CHAPTER 5 INTRODUCTION TO REACTIONS IN AQUEOUS SOLUTIONS PRACTICE EXAMPLES 1A
(E) In determining total Cl , we recall the definition of molarity: moles of solute per liter of
solution. 0.438 mol NaCl 1 mol Cl = 0.438 M Cl 1 L soln 1 mol NaCl 0.0512 mol MgCl2 2 mol Cl from MgCl2 , Cl = = 0.102 M Cl 1 L soln 1 mol MgCl2
from NaCl, Cl =
Cl total = Cl from NaCl + Cl from MgCl2 = 0.438 M + 0.102 M = 0.540 M Cl 1B
2A
(E) (a)
1.5 mg F1 g F 1 mol F = 7.9 10-5 M F L 1000 mg F 18.998 g F
(b)
1.00 106 L
7.9 105 mol F1 mol CaF2 78.075 g CaF2 1 kg = 3.1 kg CaF2 1 mol CaF2 1000 g 1L 2 mol F
(E) In each case, we use the solubility rules to determine whether either product is insoluble. The ions in each product compound are determined by simply “switching the partners” of the reactant compounds. The designation “(aq)” on each reactant indicates that it is soluble. (a)
Possible products are potassium chloride, KCl, which is soluble, and aluminum hydroxide, Al OH 3 , which is not. Net ionic equation: Al3+ aq + 3 OH aq Al OH 3 s
(b) Possible products are iron(III) sulfate, Fe 2 SO 4 3 , and potassium bromide, KBr, both of which are soluble. No reaction occurs. (c)
2B
(E) (a)
Possible products are calcium nitrate, Ca(NO3)2, which is soluble, and lead(II) iodide, PbI 2 , which is insoluble. The net ionic equation is: Pb 2+ aq + 2 I aq PbI 2 s
Possible products are sodium chloride, NaCl, which is soluble, and aluminum phosphate, 3 AlPO 4 , which is insoluble. Net ionic equation: Al3+ aq + PO 4 aq AlPO 4 s
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Chapter 5: Introduction to Reactions in Aqueous Solutions
(b)
Possible products are aluminum chloride, AlCl 3 , which is soluble, and barium sulfate, BaSO 4 , which is insoluble. Net ionic equation: Ba 2+ aq + SO 4
(c)
2
aq
BaSO 4 s
Possible products are ammonium nitrate, NH 4 NO 3 , which is soluble, and lead (II) carbonate, PbCO 3 , which is insoluble. Net ionic equation: Pb 2+ aq + CO3
2
aq
PbCO3 s
3A
(E) Propionic acid is a weak acid, not dissociated completely in aqueous solution. Ammonia similarly is a weak base. The acid and base react to form a salt solution of ammonium propionate. NH 3 aq + HC3 H 5 O 2 aq NH 4 aq + C3 H 5 O 2 aq
3B
(E) Since acetic acid is a weak acid, it is not dissociated completely in aqueous solution (except at infinite dilution); it is misleading to write it in ionic form. The products of this reaction are the gas carbon dioxide, the covalent compound water, and the ionic solute calcium acetate. Only the latter exists as ions in aqueous solution. CaCO3 s + 2 HC 2 H 3 O 2 aq CO 2 g + H 2 O l + Ca 2+ aq + 2 C 2 H 3 O 2 aq
4A
(M) (a) This is a metathesis or double displacement reaction. Elements do not change oxidation states during this reaction. It is not an oxidation–reduction reaction. (b) The presence of O2(g) as a product indicates that this is an oxidation–reduction reaction. Oxygen is oxidized from O.S. = -2 in NO3- to O.S. = 0 in O2(g). Nitrogen is reduced from O.S. = +5 in NO3- to O.S. = +4 in NO2.
4B
(M) Vanadium is oxidized from O.S. = +4 in VO2+ to an O.S. = +5 in VO2+ while manganese is reduced from O.S. = +7 in MnO4- to O.S. = +2 in Mn2+.
