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Chapter 4
The Major Classes of Chemical Reactions
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The Major Classes of Chemical Reactions
4.1 The Role of Water as a Solvent
4.2 Writing Equations for Aqueous Ionic Reactions 4.3 Precipitation Reactions 4.4 Acid-Base Reactions 4.5 Oxidation-Reduction (Redox) Reactions 4.6 Elements in Redox Reactions
4.7 Reversible Reactions: An Introduction to Chemical Equilibrium
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Figure 4.1
Electron distribution in molecules of H2 and H2O.
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Figure 4.2
The dissolution of an ionic compound.
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Figure 4.3
The electrical conductivity of ionic solutions.
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Sample Problem 4.1 Determining Moles of Ions in Aqueous Ionic Solutions PROBLEM: How many moles of each ion are in the following solutions?
(a) 5.0 mol of ammonium sulfate dissolved in water
(b) 78.5 g of cesium bromide dissolved in water (c) 7.42x1022 formula units of copper(II) nitrate dissolved in water (d) 35 mL of 0.84 M zinc chloride PLAN: We have to relate the information given and the number of moles of ions present when the substance dissolves in water.
H2O
SOLUTION: (a) (NH4)2SO4(s) 5.0 mol (NH4)2SO4
2NH4+(aq) + SO42-(aq) = 10. mol NH4+
5.0 mol SO42-
2 mol NH4+ 1 mol (NH4)2SO4
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Sample Problem 4.1 continued
H2O
Determining Moles of Ions in Aqueous Ionic Solutions
(b) CsBr(s)
Cs+(aq) + Br-(aq)
= 0.369 mol Cs+ = 0.369 mol Br= 0.123 mol Cu2+
3
78.5 g CsBr
mol CsBr
212.8 g CsBr
= 0.369 mol CsBr
(c) Cu(NO3)2(s)
H2O
Cu2+(aq) + 2NO3-(aq)
mol Cu(NO3)2 7.42x1022 formula = 0.123 mol Cu(NO3)2 units Cu(NO3)2 6.022x1023 formula units = 0.246 mol NO
(d) ZnCl2(aq) 35 mL ZnCl2 1L 103mL
H2O
Zn2+(aq) + 2Cl-(aq) L = 2.9x110-2 mol ZnCl2 = 5.8x10-2 mol Cl-
0.84 mol ZnCl2
= 2.9x110-2 mol Zn2+
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Figure 4.4
The hydrated proton.
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Sample Problem 4.2
Determining the Molarity of H+ Ions in Aqueous
Solutions of Acids PROBLEM: Nitric acid is a major chemical in the fertilizer and explosives industries. In aqueous solution, each molecule dissociates and the H becomes a solvated H+ ion. What is the molarity of H+(aq) in 1.4M nitric acid? PLAN: Use the formula to find the molarity of H+.
+ SOLUTION: One mole of H (aq) is released per mole of nitric acid (HNO3)
H2O
HNO3(l)
H+(aq) + NO3-(aq)
1.4M HNO3(aq) should have 1.4M H+(aq).
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Writing Equations for Aqueous Ionic Reactions
The molecular equation
shows all of the reactants and products as intact, undissociated compounds.
The total ionic equation
shows all of the soluble ionic substances dissociated into ions.
The net ionic equation
eliminates the spectator ions and shows the actual chemical change taking place.
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Figure 4.5
A precipitation reaction and its equation.
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Figure 4.7
The reaction of Pb(NO3)2 and NaI.
NaI(aq) + Pb(NO3)2 (aq) PbI2(s) + NaNO3(aq) 2NaI(aq) + Pb(NO3)2 (aq)
PbI2(s) + 2NaNO3(aq) 2Na+(aq) + 2I-(aq) + Pb2+(aq) + 2NO3-(aq)
PbI2(s) + 2Na+(aq) + 2NO3-(aq)
2NaI(aq) + Pb(NO3)2(aq)
PbI2(s) + 2NaNO3(aq) double displacement reaction (metathesis)
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Predicting Whether a Precipitate Will Form
1. Note the ions present in the reactants.
2. Consider the possible cation-anion combinations.
3. Decide whether any of the ion combinations is insoluble.
See Table 4.1 (next slide) for solubility rules.
