Title: Chapter 4 Types of Chemical Reactions and Solution Stoichiometry
1Chapter 4Types of Chemical Reactions and
Solution Stoichiometry
- Water is the dissolving medium of the common
solvent Some properties - Water is bent or v-shaped
- The OH bonds are covalent
- Water is a polar molecule
- Hydration occurs when salts dissolve in water
-
2The Water Molecule is Polar
3Polar Water Molecules Interact with the Positive
and Negative Ions of a Salt
4- A solute (substance being dissolved)
- Dissolves in water or other solvent
- Changes phase if different from the solvent
- Present in lesser amount (if the same phase as
the solvent) - A solvent
- Retains its phase (if different from the solute)
- Present in greater amount (if the same phase as
the solute) - Rule for predicting solubility
- Like dissolves Like
5The Nature of Aqueous Solutions Strong and Weak
Electrolytes
- Strong electrolytes Conduct current very
efficiently. Strong electrolytes are completely
ionized when they are dissolved in water (soluble
salts, strong acids, strong bases). - Weak electrolytes Conduct only a small current.
Weak electrolytes are substances that exhibit a
small degree of ionization in water (weak acids,
weak bases). - Non electrolytes Permit no current to flow. Non
electrolytes are substances that dissolve in
water but do not produce any ions. (pure water,
sugar solution).
6BaCI2 Dissolving
7HCI (aq) is Completely Ionized
8Acetic Acid in Water
9The Composition of Solutions
- To perform stoichiometric calculation in solution
- we need-
- The nature of the reaction
- The amounts of chemicals present in the solutions
(concentration) - The concentration of a solution can be expressed
by Molarity (M). - Molarity (M) moles of solute per volume of
solution in liters - M molarity (moles of solute)/(liters of
solution) -
10- Example If you weigh out 5.84 g of NaCl and
dissolved it in 500. mL of water, what is the
molarity of the solution? - 5.84 g NaCl x (1 mol NaCl)/(58.44 g NaCl)
0.100 mol - 500. mL x 1 L/1000 mL 0.500 L
- Molarity (0.100 moles)/(0.500 L) 0.200 M
11Concentration of Ions
- Example Give the concentration of each type of
ion in the - following solution
- a. 0.50 M Co(NO3)2 b. 1 M Fe(ClO4)3
c. 1.75 M Al2(SO4)3 - a. Solution reaction Co(NO3)2(s) H2O Co2(aq)
2NO3-(aq) - Each mole of Co(NO3)2 that is dissolved, the
solution contains 1 - mol Co2 ions and 2 mol NO3- ions. Thus a
solution of 0.50 M - Co(NO3)2 contains 0.50 M Co2 and (2 x 0.50) M or
1.0 M NO3-. - b. Fe(ClO4)3(s) H2O Fe3(aq) 3ClO4-(aq)
- 1 M Fe(ClO4)3 contains 1 M Fe3 ions and 3 M
ClO4- ions. - c. Al2(SO4)3(s) H2O 2Al3(aq) 3SO42-(aq)
- 1.75 M Al2(SO4)3 contains (2 x 1.75) M or 3.50 M
Al3 ions and (3 - x 1.75) M or 5.25 M SO42- ions.
12Standard Solution
- A standard solution is a solution whose
concentration is accurately known. - Finding volume or weight
- How many grams of K2Cr2O7 are required to prepare
1.00 L of a 0.20 M solution? - M moles/L
- Moles M x L 0.200 M x 1.00 L 0.200 mol
K2Cr2O7. - 0.200 mol K2Cr2O7 x (294.20 g)/(1 mol) 58.8 g
13- What volume is needed to prepare a 1.20 M
solution from 10.0 g CaCl2 ? (mm 111.0 g) - M moles/L gt L moles/M
- 10.0 g CaCl2 x (1 mole CaCl2 )/111.0 g CaCl2)
0.090 moles - L (0.090 moles)/(1.20 M) 0.0750 L
- 75.0 mL
14Preparation of a Standard Solution
15Dilution
- The desired molarity solutions are often prepared
from concentrated stock solutions (routinely used
solutions prepared in concentrated form) by
adding water. This process is called dilution. - Dilution with water does not alter the numbers of
moles of solute present. - Moles of solute before dilution
- moles of solute after dilution
- M1V1 M2V2
16 (a) A Measuring Pipet (b) A
Volumetric (transfer) Pipet
17- Example What volume of 16 M sulfuric acid
- must be used to prepare 1.5 L of 0.10 M H2SO4
- solution?