5A
(M) Aluminum is oxidized (from an O.S. of 0 to an O.S. of +3 ), while hydrogen is reduced (from an O.S. of +1 to an O.S. of 0). Oxidation : Al s Al3+ aq + 3 e 2 Reduction:
2 H aq + 2 e +
H 2 g 3
Net equation : 2 Al s + 6 H + aq 2 Al3+ aq + 3 H 2 g 5B
(M) Bromide is oxidized (from 1 to 0), while chlorine is reduced (from 0 to 1 ). Oxidation : 2 Br aq Br2 l + 2 e Reduction: Cl 2 g + 2 e 2 Cl aq
Net equation : 2 Br aq + Cl2 g Br2 l + 2 Cl aq
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6A
(D) Step 1: Write the two skeleton half reactions. MnO 4 aq Mn 2+ aq and Fe 2+ aq Fe3+ aq
Step 2: Balance each skeleton half reaction for O (with H 2 O ) and for H atoms (with H + ). MnO 4 aq 8 H + aq Mn 2+ aq 4 H 2 O(l)
Fe 2+ aq Fe3+ aq
and
Step 3: Balance electric charge by adding electrons. MnO 4
aq 8 H + aq 5 e Mn 2+ aq 4 H 2 O(l)
and
Fe 2+ aq Fe3+ aq e
Step 4: Combine the two half reactions Fe2+ aq Fe3+ aq + e 5 MnO 4 aq + 8 H + aq + 5 e Mn 2+ aq + 4 H 2O(l)
MnO 4 aq + 8 H + aq + 5 Fe 2+ aq Mn 2+ aq + 4 H 2 O(l) + 5 Fe3+ aq
6B
(D) Step 1: Uranium is oxidized and chromium is reduced in this reaction. The “skeleton” 2+ 2 UO 2+ aq UO 2 aq and Cr2 O7 (aq) Cr 3 (aq) half-equations are:
Step 2: First, balance the chromium skeleton half-equation for chromium atoms: 2 Cr2 O7 aq 2 Cr 3+ aq Next, balance oxygen atoms with water molecules in each half-equation: 2 2 UO 2+ aq + H 2O(l) UO 2 aq and Cr2 O7 (aq) 2Cr 3 (aq) 7H 2 O(l) Then, balance hydrogen atoms with hydrogen ions in each half-equation: 2 UO 2+ aq + H 2 O(l) UO 2 aq + 2 H + aq 2
Cr2 O7 (aq) 14H (aq) 2Cr 3 (aq) 7H 2 O(l) Step 3: Balance the charge of each half-equation with electrons. 2 UO 2+ aq + H 2 O(l) UO 2 aq + 2 H + aq + 2 e Cr2 O7
2
aq +14 H + aq + 6 e 2 Cr 3+ aq + 7 H 2O(l)
Step 4: Multiply the uranium half-equation by 3 and add the chromium half-equation to it. 2 UO 2+ aq + H 2 O(l) UO 2 aq + 2 H + aq + 2 e 3
Cr2 O7
2
aq +14 H + aq + 6 e 2 Cr 3+ aq + 7 H 2O(l)
3 UO 2+ (aq)+Cr2 O 7 2- (aq)+14 H + (aq)+3 H 2 O(l) 3 UO 2 2+ (aq)+2 Cr 3+ (aq)+7 H 2 O(l)+6 H + (aq)
Step 5:
Simplify. Subtract 3 H 2 O (l) and 6 H+ (aq) from each side of the equation.
3 UO 2+ aq + Cr2 O7
2
aq + 8 H + aq 3 UO2 2 aq + 2 Cr 3+ aq + 4 H 2O(l)
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Chapter 5: Introduction to Reactions in Aqueous Solutions
7A
(D) Step 1: Write the two skeleton half-equations. 2 S(s) SO 3 (aq ) and OCl (aq ) Cl (aq ) Step 2: Balance each skeleton half-equation for O (with H 2 O ) and for H atoms (with H + ).