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Table 4.1 Solubility Rules For Ionic Compounds in Water
Soluble Ionic Compounds 1. All common compounds of Group 1A(1) ions (Li +, Na+, K+, etc.) and ammonium ion (NH4+) are soluble. 2. All common nitrates (NO3-), acetates (CH3COO- or C2H3O2-) and most perchlorates (ClO4-) are soluble.
3. All common chlorides (Cl-), bromides (Br-) and iodides (I-) are soluble, except those of Ag+, Pb2+, Cu+, and Hg22+.
Insoluble Ionic Compounds
1. All common metal hydroxides are insoluble, except those of Group 1A(1) and the larger members of Group 2A(2)(beginning with Ca 2+). 2. All common carbonates (CO32-) and phosphates (PO43-) are insoluble, except those of Group 1A(1) and NH4+. 3. All common sulfides are insoluble except those of Group 1A(1), Group 2A(2) and NH4+.
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Sample Problem 4.3
Predicting Whether a Precipitation Reaction
Occurs; Writing Ionic Equations PROBLEM: Predict whether a reaction occurs when each of the following pairs of solutions are mixed. If a reaction does occur, write balanced molecular, total ionic, and net ionic equations, and identify the spectator ions. (a) sodium sulfate(aq) + strontium nitrate(aq) (b) ammonium perchlorate(aq) + sodium bromide(aq) PLAN: write ions
SOLUTION: (a) Na2SO4(aq) + Sr(NO3)2 (aq)
2NaNO3(aq) + SrSO4(s)
2Na+(aq) +SO42-(aq)+ Sr2+(aq)+2NO3-(aq) 2Na+(aq) +2NO3-(aq)+ SrSO4(s) combine anions & cations check for insolubility Table 4.1 eliminate spectator ions for net ionic equation SO42-(aq)+ Sr2+(aq) (b) NH4ClO4(aq) + NaBr (aq) SrSO4(s) NH4Br (aq) + NaClO4(aq)
All reactants and products are soluble so no reaction occurs.
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Table 4.2 Selected Acids and Bases
Acids Strong Bases Strong
hydrochloric acid, HCl
hydrobromic acid, HBr
sodium hydroxide, NaOH
potassium hydroxide, KOH
hydroiodic acid, HI
nitric acid, HNO3 sulfuric acid, H2SO4 perchloric acid, HClO4 Weak
calcium hydroxide, Ca(OH)2
strontium hydroxide, Sr(OH)2 barium hydroxide, Ba(OH)2
Weak
hydrofluoric acid, HF
phosphoric acid, H3PO4 acetic acid, CH3COOH (or HC2H3O2)
ammonia, NH3
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Sample Problem 4.4 PROBLEM:
Writing Ionic Equations for Acid-Base Reactions
Write balanced molecular, total ionic, and net ionic equations for each of the following acid-base reactions and identify the spectator ions.
(a) strontium hydroxide(aq) + perchloric acid(aq) (b) barium hydroxide(aq) + sulfuric acid(aq)
PLAN:
SOLUTION:
reactants are strong acids and (a) Sr(OH)2(aq)+2HClO4(aq) 2H2O(l)+Sr(ClO4)2(aq) bases and therefore completely Sr2+(aq) + 2OH-(aq)+ 2H+(aq)+ 2ClO4-(aq) ionized in water 2H2O(l)+Sr2+(aq)+2ClO4-(aq) products are 2OH-(aq)+ 2H+(aq) 2H2O(l)
water spectator ions
(b) Ba(OH)2(aq) + H2SO4(aq)
2H2O(l) + BaSO4(aq)
Ba2+(aq) + 2OH-(aq)+ 2H+(aq)+ SO42-(aq) 2H2O(l)+Ba2+(aq)+SO42-(aq)
2OH-(aq)+ 2H+(aq) 2H2O(l)
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Figure 4.8
An acid-base titration.
