 mol H+, EW = MW/2
HnA (in general)
• EW = MW/n, n = # H+ in molecule of acid
KOH
• KOH  K+ + OH–
• OH– + H+  H2O
• , 1 mol KOH reacts with 1 mol H+,  EW = FW
• continued with polyhydroxides…
Equivalent weight
Fe(OH)3
• 1 mol Fe(OH)3  3OH–
• 1 mol reacts with 3 mol H+,  EW = FW/3
M(OH)n (in general)
• EW = FW/n, n = # OH– in formula unit of base
Redox:
+7
MnO4– + 5e–  Mn2+
FW MnO4–
EW = —————
5
0
Zn  Zn2+ + 2 e–
AW Zn
EW = ———
2
Example: Calculate the normality of barium
hydroxide in a solution made by dissolving 0.991
g of barium hydroxide in water and diluting to
100.0 mL.
Ba(OH)2
FW
171.32
g
mg
EW = —— = ———— = 85.66 —— = 85.66 ——
2
2
eq
meq
1 eq Ba(OH)2
0.991 g Ba(OH)2 x —————— = 0.0116 eq Ba(OH)2
85.66 g Ba(OH)2
0.0116 eq Ba(OH)2
N Ba(OH)2 = ————————— = 0.116 N Ba(OH)2
0.1000 L sol’n
Example: Describe the preparation of 250.0 mL
of 0.100 N oxalic acid solution from the solid.
Oxalic acid is H2C2O4.
• 1 mol oxalic acid  2 mol H+
• EW = 1/2 MW = 1/2(90.04) = 45.03g/eq
• 0.100 N indicates 0.100 eq OA/L sol’n
0.100 eq OA
0.2500 L x ——————
1 L sol’n
= 0.0250 eq OA
45.03 g OA
0.0250 eq OA x —————— = 1.13 g OA
1 eq OA
• Dissolve 1.13 g of oxalic acid in water and dilute
to a total volume of 250.0 mL
Example: What is the normality of a 0.300 M
arsenic acid solution?
• H3AsO4
• 1 mol H3AsO4  3 mol H+
MW
EW = ——
3
g H3AsO4 g H3AsO4
g H3AsO4
eq = ————— = ———— = 3 x ————— = 3 mol
EW H3AsO4
MW
MW H3AsO4
H3AsO4
3
eq
N = ——
V(L)
mol
M = ———
V(L)
Then,
N H3AsO4 = 3 x M H3AsO4 = 3 x 0.300 = 0.900 N
CONCLUSION: For an Acid HnA or a
Base M(OH)n
N =nxM
n = # H+ in molecule of acid or # OH– in
formula unit of base
By definition, 1 eq of anything will react
with exactly 1 eq of anything else
Equivalence Point
The point in a titration at which
eq titrant = eq titrate
meq titrant = meq titrate
Example: Calculate the concentration of
phosphoric acid in a solution given that a 25.0
mL sample of that solution required 42.2 mL
of 0.274 N potassium hydroxide for titration.
• H3PO4 + 3 KOH  K3PO4 + 3 H2O
• at Eq. Pt., meq H3PO4 = meq KOH
meq H3PO4 = meq KOH
(mL H3PO4)(N H3PO4) = ( mL KOH)(N KOH)
(25.0 mL)(X ) = (42.2 mL)(0.274 N )
X = 0.463 N
Now, N = 3x M, then M = N/3
X = 0.463/3 = 0.154 M
Example: Calculate the % oxalic acid (H2C2O4)
in a solid given that a 1.709 g sample of the
solid required 24.9 mL of 0.0998 N potassium
hydroxide for titration.
• H2C2O4 + 2KOH  K2C2O4 + 2H2O
Due to the two protons of H2C2O4 (OA),
• EW = MW/2 = 45.03g/eq = 45.03mg/meq
g OA
% OA = ————— x 100
g sample
Strategy: calculate eq of NaOH that are the same
for OA; then, calculate g of OA with eq and EW
of OA
At equivalence point,
eq OA = eq KOH
data of KOH sol’n
0.0998 eq KOH
eq OA = 0.0249 L x ——————— = 0.00249 eq
1 L sol’n
45.03 g OA
0.00249 eq OA x ——————— = 0.112 g OA
1 eq OA
0.112 g OA
% OA = ——————— x 100 = 6.55%
1.709 g sample