3 H 2O(l) + S s SO3
2
aq + 6 H +
OCl (aq) 2H Cl (aq) H 2 O(l) Step 3: Balance electric charge by adding electrons. 3 H 2 O(l) + S s SO3
2
aq + 6 H + (aq) + 4 e
OCl (aq) 2H (aq) 2e Cl (aq) H 2 O(l)
Step 4: Change from an acidic medium to a basic one by adding OH to eliminate H + . 2 3H 2 O(l) + S s + 6 OH (aq) SO3 aq + 6 H + (aq) + 6 OH (aq) + 4 e OCl aq + 2 H + (aq) + 2 OH (aq) + 2 e Cl aq + H 2 O(l) + 2 OH (aq)
Step 5: Simplify by removing the items present on both sides of each half-equation, and combine the half-equations to obtain the net redox equation. 2 {S s + 6 OH (aq) SO3 aq + 3 H 2O(l) + 4 e } 1 {OCl aq + H 2 O(l) + 2 e Cl aq + 2 OH (aq)} 2 S s + 6 OH (aq) + 2 OCl aq 2H 2 O(l) SO 3
2
aq + 3 H 2O(l) + 2 Cl aq + 4OH -
Simplify by removing the species present on both sides. 2 Net ionic equation: S s + 2 OH aq + 2 OCl aq SO3 aq + H 2 O(l) + 2 Cl aq 7B
(D) Step 1: Write the two skeleton half-equations. MnO 4 aq MnO 2 s and SO32 (aq) SO 4 2 (aq)
Step 2: Balance each skeleton half-equation for O (with H 2 O ) and for H atoms (with H + ). MnO 4 aq + 4 H + aq MnO 2 s + 2 H 2 O(l)
2
2
SO3 (aq) H 2 O(l) SO 4 (aq) 2H (aq) Step 3: Balance electric charge by adding electrons. MnO 4 aq + 4 H + aq + 3 e MnO 2 s + 2 H 2 O(l) SO3
2
aq +
H 2 O(l) SO 4
2
aq + 2 H + aq + 2 e
Step 4: Change from an acidic medium to a basic one by adding OH to eliminate H + . MnO 4
SO3
2
aq + 4
H + aq + 4 OH aq + 3 e MnO 2 s + H 2 O(l) + 4 OH aq
aq + H 2 O(l) + 2 OH aq SO4 2 aq + 2 H + aq +
166
2 OH aq + 2 e
Chapter 5: Introduction to Reactions in Aqueous Solutions
Step 5:
Simplify by removing species present on both sides of each half-equation, and combine the half-equations to obtain the net redox equation. {MnO 4 aq + 2 H 2 O(l) + 3 e MnO 2 s + 4 OH aq } 2
aq + 2 OH aq SO4 2 aq + H 2O(l) + 2 e } 3 2 2 MnO 4 aq + 3SO3 aq + 6 OH - (aq) + 4 H 2 O(l) 2 2 MnO 2 s + 3SO 4 aq + 3H 2 O(l) + 8 OH aq
{SO3
2
Simplify by removing species present on both sides. Net ionic equation:
2 MnO 4 aq + 3SO3
8A
8B
2
aq +
H 2 O(l) 2 MnO 2 s + 3SO 4
2
aq + 2 OH aq
(M) Since the oxidation state of H is 0 in H2 (g) and is +1 in both NH3(g) and H2O(g), hydrogen is oxidized. A substance that is oxidized is called a reducing agent. In addition, the oxidation state of N in NO2 (g) is +4 , while it is 3 in NH 3 ; the oxidation state of the element N decreases during this reaction, meaning that NO2 (g) is reduced. The substance that is reduced is called the oxidizing agent. (M) In Au CN 2
aq , gold has an oxidation state of +1; Au has been oxidized and,
thus, Au(s) (oxidization state = 0), is the reducing agent. In OH- (aq), oxygen has an oxidation state of -2; O has been reduced and thus, O2(g) (oxidation state = 0) is the oxidizing agent. 9A
(M) We first determine the amount of NaOH that reacts with 0.500 g KHP. 1 mol KHP 1 mol OH 1 mol NaOH n NaOH = 0.5000 g KHP = 0.002448 mol NaOH 204.22 g KHP 1 mol KHP 1 mol OH 0.002448 mol NaOH 1000 mL [NaOH] = = 0.1019 M 24.03 mL soln 1 L