Start of titration Excess of acid
Point of neutralization
Slight excess of base
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Sample Problem 4.5 PROBLEM:
Finding the Concentration of Acid from an
Acid-Base Titration You perform an acid-base titration to standardize an HCl solution by placing 50.00 mL of HCl in a flask with a few drops of indicator solution. You put 0.1524 M NaOH into the buret, and the initial reading is 0.55 mL. At the end point, the buret reading is 33.87 mL. What is the concentration of the HCl solution?
PLAN: volume(L) of base
SOLUTION: NaOH(aq) + HCl(aq) (33.87-0.55) mL x 1L
103 mL
NaCl(aq) + H2O(l) = 0.03332 L
multiply by M of base
mol of base
molar ratio
mol of acid divide by L of acid 0.03332 L X 0.1524 M = 5.078x10-3 mol NaOH Molar ratio is 1:1 5.078x10-3 mol HCl
M of acid
0.05000 L
= 0.1016 M HCl
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Figure 4.9
An aqueous strong acid-strong base reaction on the atomic scale.
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Figure 4.10
An acid-base reaction that forms a gaseous product.
Molecular equation NaHCO3(aq) + CH3COOH(aq) CH3COONa(aq) + CO2(g) + H2O(l) Total ionic equation
Na+(aq)+ HCO3-(aq) + CH3COOH(aq)
CH3COO-(aq) + Na+(aq) + CO2(g) + H2O(l)
Net ionic equation
HCO3-(aq) + CH3COOH(aq)
CH3COO-(aq) + CO2(g) + H2O(l)
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Figure 4.11
The redox process in compound formation.
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Table 4.3 Rules for Assigning an Oxidation Number (O.N.)
General rules 1. For an atom in its elemental form (Na, O2, Cl2, etc.): O.N. = 0 2. For a monoatomic ion: O.N. = ion charge 3. The sum of O.N. values for the atoms in a compound equals zero. The sum of O.N. values for the atoms in a polyatomic ion equals the ion's charge. Rules for specific atoms or periodic table groups
1. For Group 1A(1):
2. For Group 2A(2):
O.N. = +1 in all compounds
O.N. = +2 in all compounds
3. For hydrogen:
4. For fluorine: 5. For oxygen:
O.N. = +1 in combination with nonmetals
O.N. = -1 in combination with metals and boron O.N. = -1 in peroxides O.N. = -2 in all other compounds(except with F) O.N. = -1 in combination with metals, nonmetals (except O), and other halogens lower in the group
6. For Group 7A(17):
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Sample Problem 4.6 PROBLEM:
Determining the Oxidation Number of an Element
Determine the oxidation number (O.N.) of each element in these compounds: (a) zinc chloride (b) sulfur trioxide (c) nitric acid
PLAN:
The O.N.s of the ions in a polyatomic ion add up to the charge of the
ion and the O.N.s of the ions in the compound add up to zero.
SOLUTION: (a) ZnCl2. The O.N. for zinc is +2 and that for chloride is -1. (b) SO3. Each oxygen is an oxide with an O.N. of -2. Therefore the O.N. of sulfur must be +6. (c) HNO3. H has an O.N. of +1 and each oxygen is -2. Therefore the N must have an O.N. of +5.
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Figure 4.12
Highest and lowest oxidation numbers of reactive main-group elements.
