Download Chapter 4: Types of Chemical Reactions and Solution Stoichiometry

Chapter 4: Types of Chemical Reactions and
Solution Stoichiometry
4.1 Water, the Common Solvent
4.2 The Nature of Aqueous Solutions: Strong and Weak Electrolytes
4.3 The Composition of Solutions (MOLARITY!)
4.4 Types of Chemical Reactions
4.5 Precipitation Reactions
4.6 Describing Reactions in Solution
4.7 Selective Precipitation
4.8 Stoichiometry of Precipitation Reactions
4.9 Acid-Base Reactions
4.10 Oxidation-Reduction Reactions
4.11 Balancing Oxidation-Reduction Equations
4.12 Simple Oxidation-Reduction Titrations
Figure 4.1: A space-filling model of the water molecule.
The expression ( aq ) is used to indicate a
solvated anion or cation (or molecule)
Figure 4.2: Polar water molecules
interact with the positive and negative
ions of a salt, assisting with the
dissolving process.
The Role of Water as a Solvent: The solubility of
Ionic Compounds
Electrical conductivity - The flow of electrical current in a solution is a
measure of the solubility of ionic compounds or a
measurement of the presence of ions in solution.
Electrolyte - A substance that conducts a current when dissolved in
water. Soluble ionic compound dissociate completely and
may conduct a large current, and are called Strong
Electrolytes.
NaCl(s) + H2O(l)
Na+(aq) + Cl -(aq)
When Sodium Chloride dissolves into water the ions become solvated,
and are surrounded by water molecules. These ions are called “aqueous”
and are free to move through out the solution, and are conducting
electricity, or helping electrons to move through out the solution
Electrical Conductivity of Ionic Solutions
Figure 4.5:
HCL (aq) is
completely
ionized.
Figure 4.6:
An aqueous
solution of
sodium
hydroxide.
Strong Electrolytes
• Produce ions in aqueous solution and conduct
electricity well.
• Strong electrolytes are soluble salts, strong acids and
strong bases.
• Strong acids produce H+ ions when they dissolve in
water.
• HCl, … , HNO3 and H2SO4 are strong acids
HNO3(aq) → H+(aq) + NO3-(aq)
• NaOH and KOH are strong bases:
NaOH(s) → Na+(aq) + OH-(aq)
All of the above species are ionized nearly 100%
Weak bases like NH3 and weak acids like acetic acid don’t
conduct electricity well
Figure 4.7:
Acetic acid
(HC2H3O2)
exists in water
mostly as
undissociated
molecules.
Figure 4.8:
The reaction of
NH3 in water.
Weak Electrolytes
• Produce relatively few ions in aqueous solution
• The most common weak electrolytes are weak acids and
weak bases.
• Acetic acid is a typical weak acid:
HC2H3O2(aq)
⇌
H+(aq) + C2H3O2-(aq)
• Ammonia is a common weak base:
NH3(aq) + H2O(l)
⇌
NH4+(aq) + OH-(aq)
Both of these species are ionized only ~1%
Because of hydrogen bond formation,
water boils at a much higher
temperature than CH4 (90 K), which
has a comparable molecular mass.
Nonelectrolytes
Dissolve in water but produce no ions in
solution.
Nonelectrolytes do not conduct electricity
because they dissolve as whole molecules,
and produce no ions.
Common nonelectrolytes include ethanol and
table sugar (sucrose, C12H22O11)
The Solubility of Covalent Compounds in Water
Polar covalent compounds are very soluble in water. They have -OH
groups that can form hydrogen bonds with water. Examples are
compounds table sugar, sucrose (C12H22O11), ethanol (C2H5-OH),
ethylene glycol (C2H6O2) in antifreeze, and methanol (CH3-OH).
These also are written with “(aq)” when dissolved in water
(i.e., aqueous).
Example: C2H5OH(aq)
Nonpolar covalent compounds can’t form hydrogen bonds and have
little or no interactions with water molecules. Examples are the
hydrocarbons in gasoline and oil, which don’t mix with water
Octane = C8H18
and / or
Benzene = C6H6
Interaction of Water and Ethanol
Carbohydrates
Molecules that contain carbon and water!
H
CH2OH
C
HO
H
C
OHH C
C
OH
H
Sucrose
CxH2yOy
O
H
C
O
CH2OH O
H
C
C
OH
H
CH2OH
C
C
H
OH
C12H22O11 , C12(H2O)11 a disaccharide
Determining Moles of Ions in Aqueous
Solutions of Ionic Compounds - I
Problem: How many moles of each ion are in each of the following:
a)
b)
c)
d)
4.0 moles of sodium carbonate dissolved in water
46.5 g of rubidium fluoride dissolved in water
5.14 x 1021 formula units of iron (III) chloride dissolved in water
7.8 moles of ammonium sulfate dissolved in water
H2O
a) Na2CO3 (s)
moles of
Na+
= 4.0 moles Na2CO3 x
2 Na+(aq) + CO3-2(aq)
2 mol Na+
1 mol Na2CO3
= 8.0 moles Na+ and 4.0 moles of CO3-2 are present
Determining Moles of Ions in Aqueous
Solutions of Ionic Compounds - II
b)
RbF(s)
H2O
Rb+(aq) + F -(aq)
moles of RbF = 46.5 g RbF x
1 mol RbF
104.47 g RbF
= 0.445 moles RbF
thus, 0.445 mol Rb+ and 0.445 mol F - are present
H2O
c) FeCl3 (s)
Fe+3(aq) + 3 Cl -(aq)
moles of FeCl3 = 9.32 x 1021 formula units
x
= 0.0155 mol FeCl3
moles of
Cl -
1 mol FeCl3
6.022 x 1023 formula units FeCl3
3
mol
Cl
= 0.0155 mol FeCl3 x
= 0.0465 mol Cl 1 mol FeCl3
and 0.0155 mol Fe+3 are also present.
Determining Moles of Ions in Aqueous
Solutions of Ionic Compounds - III
d) (NH4)2SO4 (s)
H2O
2 NH4+(aq) + SO4- 2(aq)
+
2
mol
NH
4 = 15.6 mol NH +
Moles of NH4 = 7.8 moles (NH4)2SO4 x
4
1 mol(NH4)2SO4
+
and 7.8 mol SO4- 2 are also present.
Molarity (Concentration of Solutions) = M
M=
Moles of Solute =
Liters of Solution
mol
L
solute = material dissolved into the solvent
In sea water, water is the solvent, and NaCl, MgCl2, etc are the solutes.
In brass, copper is the solvent (90%), and zinc is the solute (10%).
Determining Moles of Ions in Aqueous
Solutions of Ionic Compounds – I+
Example problem:
a)
b)
c)
d)
e)
How many moles of each ion are in each of the following:
4.0 moles of sodium carbonate dissolved in water (Na: IA)
46.5 g of rubidium fluoride dissolved in water (Rb: IA)
9.32 x 1021 formula units of iron (III) chloride dissolved in water
7.8 moles of ammonium sulfate dissolved in water
75.0 mL of 0.56 M scandium bromide dissolved in water (Sc:IIIB)
We already did a) through d) above. Let’s do e):
Determining Moles of Ions in Aqueous
Solutions of Ionic Compounds - IV
e) ScBr3(s)
H2O
Sc3+(aq) + 3 Br -(aq)
Converting from volume to moles:
Moles of ScBr3 = 75.0 mL x (0.56mol ScBr3 / L) x (1 L / 1000 mL) =
0.042 mol ScBr3
Moles of Br- = 0.042 mol ScBr3 x (3 mol Br-/ mol ScBr3) = 0.13 mol
BrMoles of Sc3+ = 0.042 mol ScBr3 x (1 mol Sc3+ / mol ScBr3) = 0.042
mol Sc3+
Molarity (Concentration of Solutions) = M
M=
Moles of Solute =
Liters of Solution
mol
L
solute = material dissolved into the solvent
For electrolytes, we describe the molarity of both the compound itself
and of each of the ions it produces:
e.g.,
Na2CO3 (s) → 2 Na+(aq) + CO3-2(aq) = “Na2CO3 (aq)”
The Na+ molarity or concentration is 2x that of the other 2 species here.