- V1 Volume before dilution ?
- M1 Concentration before dilution 16 M
- V2 Volume after dilution 1.5 L
- M2 Concentration after dilution 0.10 M
- M1V1 M2V2
- V1 M2V2 / M1 (0.10 M x 1.5 L)/ 16 M
- 0.0094 L 9.4 mL
18Types of Chemical Reactions
- Solution Reactions are classified as follows
- Precipitation reactions
- Acid base reactions
- Oxidation reduction reactions
- Other classification
- Formation reaction
- Decomposition reaction
- Single replacement
- Double replacement
19Precipitation Reactions
- When two solutions are mixed, an insoluble
substance (solid) sometimes forms and separates
from the solution. Such a reaction is called a
precipitation reaction, and the solid that forms
is called a precipitate. - Example K2CrO4(aq) Ba(NO3)2(aq) ? Products
(Soluble) (Soluble) - Contains the ions K, CrO42-, Ba2, NO3-
- Product must contain both anions and cations
- Most ionic products contain one type of cation
and one type of anion
20- Precipitation Reactions continued
- The possible combinations are-
- K2CrO4, KNO3, BaCrO4, Ba(NO3)2
- (reactant) (reactant)
- The only real possibility for the solid yellow
- products are- KNO3 or BaCrO4
- Both K and NO3- ions are colorless but CrO42-
- ion is yellow, so the yellow solid is BaCrO4.
- K2CrO4(aq) Ba(NO3)2(aq) ? BaCrO4(s) 2KNO3(aq)
21The Reaction of K2CrO4(aq) and Ba(NO3)2(aq)
22Describing Reactions in Solutions
- Regular balanced equation or molecular equation
(gives the overall reaction stoichiometry but not
the actual forms of reactants and products) - K2CrO4(aq) Ba(NO3)2(aq) ? BaCrO4(s)
2KNO3(aq) - Complete ionic equation (represents as ions all
reactants and products that are strong
electrolyte) - 2K(aq) CrO42-(aq) Ba2(aq) 2NO3-(aq)
- ? BaCrO4(s) 2K(aq) 2NO3-(aq)
- Net ionic equation (Ions participate in
reaction) - Ba2(aq) CrO42-(aq) ? BaCrO4(s)
23- Spectator ions The ions that do not participate
directly in the reaction, present in solution
both before and after the reaction. K and NO3-
ions. - 2K(aq) CrO42-(aq) Ba2(aq) 2NO3-(aq)
- ? BaCrO4(s) 2K(aq) 2NO3-(aq)
24Simple Rules for Solubility
- 1. Most nitrate (NO3?) salts are soluble.
- 2. Most alkali (group 1A) salts and NH4 are
soluble. - 3. Most Cl?, Br?, and I? salts are soluble (NOT
Ag, Pb2, Hg22) - 4. Most sulfate salts are soluble (NOT BaSO4,
PbSO4, HgSO4, CaSO4) - 5. Most OH? salts are only slightly soluble
(NaOH, KOH are soluble, Ba(OH)2, Ca(OH)2 are
marginally soluble) - 6. Most S2?, CO32?, CrO42?, PO43? salts are only
slightly soluble.
25Stoichiometry of Precipitation Reactions
- Step 1 Identify the species present in the
combined solution, and determine what reaction
occurs. - Step 2 Write the balanced net ionic equation
for the reaction. - Step 3 Calculate the moles of the reactants
(use volume and molarity). - Step 4 Determine which reactant is limiting.
- Step 5 Calculate the moles of product or
products, as required. - Step 6 Convert to grams or other units, as
required.