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Figure 4.13
A summary of terminology for oxidation-reduction (redox) reactions.
etransfer Y or shift of electrons Y gains electron(s) Y is reduced
Y is the oxidizing agent Y decreases its oxidation number
X
X loses electron(s) X is oxidized
X is the reducing agent X increases its oxidation number
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Sample Problem 4.7
Recognizing Oxidizing and Reducing Agents
PROBLEM: Identify the oxidizing agent and reducing agent in each of the following:
(a) 2Al(s) + 3H2SO4(aq) (b) PbO(s) + CO(g) (c) 2H2(g) + O2(g) Al2(SO4)3(aq) + 3H2(g) Pb(s) + CO2(g) 2H2O(g)
PLAN: Assign an O.N. for each atom and see which atom gained and which atom lost electrons in going from reactants to products. An increase in O.N. means the species was oxidized (and is the reducing agent) and a decrease in O.N. means the species was reduced (is the oxidizing agent).
SOLUTION: 0 +1 +6 -2 +3 +6 -2 0
(a) 2Al(s) + 3H2SO4(aq)
Al2(SO4)3(aq) + 3H2(g)
The O.N. of Al increases; it is oxidized; it is the reducing agent.
The O.N. of H decreases; it is reduced; H2SO4 is the oxidizing agent.
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Sample Problem 4.7 continued +2 -2
Recognizing Oxidizing and Reducing Agents +2 -2 0 +4 -2
(b) PbO(s) + CO(g)
Pb(s) + CO2(g)
The O.N. of C increases; it is oxidized; CO is the reducing agent.
The O.N. of Pb decreases; it is reduced; PbO is the oxidizing agent. 0 0 +1 -2
(c) 2H2(g) + O2(g)
2H2O(g)
The O.N. of H increases; it is oxidized; it is the reducing agent.
The O.N. of O decreases; it is reduced; it is the oxidizing agent.
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Sample Problem 4.8
Balancing Redox Equations by the Oxidation
Number Method PROBLEM: Use the oxidation number method to balance the following equations: (a) Cu(s) + HNO3(aq) Cu(NO3)2(aq) + NO2(g) + H2O(l) (b) PbS(s) + O2(g) SOLUTION: 0 +1 +5 -2 PbO(s) + SO2(g) +2 +5 -2 +4 -2 +1 -2
(a) Cu(s) + HNO3(aq)
Cu(NO3)2(aq) + NO2(g) + H2O(l)
O.N. of Cu increases because it loses 2e -; it is oxidized and is the reducing agent. O.N. of N decreases because it gains1e -; it is reduced and is the oxidizing agent.
loses 2ebalance other ions
Cu(s) + HNO3(aq)
Cu(NO3)2(aq) + NO2(g) + H2O(l) gains 1ex2 to balance e-
balance unchanged polyatomic ions
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Sample Problem 4.8 continued
Balancing Redox Equations by the Oxidation Number Method
Cu(s) + 2 HNO3(aq) 4
Cu(NO3)2(aq) + 2 NO2(g) + 2 H2O(l) +2 -2 +4 -2
2
+2 -2 0
(b) PbS(s) + O2(g)
PbO(s) + SO2(g)
loses 6ePbS(s) + 3/2 O2(g) PbO(s) + SO2(g)
gains 2e- per O; need 3/2 O2 to make 3O2total of a 6e- exchange Multiply by 2 to have whole number coefficients. 2PbS(s) + 3O2(g) 2PbO(s) + 2SO2(g)
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Figure 4.14
A redox titration.
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Sample Problem 4.9
Finding an Unknown Concentration by a Redox
Titration 2+ PROBLEM: Calcium ion (Ca ) is required for blood to clot and for many other cell processes. An abnormal Ca2+ concentration is indicative of disease. To measure the Ca2+ concentration, 1.00mL of human blood was treated with Na2C2O4 solution. The resulting CaC2O4 precipitate was filtered and dissolved in dilute H 2SO4. This solution required 2.05mL of 4.88x10-4M KMnO4 to reach the end point. The unbalanced equation is KMnO4(aq) + CaC2O4(s) + H2SO4(aq) MnSO4(aq) + K2SO4(aq) + CaSO4(s) + CO2(g) + H2O(l) (a) Calculate the amount (mol) of Ca2+. (b) Calculate the amount (mol) of Ca2+ ion concentration expressed in units of mg Ca2+/100mL blood.