Preparing a Solution - I
Example problem: A solution of sodium phosphate
is prepared by dissolving 3.95 g of sodium
phosphate in water and diluting it to 300.0 mL. What
is the molarity, M, of the salt and each of the ions?
Strategy
(1) Write chemical equation showing process.
(2) Calculate moles of each species.
(3) Divide # moles by # L water to obtain molarity.
(1) Na3PO4 (s)
H2O(solvent)
→
3 Na+ (aq) + PO4-3 (aq)
(2) Calculate moles of each species.
Mol wt of Na3PO4 = 163.94 g / mol
mol Na3PO4 = 3.95 g / 163.94 g/mol = 0.0241 mol
mol Na+ = 3 x mol Na3PO4 = 0.0723 mol
mol PO4-3 = mol Na3PO4 = 0.0241 mol
(3) Divide # moles by # L water to obtain molarity.
M(Na3PO4) = 0.0241 mol Na3PO4 / 0.3000 L
= 0.0803 M
M(PO4-3) = M(Na3PO4) = 0.0803 M
M(Na+ ) = 3 x M(Na3PO4) = 0.241 M
Figure 4.9: Steps involved in the
preparation of a standard solution.
Like Example 4.4 (P 97)
A Chemist must prepare a 1.00 L of a 0.375 M solution of Ammonium
Carbonate. What mass of (NH4)2CO3 must be weighed out to prepare
this solution?
First, determine the moles of Ammonium Carbonate required:
1.00 L x
0.375 mol (NH4)2CO3
= 0.375 mol (NH4)2CO3
L solution
This amount can be converted to grams by using the molar mass:
94.07 g (NH4)2CO3
0.375 mol (NH4)2CO3 x
= 35.276 g (NH4)2CO3
mol (NH4)2CO3
Or, to make 1.00L of solution, one must weigh out 35.3 g of
(NH4)2CO3, put this into a 1.00 L volumetric flask, and add water
to the mark on the flask.
Like Example 4.3 (P 95)
An isotonic solution, one with the same ionic content as blood is about
0.14 M NaCl. Calculate the volume of blood that would contain 2.5 mg
of NaCl?
Find the moles in 2.5 mg NaCl:
1 g NaCl x
2.5 mg NaCl x
1000 mg NaCl
1 mol NaCl = 4.28 x 10-5 mol
58.45g NaCl
NaCl
What volume of 0.14 M NaCl would contain this amount of NaCl
(4.28 x 10-5 mol NaCl): V x M = moles
0.14 M NaCl
Vx
= 4.28 x 10-5 mol NaCl
L solution
Solving for Volume gives:
-5
V = 4.28 x 10 mol NaCl = 3.06 x 10-4 L
0.14 mol NaCl
or 0.306 mL of Blood!
L solution
Dilution of Solutions
Dilute 25.00 mL of 0.0400 M KMnO4 to a final volume
of 500. mL. What is the resulting molarity (M) of the
diluted solution?
A Strategy for calculating final concentration:
The number of moles of solute is the same before and
after dilution
V1 x M1 = moles solute = V2 x M2
V1M1 = V2 M 2
V1M1
M2 =
V2
25.00mL × 0.0400M
= 0.00200 M
M2 =
500.mL
Make a Standard Solution of 0.00100 M Potassium Permanganate
Potassium Permanganate is KMnO4 and has a molecular mass
of 158.04 g / mole
Step 1: Prepare a 0.0400 M solution. For example, dissolve
1.00x10-2 moles of KMnO4 into water to a final volume of ¼ L
(250.00 mL) of solution.
158.04 g KMnO4 = 1.58 g KMnO
0.0100 mol KMnO4 x
4
mole KMnO4
Molarity = 0.0100 moles KMnO4 = 0.0400 mol/L = 0.0400 M
0.250 L
Molarity of K+ ion = [K+] ion = [MnO4-] ion = 0.0400 M
Step 2: Dilute some of it to 0.00100 M.
Dilution of Solutions
• Take 25.00 mL of the 0.0400 M KMnO4
• Dilute the 25.00 mL to 1.000 L - What is the
resulting Molarity of the diluted solution?
• # moles = Vol x M
• 0.0250 L x 0.0400 M = 0.00100 moles
• 0.00100 mol / 1.00 L = 0.00100 M
Figure 4.10:
(a) A measuring pipette
or graduated pipette
(b) A volumetric pipette:
delivers one specific
volume only.
Figure 4.11: (a) A measuring pipette
(b) Water is added to the flask.
(c) The resulting molarity is M2 = M1 x V1/V2.
V1
V2 =
TYPES of CHEMICAL REACTIONS
in LIQUID SOLUTIONS:
• Precipitation reactions
• Acid – Base reactions
• Reduction – Oxidation (REDOX) reactions
PRECIPITATION REACTIONS:
a solid compound or “precipitate” forms from two (or
more) solutes
DEMO:
Pb(NO3)2(aq) + 2 KI(aq) → PbI2(s) + 2 KNO3(aq)
Lead iodide is a gold precipitate
Show all ions explicitly:
Pb2+(aq) + 2 NO3-(aq) + 2 K+(aq) + 2 I-(aq)→
PbI2(s) + 2 K+(aq) + 2 NO3-(aq)
Remove spectator ions:
Pb2+(aq) + 2 I-(aq) → PbI2(s)
( same as with KI )
DEMO: Purple precipitate
CoCl2(aq)+ Na2CO3(aq)
→ CoCO3(s) + 2 NaCl(aq)
A note on this one: You want an excess of carbonate in the cylinder first, then pour the cobalt in.
Otherwise, the background is pink!
Go HUSKIES!!
Figure 4.12:
When yellow
aqueous potassium
chromate is added
to a colorless
barium nitrate
solution, yellow
barium chromate
precipitates.
Figure 4.13:
Reactant
solutions: (a)
Ba(NO3)3(aq)
Figure 4.13:
Reactant
solutions: (b)
K2CrO4(aq).
Figure 4.14: Reaction of K2CrO4 (aq) and
Ba(NO3)2 (aq).
Table 4.1 Simple Rules for Solubility
of Salts in Water (I will give you this table on tests.)
1. Most nitrate (NO3-) salts are soluble.
2. Most salts of Na+, K+, and NH4+ are soluble.
3. Most chloride salts are soluble. Notable exceptions are AgCl,
PbCl2, and Hg2Cl2.
4. Most sulfate salts are soluble. Notable exceptions are BaSO4,
PbSO4, and CaSO4.
5. Most hydroxide salts are only slightly soluble. The important
soluble hydroxides are NaOH, KOH, and
Ca(OH)2 (marginally soluble).
6. Most sulfide (S2-), carbonate (CO32-), and phosphate (PO43-)
salts are only slightly soluble.
Predicting if a precipitate forms, and which?
Pb(NO3)2(aq) + NaCl(aq) →
Pb+2(aq) + 2 NO3- (aq) + Na+ (aq) + Cl-(aq)
RULE: If any of the possible new species formed by
combining anions with cations is insoluble,
then that precipitate will form.
USE TABLE 4.1
(I’ll give you this Table for the tests and quiz.)
In this case, PbCl2 is insoluble, and a precipitate forms.
Note: Like the demo with PbI2(s) above.
Precipitation Reactions: Will a Precipitate form?
Example: If a solution containing potassium chloride is added to a
solution containing ammonium nitrate, will a precipitate form?
KCl(aq) + NH4NO3(aq) → K+(aq) + Cl-(aq) + NH4+(aq) + NO3-(aq)
Possible reaction products are KCl and NH4NO3, NH4Cl and KNO3.
All are soluble, so there is no precipitate.
KCl(aq) + NH4NO3 (aq) = No Reaction!
Example: If a solution containing sodium sulfate is added to a solution
containing barium nitrate, will a precipitate form?
Na2SO4 (aq) + Ba(NO3)2 (aq) →
Ba2+(aq) + SO42- (aq) + 2 Na+(aq) + 2NO3- (aq)
Barium sulfate is insoluble; therefore its precipitate will form.