26- Example Calculate the mass of solid NaCl that
must be added to 1.50 L of a 0.100 M AgNO3
solution to precipitate all the Ag ions in the
form of AgCl. - Ag NO-3 Na Cl-
- Ag(aq) Cl-(aq) ? AgCl(s)
- 1.50 L x (0.100 mol Ag) 0.150 mol Ag
- Ag and Cl- react in a 11 ratio, 0.150 mol Cl-
ions and 0.150 mol NaCl x (58.45 g NaCl)/(1 mol
NaCl) 8.77 g NaCl
27Acid-Base Reactions
- Arrhenius
- Acid is a substance that produces H ions when
dissolved in water - Base is substances that produces OH- ions
- Bronsted-Lowry
- An acid is a proton donor
- A base is a proton acceptor
- Balanced equation HCl NaOH ? NaCl H2O
- Complete ionic equation H Cl- Na OH- ?
- Na Cl- H2O
- Net ionic equation H OH- ?H2O
-
28Neutralization Reaction
- Example What volume of a 0.100 M HCl solution
is needed to neutralize 25.0 mL of 0.350 M NaOH? - The species available for reaction are H,
Cl-, Na, OH- - NaCl is soluble, so Na, Cl- are spectator ions,
net ionic equation, H(aq) OH-(aq) ? H2O(l) - Moles of OH- ions 25.0 mL x (1L)/(1000 mL) x
(0.350 mol OH- )/(L NaOH) 8.75 x 10-3 mol OH- - Hand OH- ions react in a 11 ratio, 8.75 X
10-3 mol H ions is required to neutralize OH-
ions present. - V x (0.100 mol H)/L 8.75 x 10-3 mol H
- V (8.75 x 10-3 mol H)/(0.100 mol H)/L
- 8.75 x 10-2 L x (1000 mL)/1L 87.5 mL
29Acid-Base Titration
- Acid-base reaction is called a neutralization
reaction (the acid and base are destroyed to
leave a neutral solution). - Volumetric analysis is a technique for
determining the amount of a certain substance by
doing a titration. - A titration involves delivery of a measured
volume of a solution of known concentration (the
titrant) into a solution containing the substance
being analyzed (the analyte). - Equivalence point or stoichiometric point
enough titrant added to react exactly with the
analyte. - Indicator A substance added at the beginning of
the titration that changes color at or near the
equivalence point. - End point The point where the indicator actually
changes color.
30Oxidation-Reduction Reactions
- In oxidation-reduction (redox) reaction, one
species loses electrons while another species
gains electrons. - Both processed must occur simultaneously. We
cant have a net ionic reaction that contain only
oxidation or only reduction. - The number of electron lost by the oxidation
process must equal the number gained by the
reduction process. We cannot create or destroy
electrons. - Cu ? Cu2 2e- (oxidation)
- 2Ag 2e- ? 2Ag (reduction)
- Cu 2Ag ? Cu2 2Ag (net ionic reaction)
31Oxidation States
- Oxidation states or oxidation numbers provides a
way to keep track of electrons (accounting for
electrons) in redox reactions. - Rules for Assigning Oxidation States
- The oxidation state of an atom in an element is O
(element in its elemental form is O). - The oxidation state of a monatomic ion is the
same as its charge. - In its compound, fluorine is always assigned an
oxidation state of 1.
32- Oxidation States continued
-
- 4. Oxygen is assigned an oxidation state 2 in
covalent compound (except in peroxides where it
is 1) - 5. Hydrogen is assigned an oxidation state 1 in
its covalent compounds with non metals. - 6. Sum of the oxidation states must be zero for
an electrically neutral compound, for an ion
equals the charge of the ion.
33Assigning Oxidation States
- Example Assign oxidation states to all atoms in
the - following.