PLAN:
(a)
volume of KMnO4 soln multiply by M mol of KMnO4
mol of Ca2+ ratio of elements in formula mol of CaC2O4
molar ratio
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Sample Problem 4.9 continued 2.05mL soln L 103 mL 1.00x10-6mol KMnO4 2.50x10-6 mol CaC2O4 PLAN: (b)
Finding an Unknown Concentration by a Redox Titration SOLUTION: 4.88x10-4mol KMnO4 L 5mol CaC2O4 = 2.50x10-6 mol CaC2O4 = 1.00x10-6mol KMnO4
2mol KMnO4
1mol Ca2+ = 2.50x10-6 mol Ca2+
1mol CaC2O4
mol Ca2+/1mL blood multiply by 100 mol Ca2+/100mL blood multiply by M g Ca2+/100mL blood 10-3g = 1mg =10.0mg Ca2+/100mL blood SOLUTION: 2.50x10-6 mol Ca2+ x100 =2.50x10-4 mol Ca2+ 1mL blood 100mL blood
2.50x10-4 mol Ca2+ 40.08g Ca2+ mg 100mL blood mol Ca2+ 10-3g
mg Ca2+/100mL blood
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Figure 4.15 Combining elements to form an ionic compound.
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Figure 4.16
Decomposing a compound to its elements.
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Figure 4.17
An active metal displacing hydrogen from water.
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Figure 4.18
The displacement of H from acid by nickel.
O.N. increasing oxidation occurring reducing agent
O.N. decreasing reduction occurring oxidizing agent
0
+1
+2
0
Ni(s) + 2H+(aq)
Ni2+(aq) + H2(g)
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Figure 4.19
Displacing one metal with another.
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Figure 4.20
The activity series of the metals.
strength as reducing agents
Li K Ba Ca Na Mg Al Mn An Cr Fe Cd Co Ni Sn Pb H2 Cu Hg Ag Au
can displace H from water
can displace H from steam
can displace H from acid
cannot displace H from any source
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Sample Problem 4.10 PROBLEM:
Identifying the Type of Redox Reaction
Classify each of the following redox reactions as a combination, decomposition, or displacement reaction, write a balanced molecular equation for each, as well as total and net ionic equations for part (c), and identify the oxidizing and reducing agents: magnesium nitride (aq)
(a) magnesium(s) + nitrogen(g) (b) hydrogen peroxide(l)
water(l) + oxygen gas
(c) aluminum(s) + lead(II) nitrate(aq)
aluminum nitrate(aq) + lead(s)
PLAN: Combination reactions produce fewer products than reactants. Decomposition reactions produce more products than reactants.
Displacement reactions have the same number of products and reactants.
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Sample Problem 4.10 continued
Identifying the Type of Redox Reaction
(a) Combination
0 3Mg(s) +
0 N2(g)
+2 -3 Mg3N2 (aq)
Mg is the reducing agent; N2 is the oxidizing agent. (b) Decomposition +1 -1 H2O2(l) +1 -2 0 or
H2O(l) + 1/2 O2(g)
2 H2O2(l)
H2O2 is the oxidizing and reducing agent. (c) Displacement 0 +2 +5 -2
2 H2O(l) +
O2(g)
+3 +5 -2
0
Al(s) + Pb(NO3)2(aq)
Al(NO3)3(aq) + Pb(s)
2Al(s) + 3Pb(NO3)2(aq)
2Al(NO3)3(aq) + 3Pb(s)
Pb(NO3)2 is the oxidizing and Al is the reducing agent.
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Figure 4.21
The equilibrium state.
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