Aqueous
KOH and
Fe(NO3)3 are
mixed.
Solid
Fe(OH)3
forms.
Figure 4.16: Photos and molecular-level
representations illustrating the reaction of
KCl(aq) with AgNO3(aq) to form AgCl(s).
Quantitative Precipitation Problems:
Calculate the mass of solid sodium iodide that must be added to
2.50 L of a 0.125 M lead nitrate (Pb(NO3)2) solution to precipitate
all of the lead as PbI2 (s)!
The chemical equation for the reaction is:
Pb(NO3)2 (aq) + 2 NaI(aq)
PbI2 (s) + 2 NaNO3 (aq)
The moles of sodium iodide needed to precipitate PbI2 is twice the
lead ions. The number of moles of sodium iodide needed is:
2.50 L
2+
0.125
Mol
Pb(NO
)
1
Mol
Pb
3
2
x
x
1.0 L soln.
1 Mol Pb(NO3)2
The mass of sodium iodide is:
0.625 mol I- x 1 mol NaI x
1 mol I-
2
mol
I
x
= 0.625 mol I1 mol Pb2+
149.9 g NaI
= 93.68 g NaI
1 mol NaI
Quantitative Precipitation Problems:
Calculate the mass of solid sodium iodide that must be added to
2.50 L of a 0.125 M lead nitrate (Pb(NO3)2) solution to precipitate
all of the lead as PbI2 (s)!
The chemical equation for the reaction is:
Pb(NO3)2 (aq) + 2 NaI(aq)
PbI2 (s) + 2 NaNO3 (aq)
The moles of sodium iodide needed to precipitate PbI2 is twice the
lead ions. The number of moles of sodium iodide needed is:
2.50 L
2+
0.125
mol
Pb(NO
)
1
mol
Pb
3
2
x
x
1.0 L soln.
1 mol Pb(NO3)2
1 mol NaI
2
mol
I
x
x
2+
1 mol I1 mol Pb
An alternate method:
2.50 L x 0.125 mol Pb(NO3)2 x 2 mol NaI
1.0 L soln.
1 mol Pb(NO3)2
=
Like Example 4.9 (P 112)
What mass of Pb2+ could by precipitated from a solution by the
addition of 0.785 L of 0.0015 M Sodium Iodide solution?
Find the stoichiometric relationship from the chemical equation:
Pb2+(aq) + 2 I-(aq)
PbI2 (s)
It will take twice the iodide ion to precipitate the Lead ions:
Moles I - = VNaI x MNaI = 0.785 L x 0.0015 mol I- = 0.00118 mol IL
2+
1
mol
Pb
2+
Moles Lead ion = 0.00118 mol I- x
=
0.000590
mol
Pb
2 mol IMass of Lead = 207.2 g Pb x 0.000590 moles = 0.122 g Pb
mol Pb
Writing Equations:
a) Calcium Nitrate and Sodium Sulfate solutions are added together.
Molecular Equation
Ca(NO3)2 (aq) + Na2SO4 (aq)
CaSO4 (s) +2 NaNO3 (aq)
Total Ionic Equation
Ca2+(aq)+2 NO3-(aq) + 2 Na+(aq)+ SO4-2(aq) CaSO4 (s) + 2 Na+(aq+) 2 NO3-(aq)
Net Ionic Equation
Ca2+(aq) + SO4-2(aq)
“Spectator Ions” here are Na+ and NO3-
CaSO4 (s)
DEMO:
0.1 M Potassium ferrocyanide = 4 K+(aq) + Fe(CN)6-4(aq)
0.1 M Ferric chloride = Fe3+(aq) + 3 Cl-(aq)
4 K+(aq) + Fe(CN)6-4(aq) + Fe+3 (aq) + 3 Cl-(aq) →
K[Fe(CN)6](s) + 3 K+(aq) + 3 Cl-(aq)
Prussian Blue
(very dark purplish blue)
Remove spectator ions:
K+(aq) + Fe(CN)6-4(aq) + Fe+3 (aq) → K[Fe(CN)6](s)
Limiting reagent
problems in
precipitation reactions.
Solve as usual, using full
equation or only net ionic
equation.
÷ stoich. coeff. to get
equivalents.
Amounts:
Just use molarities and
volumes to get moles
(when moles or grams
are not given).
Example problem: Lead has been used as a glaze for pottery for years, and if not fired
properly is leachable from the pottery. Vinegar is used in leaching tests, followed by lead
precipitation as a sulfide.
If 257.8 ml of a 0.0468 M solution of lead(II) acetate (Pb(C2H3O2)2) is added to
156.00 mL of a 0.095 M solution of sodium sulfide, what mass of solid lead
sulfide will be formed? This is a limiting-reactant problem because the amounts of
two reactants are given.
Strategy:
(1)
(2)
(3)
(4)
Write the balanced equation.
Determine the limiting reactant.
Calculate the moles of product.
Convert moles of product to mass of the product using molar mass.
(1) Write the balanced equation
Pb(C2H3O2)2 (aq) + Na2S (aq)
→
PbS (s) + 2 Na(C2H3O2) (aq)
(2) Determine the limiting reactant.
Mol Pb(C2H3O2)2 = V x M = 0.2578 L x (0.0468 mol/L) = 0.0121 mol
Mol Na2S = V x M = 0.15600 L x (0.095 mol/L) = 0.0148 mol
Mol Pb(C2H3O2)2 / 1 mol = 0.0121 eq.
Mol Mol Na2S / 1 mol = 0.0148 eq.
Therefore lead(II) acetate is the limiting reactant.
(3) Calculate the moles of product.
0.00121 mol Pb(C2H3O2)2 x (1 mol PbS / mol Pb(C2H3O2)2) =
0.0121 mol PbS
(4) Calculate mass of product
0.0121 mol PbS x 239.3g PbS/mol PbS = 2.90 g PbS
Precipitation problem: When aqueous silver nitrate and sodium
chromate solutions are mixed, solid silver chromate forms in a
solution of sodium nitrate. If 257.8 mL of a 0.0468 M solution of
silver nitrate is added to 156.00 mL of a 0.0950 M solution of
sodium chromate, what mass of silver chromate (M = 331.8 g/mol)
will be formed?
This is a limiting-reactant problem because the amounts
of two reactants are given.
Strategy:
(1) Write the balanced equation.
(2) Calculate the number of moles of each reactant.
(3) Determine the limiting reactant.
(4) Calculate the moles of product.
(5) Convert moles of product to mass of the product using molar mass.
Molecular equation,
Total ionic equation,
and net ionic
equation
Precipitation problem: When aqueous silver nitrate and sodium chromate
solutions are mixed, solid silver chromate forms in a solution of sodium nitrate.
If 257.8 mL of a 0.0468 M solution of silver nitrate is added to
156.00 mL of a 0.0950 M solution of sodium chromate, what mass of silver
chromate (M = 331.8 g/mol) will be formed?
•
Write the balanced equation
2AgNO3(aq) + Na2CrO4(aq)
→
Ag2CrO4(s) + 2 NaNO3(aq)
(2) Determine the number of moles of each reactant
Mol AgNO3 = V x M = 0.2578 L x (0.0468 mol/L) = 0.0121 mol
Mol Na2CrO4 = V x M = 0.15600 L x (0.095 mol/L) = 0.0148 mol
2) Determine the limiting reactant.
0.0121 mol AgNO3 / (2 mol/eq) = 0.00605 eq. AgNO3
0.0148 mol Na2CrO4 / (1 mol/eq) = 0.0148 eq. Na2CrO4
Therefore silver nitrate is the limiting reactant.
2AgNO3(aq) + Na2CrO4(aq)
→
Ag2CrO4(s) + 2 NaNO3(aq)
(3) Calculate the moles of product.