- a. CO2 ? O -2 (for each oxygen) C 4
- SF6 ? F -1 (for each fluorine) S 6
- NO3- ? O -2 (each oxygen) N 5
- MnO4- ? O -2 (each oxygen) Mn 7
- HSO3- ? O -2 (each oxygen) H 1, S 4
- H2O ? H 1 (each) O -2
- Li3N ? Li 1 (each) N -3
-
34Oxidation-Reduction
- Oxidation is an increase in oxidation state or a
loss of electron. - Reduction is a decrease in oxidation state or a
gain of electron. - Oxidizing Agent is the electron acceptor
(reduced). - Reducing Agent is the electron donor (oxidized)
- PbO(s) CO(g) ? Pb(s) CO2(g)
-
- 2 -2 2 -2 0 4 -2 (each)
- Pb ? reduced, C ? oxidized, PbO ? oxidizing
agent, - CO ? reducing agent
35A Summary of an Oxidation-Reduction Process
36Balancing Oxidation-Reduction Equations
- Redox reactions in acidic solution
- Step 1 Write separate equation for the oxidation
and - reduction half reactions.
- Step 2 For each half reaction-
- Balance all the elements except hydrogen and
oxygen - Balance oxygen using H2O
- Balance hydrogen using H
- Balance the charge using electrons
- Step 3 If necessary, multiply one or both
balanced half- reactions by an integer to
equalize the number of electrons transferred - Step 4 Add the half reactions and cancel
identical species - Step 5 Check that the elements and charges are
balanced -
37Balancing Oxidation-Reduction Reactions
- Example Balance the equation for the reaction
- between permanganate and iron (II) ion in acidic
- solution
- MnO4-(aq) Fe2(aq) Acid Fe3(aq) Mn2(aq)
- Half reactions MnO4- ? Mn2 (reduction)
- Fe2 ? Fe3 (oxidation)
- Balance half reactions MnO4- ? Mn2
- Manganese is balanced
- Balance oxygen by adding 4H2O to the right side
of the equation
38- ..continued..
- c. Balance hydrogen by adding 8H to the left
side - 8H(aq) MnO4-(aq) ? Mn2(aq) 4H2O
- Equalize the charge by adding electrons
- 5e- 8H(aq) MnO4-(aq) ? Mn2(aq) 4H2O(l)
- Fe2(aq) ? Fe3(aq) e- x 5
- 5 Fe2(aq) ? 5 Fe3(aq) 5e-
-
- Add half reactions
- 5Fe2(aq) MnO-4(aq) 8H(aq)
- ? 5Fe3(aq) Mn2(aq) 4H2O(l)
- Check that elements and charges are balanced.
39Balancing Oxidation-Reduction Reactions in Basic
Solution
- Step 1 Use the half reaction method as
- specified for acidic solution.
- Step 2 Add a number of OH- ions that is equal
- to the number of H ions (to both
sides). - Step 3 Form H2O by combining H and OH-.
- Step 4 Check that elements and charges are
balanced.
40- Example
- ClO- CrO2- ? Cl - CrO42- (basic solution)
- 1 3 -1 6
- First half reaction
- ClO- 2e- ? Cl-
- ClO- 2e- ? Cl- H2O
- ClO- 2e- 2H ? Cl- H2O
- Second half reaction
- CrO2- ? CrO42-
- CrO2- ? CrO42- 3e-
- CrO2- 2H2O ? CrO42- 3e-
- CrO2- 2H2O ? CrO42- 3e- 4H
41- Continued
- Multiply first half reaction by 3 and second half
reaction - by 2 then add two half reactions.
- 3ClO- 6e- 6H ? 3Cl- 3H2O
- 2CrO2- 4H2O ? 2CrO42- 6e- 8H
- 3ClO- 2CrO2- H2O ? 3Cl- 2CrO42- 2H
- Add OH- to both sides to remove H
- 3ClO- 2CrO2- H2O 2OH- ? 3Cl- 2CrO42-
2H 2OH- - 3ClO- 2CrO2- H2O 2OH- ? 3Cl- 2CrO42-
2H2O - 3ClO- 2CrO2- 2OH- ? 3Cl- 2CrO42- H2O
42Summary
- Like dissolves Like
- Strong, weak and non electrolytes
- Molarity moles of solute/Liters of solution
- Standard Solution concentration accurately known
- Dilution M1V1 M2V2
- Precipitation, acid-base and oxidation reduction
reactions - Spectator ions
- Acids produces H or proton donor
- Bases produces OH- and proton acceptor
- Acid-Base titration Neutralization reaction
- Oxidation-Reduction reactions
- Oxidation States
- Balancing redox equations