Moles Ag2CrO4(s)
= 0.0121 moles AgNO3 x (1 mol Ag2CrO4 / 2 mol AgNO3 )
= 0.0121 mol/2 = 0.00605 mol Ag2CrO4
(4) Calculate mass of product
0.00605 mol Ag2CrO4 x (331.8 g Ag2CrO4 /mol Ag2CrO4)
= 2.01 g Ag2CrO4(s)
Like Example 4.8 (P 111)
When aqueous solutions of silver nitrate and sodium chloride are
mixed, silver chloride is precipitated. What mass of silver chloride
would be formed by the addition of 75.00 mL to 3.17 M NaCl and
128 mL of 2.44 M silver nitrate?
The stoichiometric relationship comes from the chemical equation:
__AgNO3 (aq) + __ NaCl(aq)
__ AgCl(s) + _________
mol Ag+ =
mol Cl- =
__________ is limiting, and we use it to calculate the mass of AgCl:
Mass AgCl =
Like Example 4.8 (P 111)
When aqueous solutions of silver nitrate and sodium chloride are
mixed, silver chloride is precipitated. What mass of silver chloride
would be formed by the addition of 75.00 ml to 3.17 M NaCl and
128 ml of 2.44 M silver nitrate?
The stoichiometric relationship comes from the chemical equation:
AgNO3 (aq) + NaCl(aq)
AgCl(s) + NaNO3 (aq)
There is a one to one relationship, therefore the number of moles are
the same, but which is in the lowest quantity?
VAgNO3 x MAgNO3 = 0.128 L x 2.44 M = 0.312 mol Ag+
VNaCl x MNaCl = 0.07500 L x 3.17 M = 0.238 mol ClSince the chloride ion is smaller, it is limiting, and we use it to
calculate the mass of AgCl, since we can only obtain 0.238 mol of
AgCl:
Mass AgCl = 0.238 mol x 143.35 g AgCl/ mol = 34.1 g
Limiting reagent
problems in solution:
Only difference is:
using molarity and / or
volume to connect to
moles via
Mx = (moles of x) / V.
The gravimetric procedure.(P111-112)
Determining an unknown amount by precipitation:
Finding the Ca concentration in rock!
Full balanced equation not needed.
All you need to know is that ALL the Ca ends up
as CaC2O4 . H2O (s).
.H O
1
mol
CaC
O
2
4
2
-3 mol
0.2920 g CaC2O4 . H2O X
=
1.998
x
10
.
146.12g CaC2O4 H2O
of CaC2O4 .H2O
2+
40.08 g Ca
-3
2+
-2g Ca2+
=
8.009
x
10
1.998 x 10 mol Ca X
2+
1 mol Ca
Mass % Ca is:
8.009 x 10-2 g Ca2+
0.4367 g rock
x 100% = 18.34% Ca by mass
Acid-Base
Reactions
Acids = Covalent molecules which lose
H+ cations (= protons) to water molecules in solution
When gaseous hydrogen iodide dissolves in water, the water molecules
help the HI dissociate into H+ and I- ions, both solvated by water.
We can write the hydrogen ion in water as either H+(aq) or H3O+(aq).
They mean the same thing. A molecule which has a H atom that is easily
lost to the water solution as H+ (or proton) is called an “acid”, and the
resulting solution is called an “acidic” solution. The acid is said to be
“deprotonated” in solution:
HI(g)
H+(aq) + I -(aq)
HI(g) + H2O(l)
H3O+(aq) + I -(aq)
The water (H2O) could also be written above the arrow indicating that
the solvent was water in which the HI was dissolved:
HI(g)
H2O
H+(aq) + I -(aq)
H3O+ is called the hydronium ion
Strong Acids and the Molarity of H+ Ions in
Aqueous Solutions of Acids
Problem: In aqueous solutions, each molecule of sulfuric acid will
loose two protons to yield two H+ ions, and one sulfate ion. NOT TRUE
What is the molarity of the sulfate and Hydronium ions in a solution
prepared by dissolving 155 g of concentrated sulfuric acid into
sufficient water to produce 2.30 Liters of acid solution?
Plan: Determine the number of moles of sulfuric acid, divide the moles
by the volume to get the molarity of the acid and the sulfate ion. The
hydronium ions concentration will be twice the acid molarity.
Solution: Two moles of H+ are released for every mole of acid:
H2SO4 (l) + 2 H2O(l)
2 H3O+(aq) + SO4- 2(aq)
Moles H2SO4 =
Molarity of SO4- 2 =
Molarity of H+ =
Strong Acids and the Molarity of H+ Ions in
Aqueous Solutions of Acids
Problem: In aqueous solutions, each molecule of sulfuric acid will
loose one proton to yield one H+ ion and one HSO4- ion. (Approx.)
What is the molarity of the HSO4- and Hydronium ions in a solution
prepared by dissolving 155 g of concentrated sulfuric acid into
sufficient water to produce 2.30 Liters of acid solution?
Plan: Determine the number of moles of sulfuric acid, divide the moles
by the volume to get the molarity of the acid and the HSO4-1 ion. The
hydronium ion’s concentration will be the same molarity.
Solution: One mole of H+ is released for every mole of acid:
H2SO4 (l) + H2O(l)
H3O+(aq) + HSO4-(aq)
Moles H2SO4 = 155 g H2SO4 x 1 mole H2SO4 = 1.58 moles H2SO4
98.09 g H2SO4
-2
1.58
mol
SO
4
Molarity of HSO4- =
= 0.687 Molar in SO4- 2
2.30 L solution
Molarity of H+ = 0.687 Molar in H+ (or H3O+) Approximately true only.
Bases
An Acid is a substance that produces H+ (H3O+) ions when dissolved
in water, and is a proton donor
A Base is a substance that produces OH - ions when dissolved in water:
Example: NaOH(aq) → Na+(aq) + OH-(aq)
The OH- ion react with the H+ ions (if an acid is present) to produce
water, H2O, and is therefore a proton acceptor.
Acids and Bases are electrolytes. Their strength is categorized in
terms of their degree of dissociation in water to make hydronium or
hydroxide ions. Strong acids and bases dissociate completely, and
are strong electrolytes. Weak acids dissociate partially (some small
% of the molecules dissociate) and are weak electrolytes.
Selected Acids and Bases
Acids
Bases
Strong: H+(aq) + A-(aq)
Strong: M+(aq) + OH-(aq)
(100% dissociates in solution)
Hydrochloric, HCl
Sodium hydroxide, NaOH
Hydrobromic, HBr
Potassium hydroxide, KOH
Hydroiodoic, HI
Calcium hydroxide, Ca(OH)2
Nitric acid, HNO3
Strontium hydroxide, Sr(OH)2
Sulfuric acid, H2SO4
Barium hydroxide, Ba(OH)2
Perchloric acid, HClO4
Weak
Weak
(small % dissociates, most intact)
Hydrofluoric, HF
Ammonia, NH3
Phosphoric acid, H3PO4
accepts proton from water to make
Acetic acid, CH3COOH
NH4+(aq) and OH-(aq)
(or HC2H3O2)
Acid - Base Reactions : Neutralization Rxns.
The generalized neutralization reaction
between an Acid and a Base is:
HA(aq) + MOH(aq)
MA(aq) + H2O(l)
(Here “M” stands for anything.)
Acid
+ Base
→
Salt
+ Water
The salt product can either be
dissolved as ions or form a
precipitate.
DEMO
Carbonic acid (aq) + NaOH (aq) → Na bicarbonate + water
(makes CO2(gas))
Writing Balanced Equations for
Neutralization Reactions - I
Problem: Write balanced chemical reactions (molecular, total ionic, and
net ionic) for the following Chemical reactions:
a) Calcium Hydroxide(aq) and Hydroiodic acid(aq)
b) Lithium Hydroxide(aq) and Nitric acid(aq)
c) Barium Hydroxide(aq) and Sulfuric acid(aq)
Plan: These are all strong acids and bases, therefore they will make
water and the corresponding salts.
Solution:
a)
Ca(OH)2 (aq) + 2HI(aq)
CaI2 (aq) + 2H2O(l)
Ca2+(aq) + 2 OH -(aq) + 2 H+(aq) + 2 I -(aq)
Ca2+(aq) + 2 I -(aq) + 2 H2O(l)
2 OH -(aq) + 2 H+(aq)
2 H2O(l)
Writing Balanced Equations for
Neutralization Reactions - II
b) LiOH(aq) + HNO3
LiNO3 (aq) + H2O(l)
(aq)
Li+(aq) + OH -(aq) + H+(aq) + NO3-(aq)
Li+(aq) + NO3-(aq) + H2O(l)
OH -(aq) + H+(aq)
c)
Ba(OH)2 (aq) + H2SO4 (aq)
H2O(l)
BaSO4 (s) + 2 H2O(l)
Ba2+(aq) + 2 OH -(aq) + 2 H+(aq) + SO42-(aq)
BaSO4 (s) + 2 H2O(l)
Ba2+(aq) + 2 OH -(aq) + 2 H+(aq) + SO42-(aq)
BaSO4 (s) + 2 H2O(l)
Burette
Like Example 4.10 (P 115)
What volume of 0.468 M H2SO4 is needed to neutralize
215.00 mL of a 0.125 M LiOH solution?
Calculate the number of moles of base:
Vbase x Mbase =
From the balance equation find the moles of acid needed:
__ LiOH(aq) + __ H2SO4 (aq)
__ H2O(l) + __ Li2SO4 (aq)
Use their ratio of stoichiometric coefficients to convert
mol LiOH to mol H2SO4. Then use it molarity to get volume.
Volume of acid:
Moles acid
=
Vacid =
Macid
Like Example 4.10 (P 115)
What volume of 0.468 M H2SO4 is needed to neutralize
215.00 mL of a 0.125 M LiOH solution?
Calculate the number of moles of base:
Vbase x Mbase = 0.21500 L x 0.125 M = 0.0268 mol LiOH
From the balance equation find the moles of acid needed:
2 LiOH(aq) + H2SO4 (aq)
2 H2O(l) + Li2SO4 (aq)
0.0268 mol LiOH x (1 mol H2SO4 / 2 mol LiOH) = 0.0134 mol H2SO4
Volume of acid:
Moles acid
0.0134 mol
= 0.468 mol
Vacid =
Macid
L
= 0.0286 L H2SO4
On Exam Day:
• Seat according to seating chart- check at home the
day before the exam!!
• If you arrive late on exam day- see TA nearest the
door.
• Bring a photo ID to the exam.
Finding some unknown Concentration of Acid
by an Acid - Base Titration with known base
Volume (L) of base (difference in
burette readings) needed to titrate
M (mol/L) of base
Moles of base needed to titrate
molar ratio
Moles of acid which were titrated
volume (L) of acid
M (mol/L) of original acid soln.
Potassium Hydrogenphthalate KHC8H4O4
(called “KHP” for short)
O
O
C
C
C
O
K+
O K+
O
O
O
H
C
O
An easy-to-weigh acid: Convenient for making a
standard acid concentration for doing titrations
H+
Example: A titration is performed between sodium hydroxide and
KHP(204.2 g/mol) to “standardize” a base solution (i.e., to determine its
exact concentration for use in later titrations). You place 50.00 mg
of solid KHP in a flask with a little water and a few drops of an
indicator. At the beginning of the titration, the burette reading of the base
solution is 43.87 mL and its final burette reading is 10.55 mL .
The reaction is:
HKC8H4O4(aq) +Na+(aq) +OH-(aq) → KC8H4O4-(aq) +Na+(aq) +H2O(l)
H+(aq) + OH-(aq) → H2O(l)
What is the concentration of the base (i.e., of the NaOH soln.)?
Strategy:
(1)Use the molar mass of KHP to calculate the number of moles
of the acid.
(2) From the balanced chemical equation, calculate moles of base
(3) From the difference in the burette readings, and the definition of
molarity, calculate the molarity of the base.
(1)Use the molar mass of KHP to calculate the number of moles of
the acid.
50.00 mg
1g
moles KHP =
×
= 0.0002449 mol
-1
204.2 g mol 1000 mg
(2) From the balanced chemical equation, calculate moles of base
→ KC8H4O4-(aq) )+Na+(aq)+H2O(l)
HKC8H4O4(aq)+Na+(aq)+OH-(aq)
0.0002449 mol KHP x (1 mol OH- / 1 mol KHP) = 0.0002449 mol OH(3) From the difference in the burette readings (V of NaOH), and the
definition of molarity, calculate the molarity of the base.
moles of NaOH 0.0002449 mol
molarity of NaOH =
=
= 0.007349 M
V NaOH
0.3332 L
Example: Aluminum hydroxide reacts with hydrochloric acid
according to the balanced equation
Al(OH)3 (s) + 3 HCl (aq)
→ 3 H2O(l) + AlCl3 (aq)
What volume of 1.50 M HCl(aq) is required to neutralize
10.0 g Al(OH)3(s)?
Strategy:
(1) Calculate moles of Al(OH)3(s).
(2) Calculate moles of HCl needed using balanced equation
(3) Calculate volume HCl from (2) and known molarity
(1) Calculate moles of Al(OH)3(s).
10.0 g Al ( OH )3
= 0.128 mol Al ( OH )3
-1
78.00 g mol
(2) Calculate moles of HCl needed using balanced equation
Al(OH)3 (s) + 3 HCl (aq)
→ 3 H2O(l) + AlCl3 (aq)
3 mol HCl
0.128 mol Al ( OH )3 ×
= 0.385 mol HCl
mol Al ( OH )3
(3) Calculate volume HCl from (2) and known molarity
mol HCl 0.385 mol HCl
V=
=
= 0.256 L
-1
molarity
1.50 mol L
Chapter 4: Types of Chemical Reactions and
Solution Stoichiometry
4.1 Water, the Common Solvent
4.2 The Nature of Aqueous Solutions: Strong and Weak Electrolytes
4.3 The Composition of Solutions
4.4 Types of Chemical Reactions
4.5 Precipitation Reactions
4.6 Describing Reactions in Solution
4.7 Selective Precipitation
4.8 Stoichiometry of Precipitation Reactions
4.9 Acid-Base Reactions
NOW:
4.10 Oxidation-Reduction Reactions (Redox rxns.)
4.11 Balancing Oxidation-Reduction Equations
4.12 Simple Oxidation-Reduction Titrations
Figure 4.19: Reaction of solid sodium and
gaseous chlorine to form solid sodium
chloride: an oxidation – reduction reaction.
Oxidation - A substance gives up electrons to another
substance. Mg + ½ O2 → Mg2+ + O2- Mg is oxidized
Reduction - A substance accept electrons from another
substance. Mg + ½ O2 → Mg2+ + O2- O is reduced
In a chemical reaction, the total number of electrons
and the number of charges are conserved. It is
convenient to assign fictitious charges to the
atoms in a molecule and call them “oxidation
states” or “oxidation numbers” (ON).
Oxidation numbers are chosen so that (a) charge are
conserved, and (b) in ionic compounds the sum of
oxidation numbers on the atoms coincides with the
charge on the ion.
(also called Oxidation Numbers)
(i.e., except w/ O and higher halogens)
The Periodic Table of the Elements
Most Probable Oxidation State
0
+1
+3 +_4 - 3
H +2
Li Be
B C N
+1 + 2 Al Si P
Na Mg +3 +4 +5
K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As
Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb
Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi
Fr Ra Ac Rf Db Sg Bh Hs Mt Ds
+3
+3
-2 -1
He
O F Ne
S Cl Ar
Se Br Kr
Te I Xe
Po At Rn
Ce Pr Nd PmSm Eu Gd Tb Dy Ho Er TmYb Lu
Th Pa U Np Pu AmCmBk Cf Es FmMd No Lr
Determine the oxidation number (ON) of each element
in the following compounds.
Ex 1: Iron III Chloride Ex 2: Nitrogen Dioxide
Ex 3: Sulfuric acid
Strategy: We apply the rules in Table 4.3, always
making sure that the ON values in a compound add up to
zero, and in a polyatomic ion, to the ion’s charge.
Solution:
Ex 1: FeCl3 This compound is composed of monoatomic
ions. The ON of Cl- is -1, for a total of -3. Therefore the
ON of Fe is +3.
Ex 2: NO2
The ON of oxygen is -2 for a total of -4. Since the ON in
a compound must add up to zero, the ON of N is +4.
Ex 3: H2SO4
The ON of H is +1, so the SO42- group must sum to -2.
The ON of each O is -2 for a total of -8. Therefore S has
the ON +6.
An aside: What is the ON for S in the SO32- ion?
Examples (cont.):
Ex 4: BaI2
I = -1
Ba = +2
Ex 5: NH4NO3
H = +1
O = -2
N = -3
N = +5
(i.e., except w/ O and higher halogens)
Recognizing Oxidizing and Reducing Agents - I
Problem: Identify the oxidizing and reducing agent in each Rxn:
a) Zn(s) + 2 HCl(aq)
ZnCl2 (aq) + H2 (g)
b) S8 (s) + 12 O2 (g)
8 SO3 (g)
c) NiO(s) + CO(g)
Ni(s) + CO2 (g)
Plan: First we assign an oxidation number (O.N.) to each atom (or ion)
based on the rules in Table 4.3. Then apply RULE:
RULE: The reactant is the reducing agent if it contains an atom that
is oxidized (O.N. increased in the reaction). The/reactant is the
oxidizing agent if it contains an atom that is reduced ( O.N. decreased).
Solution:
a) Assigning oxidation numbers:
-1
+1
0
Zn(s) + 2 HCl(aq)
-1
0
+2
ZnCl2 (aq) + H2 (g)
HCl is the oxidizing agent, and Zn is the reducing agent!
Recognizing Oxidizing and Reducing Agents - II
b) Assigning oxidation numbers:
0
0
S8 (s) + 12 O2 (g)
+6
S [0]
-2
8 SO3 (g)
S[+6]
S is Oxidized
O[0]
O[-2]
O is Reduced
S8 is the reducing agent and O2 is the oxidizing agent
c) Assigning oxidation numbers:
Ni[+2]
Ni[0]
Ni is Reduced
C[+2]
-2
+2
NiO(s)
-2
+2
+ CO(g)
0
Ni(s)
+4 -2
+ CO2 (g)
C[+4]
C is oxidized
CO is the reducing agent and NiO is the oxidizing agent
Activity Series of the Metals
Strongly
reducing
Weakly
reducing
Li
K
Ba
Ca
Na
Li+ + eK+ + eBa2+ + 2 eCa2+ + 2 eNa+ + e-
Mg
Al
Mn
Zn
Cr
Fe
Mg2+ + 2eAl3+ + 3eMn2+ + 2eZn2+ + 2eCr3+ + 3eFe2+ + 2e-
Co
Ni
Sn
Co2+ + 2eNi2+ + 2eSn2+ + 2e-
H2
2 H+ + 2e-
Cu
Ag
Hg
Pt
Au
Cu2+ + 2eAg+ + eHg2+ + 2ePt2+ + 2eAu3+ + 3e-
These elements react rapidly with aqueous H+ ions
(acid) or with liquid H2O to release H2 gas.
These elements react with aqueous H+ ions or with
steam to release H2 gas.
These elements react with aqueous H+ ions to
release H2 gas.
These elements do not react with aqueous H+ ions
to release H2 gas.
Examples of Activity Series Predictions
Cu(s) + 2 Ag+(aq)
Cu2+(aq) + 2 Ag(s)
2 Fe(s) + 3 Cu2+(aq)
2 Fe3+(aq) + 3 Cu(s)
Mg(s) + Zn2+(aq)
Mg2+(aq) + Zn(s)
2 K(s) + Sn2+(aq)
2 K+(aq) + Sn(s)
Pt(s) + Ni2+(aq)
N. R.
2 Al(s) + 6 H+(aq)
2 Al3+(aq) + 3 H2 (g)
Au(s) + H+(aq)
N. R.
Balancing REDOX Equations:
The oxidation states (number) method
Step 1) Assign oxidation numbers to all elements in the equation.
Step 2) From the changes in oxidation numbers, identify the oxidized
and reduced species.
Step 3) Compute the number of electrons lost in the oxidation and
gained in the reduction from the oxidation number changes.
Draw tie-lines between these atoms to show electron changes.
Step 4) Choose coefficients for these species to make the electrons lost
equal the electrons gained (or total increase in ON = total
decrease in ON), remembering to first balance the numbers of
atoms of those elements which change ON (red = my addition).
Step 5) Complete the balancing by inspection.
REDOX Balancing using Ox. No. Method - I
+2 e- per O
0
___ H2 (g) +___ O2 (g)
0
- 1 e- per H
-2
___ H2O(g)
+1
electrons lost must = electrons gained
Two O atoms each gain 2 eTherefore need 4 H atoms give these 4 e-s!
REDOX Balancing using Ox. No. Method - I
2x +2 e-
0
___
2 H2 (g) +_1_ O2 (g)
0
4x -1 e-
-2
___
2 H2O(g)
+1
electrons lost must = electrons gained
Two O atoms each gain 2 eTherefore need 4 H atoms give these 4 e-s!
And we are balanced! (not necessarily emotionally)
DEMO
Balance the thermite reaction:
__ Fe2O3(s) + __ Al(s) → __ Al2O3(s) + __ Fe(l)
Assign ON to each element
Reactants
O: -2 Al:0
Fe:+3
Products
O:-2 Al:+3
Fe:0
Identify the elements that are oxidized and reduced
Fe3+ →Fe0 : Iron is reduced (ON decreases by 3)
Al0 →Al3+ : Aluminum is oxidized (ON increases by 3)
Make ON total increase = ON total loss
Fe2O3 + 2Al → Al2O3 + 2Fe
Thermite reaction
_ Fe2O3(s) + __Al(s) → _ Al2O3(s) + _ Fe(l)
+3
0 →
+3
- 3e- per Al
+3e- per Fe
0
Thermite reaction
Fe2O3(s) + 2 Al(s) → Al2O3(s) + 2 Fe(l)
+3
0 →
+3
2x(- 3e- per Al)
2x(+3e- per Fe)
BALANCED!
0
Problem: Calculate the mass of metallic Iron that must be
added to 500.0 liters of a solution containing
0.00040 M Pt2+(aq) ions in solution to reclaim all Pt
via: __ Fe(s) + __ Pt2+(aq) → __ Fe3+(aq) + __ Pt(s)
Solution:
V x M = # moles
500.0 L x 0.00040 mol Pt2+/ L = 0.20 mol Pt2+
Balance the equation to detn. how many
moles of Fe needed for every mole of Pt.
0.20 mol Pt2+ x
=
Problem: Calculate the mass of metallic Iron that must be
added to 500.0 liters of a solution containing
0.00040M of Pt2+(aq) ions in solution to reclaim all Pt
via: 2 Fe(s) + 3 Pt2+(aq) → 2 Fe3+(aq) + 3 Pt(s)
Solution:
V x M = # moles
500.0L x 0.00040 mol Pt2+/ L = 0.20 mol Pt2+
Fe(s) → Fe3+ + 3 ePt2+ + 2 e- → Pt(s)
Need 2 moles of Iron for every 3 moles of Platinum
0.20 mol
Pt2+
2 mol Fe
x
= 0.133 mol Fe
2+
3 mol Pt
0.133 mol Fe x
55.85 g Fe
= 7.4 g Fe
mol Fe
REDOX Balancing Using Ox. No. Method - II
+2
-1e- per Fe
+3
Fe+2(aq) + MnO4-(aq) + H+(aq)
Fe+3(aq) + Mn+2(aq) + H2O(aq)
+5 e- per Mn
+2
+7
Balance the number of each redox element, and then # electrons:
Multiply Fe+2 & Fe+3 by five to balance the electrons gained by Mn:
5 Fe+2(aq) + MnO4-(aq) + H+(aq)
5 Fe+3(aq) + Mn+2(aq) + H2O(aq)
Balance O: Need 4 on H2O on right to balance 4 O from the MnO4-.
Balance H: Need 8 on H+ on the left to balance 8 H in these 4 H2O s.
5 Fe+2(aq) + MnO4-(aq) +8 H+(aq)
5 Fe+3(aq) + Mn+2(aq) +4 H2O(aq)
BALANCED!
Note: Add 8 water molecules to both sides if you prefer to express the 8 H+ ions instead as H3O+ ions.
REDOX Balancing Using Ox. No. Method - II
+2
-1e- per Fe
+3
Fe+2(aq) + MnO4-(aq) + H3O+(aq)
+7
Fe+3(aq) + Mn+2(aq) + H2O(aq)
+5 e- per Mn
+2
5 Fe+2(aq) + MnO4-(aq) +8 H+(aq)
5 Fe+3(aq) + Mn+2(aq) +4 H2O(aq)
5 Fe+2(aq) + MnO4-(aq) +8 H3O+(aq)
5 Fe+3(aq) + Mn+2(aq) +12 H2O(aq)
Balancing redox eqn using half-cell method in acidic solutions
Cu(s) +HNO3(aq) → Cu2+(aq) + NO(g)
Identify half-reactions, one is ox, other is red
Cu(s) → Cu2+(aq)
0 → +2
HNO3(aq) → NO(g)
+5 → +2
Balance all atoms that are neither H nor O
OK as is
Balance O by adding H2O to side deficient in O
Cu(s) → Cu2+(aq)
HNO3(aq) → NO(g) + 2H2O
Balance H by adding H+ to side deficient in H
Cu(s) → Cu2+(aq)
3H+ +HNO3(aq) → NO(g) + 2H2O
Balance charge by adding e- to side that has + charge
Cu(s) → Cu2+(aq) + 2e3H+ + HNO3(aq) + 3e- → NO(g) + 2H2O
Multiply each eq by factors so electrons cancel out
3x(Cu(s) → Cu2+(aq) + 2e-)
2x(3H+ + HNO3(aq) + 3e- → NO(g) + 2H2O)
Add equations and cancel spectators (none here)
3Cu(s)+ 6H+ + 2HNO3(aq) → 3Cu2+(aq) + 2NO(g) + 4H2O
Balancing redox
equations in
basic solutions:
First balance in
acidic solution
3Cu(s)+ 6H+(aq)+ 2HNO3(aq) → 3Cu2+(aq) + 2NO(g) + 4H2O
+6 OH-(aq)
+6 OH-(aq)
3Cu(s)+ 6H2O(l)+ 2HNO3(aq) → 3Cu2+(aq) + 2NO(g) + 4H2O
+6 OH-(aq)
3Cu(s)+ 2H2O(l)+ 2HNO3(aq) → 3Cu2+(aq) + 2NO(g) + 4H2O
+6 OH-(aq)
The following redox equation is balanced in acidic
solution. Balance it for a basic solution.
Cr2O72-(aq)+2NO(g)+6H+(aq)→2Cr3+(aq)+2NO3- (aq)
+3H2O(l)
Add OH- equal to number of H+ on both sides
6OH-(aq) +Cr2O72-(aq)+2NO(g)+6H+(aq)→2Cr3+(aq)+2NO3-+3H2O(l)
+6OH-(aq)
Combine OH- and H+ to form water to max extent possible
6H2O(l)+Cr2O72-(aq)+2NO(g)→2Cr3+(aq)+2NO3-+3H2O(l) +6OH- (aq)
Cancel H2O on both sides to max extent possible
3H2O(l)+Cr2O72-(aq)+2NO(g)→2Cr3+(aq)+2NO3- +6OH- (aq)
Balancing Redox Equations in Aqueous
Acid and Base Solutions :
ACID : You may add either H+ ( H3O+ ), or water ( H2O ) to either side
of the chemical equation.
H+ + OH -
H2O
BASE : You may add either OH -, or water to either side of the
chemical equation.
H+ + OH H+ + H2O
H2O
H3O+
REDOX Balancing by Half-Reaction Method-I
Fe+2(aq) + MnO4-(aq)
Fe+3(aq) + Mn+2(aq) [acid solution]
Identify Oxidation and Reduction Half Reactions
Fe+2(aq)
MnO4-(aq)
Fe+3(aq) + e- [oxidation half-reaction]
Mn+2(aq)
Add water as needed to balance O, then add H+ to balance H, then e-s.
MnO4-(aq) + 8 H+(aq) +5e-
Mn+2(aq) + 4 H2O(l)
[reduction half-reaction]
Sum the two half-reactions (multiplied to balance electrons):
{ Fe+2(aq)
Fe+3(aq) +e- } x5
MnO4-(aq) + 8 H+(aq) +5eMn+2(aq) + 4H2O(l)
MnO4-(aq)+ 8 H+(aq)+5e- +5Fe+2(aq)
5Fe+3(aq)+5e- + Mn+2(aq)+ 4 H2O(l)
REDOX Balancing by Half-Reaction Method-III
MnO4-(aq) + SO32-(aq)
MnO2 (s) + SO42-(aq) [Acidic solution]
Oxidation:
SO32-(aq)
SO42-(aq)
Add water to balance O, then add protons to balance H,
and finally add electrons to balance charge.
SO32-(aq) + H2O(l)
SO42-(aq) + 2 H+(aq) + 2 e Reduction:
MnO4-(aq)
MnO2 (s)
Add water to balance O, then add protons to balance H,
and finally add electrons to balance charge.
MnO4-(aq) + 3 e- + 4H+
MnO2 (s) + 2 H2O(l)
Multiply the oxidation equation by 3, and the reduction equation by 2
to balance electrons. Cancel electrons, protons and water molecules.
3SO32-(aq) + 2MnO4-(aq) + 2H+(aq)
3 SO42-(aq) + 2MnO2 (s) + H2O(l)
REDOX Balancing using Ox. No. Method - III
+7
+ 3 e-
MnO4-(aq) + SO32-(aq)
+4
( Acidic Solution )
MnO2 (s) + SO42-(aq)
+4
- 2 e+6
To balance the electrons, we must multiply the sulfite by 3, and the
permanganate by 2. Then balance oxygen by adding water.
Then add H+ as needed.
2 MnO4-(aq) + 3 SO32-(aq) + 2H+(aq)
2 MnO2 (s) + 3 SO42-(aq) + H2O(aq)
If desired, you may write the H+ ions as hydronium ions instead:
2 MnO4-(aq)+ 3 SO32-(aq)+2 H3O+(aq)
2 MnO2 (s) + 3 SO42-(aq) +3 H2O(aq)
REDOX Balancing by Half-Reaction Method-IV
MnO4-(aq) +SO32-(aq)
MnO2(s) + SO42-(aq) [Basic solution]
balance the equation as if it were in acid, and then convert it to base:
2MnO4-(aq) + 3SO32-(aq) + 2H+(aq)
2MnO2(s) + 3SO42-(aq) + H2O(l)
To convert to base, add enough OH- to each side to cancel any H+:
2MnO4-(aq)+ 3SO32-(aq)+ 2 H+(aq) + 2 OH(aq)
2MnO2(s)+ 3SO42-(aq)+ H2O(l)+2OH-(aq)
On the reactant side, the H+ and the OH- cancel to give water.
2MnO4-(aq)+ 3SO32-(aq)+2H2O(l)
2MnO2(s)+ 3SO42-(aq)+ H2O(l)+2OH-(aq)
Cancel out the water on each side of the equation, and you are done!
2MnO4-(aq) + 3SO32-(aq) + H2O(l)
2MnO2(s) + 3SO42-(aq) +2OH-(aq)
REDOX Balancing Using Ox. No. Method-IV
Zinc metal is dissolved in nitric acid to give Zn2+ and the ammonium
ion from the reduced nitric acid. Write the balanced chemical equation!
Zn(s) + H+(aq) + NO3-(aq)
Zn2+(aq) + NH4+(aq)
Oxidation # method
- 2 e-
Zn(s) + H+(aq) + NO3-(aq)
Zn2+(aq) + NH4+(aq)
+5
+8 e-3
Multiply Zinc and Zn2+ by 4, and ammonia by unity. Since we have no
oxygen on the product side, so add 3 water molecules to the products.
It requires 10 H+ on the reactant side to balance H.
4 Zn(s) +10 H+(aq) + NO3-(aq)
4 Zn2+(aq) + NH4+(aq) + 3 H2O(l)
REDOX Balancing by Half-Reaction Method-V
Given:
Oxidation:
Zn(s) + H3O+(aq) + NO3-(aq)
Zn(s)
Zn2+(aq) + NH4+(aq)
Zn2+ + 2 e-
Reduction:
NO3-(aq)
NH4+(aq)
Other atoms (not O, H) are already balanced (N).
We need 3 waters to balance the O from the nitrate ion.
For H balance, we need 10 hydrogen ions.
Charge balance then requires 8 e-.
NH4+(aq) + 3 H2O(l)
10 H+(aq) + NO3-(aq) + 8 e -
Finally, if we are to add the two equations, we must multiply the Ox. one
by 4 to be able to cancel out the electrons, so the final balanced
equation is:
10 H+(aq) + NO3-(aq) + 4 Zn(s)
4 Zn+2(aq) + NH4+(aq) + 3 H2O(l)
REDOX Balancing by Half-Reaction Method
-VI - A
In acid, potassium dichromate reacts with ethanol(C2H5OH) to yield the
blue-green solution of Cr+3, the reaction used in “breathalyzers”.
H3O+(aq) + Cr2O72-(aq) + C2H5OH(l)
Oxidation:
Cr3+(aq) + CO2 (g) + H2O(l)
C2H5OH(l)
CO2 (g)
Balance C first. Then balance O by adding water to the reactant side.
Balance H by adding protons to the product side:
C2H5OH(l) + 3 H2O(l)
Add the electrons:
C2H5OH(l) + 3 H2O(l)
2 CO2 (g) + 12 H+(aq)
2 CO2 (g) + 12 H+(aq) + 12 e -
REDOX Balancing by Half-Reaction Method
- VI - B
Reduction:
Cr2O72-(aq)
Cr+3(aq)
Dichromate has two chromium atoms, therefore the products need to
have two Cr+3. The 7 oxygen atoms from the
dichromate need to be balanced with water on the product side.
Then add protons to the reactant side to balance H.
14H+(aq) + Cr2O72-(aq)
2 Cr+3(aq) + 7 H2O(l)
Add 6 electrons to the reactant side to balance charge:
6e - + 14 H+(aq) + Cr2O72-(aq)
2 Cr+3(aq) + 7 H2O(l)
Adding the two equations to cancel electrons:
Ox:
C2H5OH(l) + 3 H2O(l)
2 CO2 (g) + 12 H+(aq) + 12 e Rd: [6e - + 14 H+(aq) + Cr2O72-(aq)
2 Cr+3(aq) + 7 H2O(l)] x 2
C2H5OH(l) + 16 H+(aq) + 2 Cr2O72-(aq)
2 CO2 (g) + 4 Cr+3(aq) + 11 H2O(l)
REDOX Balancing by Half-Reaction Method
-VII - A
Silver is reclaimed from ores by extraction using basic Cyanide ion.
OH
Ag(s) + CN (aq) + O2 (g)
Ag(CN)2-(aq)
Oxidation:
CN-(aq) + Ag(s)
Ag(CN)2-(aq)
Since we need two cyanide ions to form the complex, add two to the
reactant side of the equation. Silver is also oxidized, so it looses an
electron, so we add one electron to the product side.
2 CN-(aq) + Ag(s)
Ag(CN)2-(aq) + e Reduction:
O2 (g)
2 H2O(l)
Add 2 water molecules to bal. O. Add 4 H+ to bal. H. Needs 4e- .
4 e - + O2 (g) + 4 H+(aq)
2 H2O(aq)
Basic: Add 4 OH- ions on each side to neutralize H+.
4 e - + O2 (g) + 2 H2O(aq)
4 OH-(l)
REDOX Balancing by Half-Reaction Method
- VII - B
Adding the Reduction equation to the Oxidation equation will require
the Oxidation one to be multiplied by 4 to eliminate the electrons.
Ox (x4)
8CN-(aq) + 4 Ag(s)
4 Ag(CN)2-(aq) + 4 e -
Rd
4 e - + O2 (g) + 2 H2O(l)
4 OH -(aq)
8 CN -(aq) + 4 Ag(s) + O2 (g) + 2 H2O(l)
4 Ag(CN)2-(aq) + 4 OH -(aq)
REDOX Balancing Using Ox. No. Method -V
0
-1 e -
Ag(s) + CN -(aq) + O2 (g)
+1
Ag(CN)2-(aq) + OH -(aq)
0
+ 2 e-2
To balance electrons we must put a 4 in front of the Ag, since each
oxygen looses two electrons, and they come two at a time! That requires
us to put a 4 in front of the silver complex, yielding 8 cyanide ions.
4 Ag(CN)2-(aq) + OH -(aq)
4 Ag(s) + 8 CN -(aq) + O2 (g)
We have no hydrogen's on the reactant side therefore we must add water
as a reactant, and since we also add oxygen, we must add two water
molecules, that well give us 4 hydroxide anions, giving us a balanced
chemical equation.
4 Ag(s) + 8 CN -(aq) + O2 (g) + 2 H2O(l)
4 Ag(CN)2-(aq) + 4 OH -(aq)
A Redox Titration Example Problem: Ca in Blood
In order to measure the Ca2+ conc. in blood, calcium oxalate was
precipitated from a 1.00 mL sample of blood. (“oxalate” = C2O42-.)
This precipitate was dissolved in a sulfuric acid solution, which
then required 2.05 mL of 4.88 x 10-4 M KMnO4 to reach the
endpoint via the rxn.:
2MnO4-(aq) + 5C2O42-(aq) + 16H+ →
2Mn 2+ (aq) + 10CO2(g) + 8H2O(l)
a) Calculate the moles of C2O42- which = moles of Ca2+ in the
blood sample.
b) Calculate the Ca2+ conc. in blood, in M then in mg / 100 mL.
Redox Titration- Calculation outline - I
Volume (L) of KMnO4 Solution
a)
M (mol/L)
Moles of KMnO4
b)
Molar ratio
Moles of CaC2O4
c)
Chemical Formula
Moles of Ca+2
Problem: Calcium Oxalate was
precipitated from blood by the
addition of Sodium Oxalate so that
calcium ion could be determined. In
the blood sample. The sulfuric acid
solution that the precipitate was
dissolved in required 2.05 ml of
4.88 x 10-4 M KMnO4 to reach the
endpoint.
a) calculate the amount (mol) of
Ca+2.
b) calculate the Ca+2 ion conc.
Plan: a) Calculate the molarity of
Ca+2 in the H2SO4 solution.
b) Convert the Ca+2 concentration
into units of mg Ca+2/ 100 ml blood.
Redox Titration - Calculation – cont.
Solution:
a) Calculate the number of moles of MnO4- used in titration.
moles MnO-4 = V×M = 2.05×10-3 L×4.88×10-4 moles L-1
=1.00 ×10-6 moles
b) Calculate the number of moles of calcium oxalate using
balanced chemical equation
25
moles
C
O
-6
22 4
1.00 ×10-6moles MnO-4 ×
=
2.50
×
10
moles
C
O
2 4
2 moles MnO4
moles Ca2+ = moles C2O42c) Calculate the concentration of Ca2+ in the blood sample.
2+
−6
moles
Ca
2.50
×
10
moles
2+
molarity Ca =
=
volume blood
1.00 ×10−3 L
= 2.50 ×10−3 M
Redox Titration - Calculation – cont.
Change this M concentration into the correct units:
a) Mol Ca+2 per 100. mL (or 0.100 L) of Blood
2.50x10-3 mol Ca+2 x 0.100 L Blood = 2.50 x 10-4 mol Ca+2
L Blood
100. mL Blood
100. mL Blood
b) mass (g) of Ca+2
Mass Ca+2 = 2.50 x 10 -4mol Ca+2 x 40.08g Ca/mol = 0.0100 g Ca+2
c) mass (mg) of Ca+2
Mass Ca+2 = 0.0100g Ca+2 x 1000mg Ca+2/g Ca+2 = 10.0 mg Ca+2
100 ml Blood