9B
(M) The net ionic equation when solid hydroxides react with a strong acid is OH- + H+ H2O. There are two sources of OH-: NaOH and Ca(OH)2. We compute the amount of OHfrom each source and add the results. moles of OH from NaOH:
= 0.235 g sample
92.5 g NaOH 1 mol NaOH 1 mol OH = 0.00543 mol OH 100.0 g sample 39.997 g NaOH 1 mol NaOH
moles of OH from Ca OH 2 : = 0.235 g sample
7.5 g Ca OH 2 100.0 g sample
1 mol Ca OH 2
74.093 g Ba OH 2
2 mol OH 1 mol Ca OH 2
= 0.00048 mol OH
total amount OH = 0.00543 mol from NaOH + 0.00048 mol from Ca OH 2 = 0.00591 mol OH -
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Chapter 5: Introduction to Reactions in Aqueous Solutions
[HCl] =
0.00591 mol OH 1 mol H + 1 mol HCl 1000 mL soln = 0.130 M 45.6 mL HCl soln 1 mol OH 1 mol H + 1 L soln
10A (M) First, determine the mass of iron that has reacted as Fe 2+ with the titrant. The balanced chemical equation provides the essential conversion factor to answer this question. Namely: 5 Fe 2+ aq MnO 4 aq 8 H aq 5 Fe 3+ aq Mn 2+ aq 4 H 2 Ol mass Fe = 0.04125 L titrant
0.02140 mol MnO 4 1 L titrant
Then determine the % Fe in the ore.
% Fe =
5 mol Fe 2+ 1 mol MnO 4
55.847 g Fe 1 mol Fe 2+
= 0.246 g Fe
0.246 g Fe 100% = 65.4% Fe 0.376 g ore
10B (M) The balanced equation provides us with the stoichiometric coefficients needed for the solution. Namely: 5 C2 O 4 2- aq 2 MnO 4 aq 16 H aq 10 CO 2 g 2 Mn 2+ aq 8 H 2 O l 1 mol Na 2 C 2 O 4
amount MnO 4 = 0.2482 g Na 2 C 2 O 4
134.00 g Na 2 C 2 O 4
= 0.0007409 mol MnO 4
1 mol C 2 O 4
2
1 mol Na 2 C 2 O 4
2 mol MnO 4 5 mol C 2 O 4
2
[KMnO 4 ] =
0.0007409 mol MnO 4 1000 mL 1 mol KMnO 4 = 0.03129 M KMnO4 23.68 mL soln 1 L 1 mol MnO 4
INTEGRATIVE EXAMPLE A.
(M) First, balance the equation. Break down the reaction of chlorate and ferrous ion as follows:
ClO3 +6H +6e Cl +3H 2 O 6 Fe 2 Fe3 e
Net reaction: ClO 3 6Fe 2 6H Cl 6Fe3 3H 2 O The reaction between Fe2+ and Ce4+ is already balanced. To calculate the moles of Fe2+ that remains after the reaction with ClO3-, determine the moles of Ce4+ that react with Fe2+: mol Ce4+ = 0.01259 L × 0.08362 M = 1.0527×10-3 mol = mol of excess Fe2+ total mol of Fe2+ = 0.0500 L × 0.09101 = 4.551×10-3 mol Therefore, the moles of Fe2+ reacted = 4.551×10-3 - 1.0527×10-3 = 3.498×10-3 mol. To determine the mass of KClO3, use the mole ratios in the balanced equation in conjunction with the molar mass of KClO3.
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Chapter 5: Introduction to Reactions in Aqueous Solutions
3.498 10
3
mol Fe
2
1 mol ClO3 1 mol KClO3 122.54 g KClO3 2 6 mol Fe 1 mol ClO3 1 mol KClO3
= 0.07144 g KClO3 %KClO3 = B.
0.07144 g 100% = 49.89% 0.1432 g
(M) First, balance the equation. Break down the reaction of arsenous acid and permanganate as follows:
5 H 3 AsO3 + H 2 O H3 AsO 4 + 2e- + 2H + 2 MnO 4 + 8H + 5e- Mn 2+ + 4H 2 O Net reaction: 5H 3AsO3 + 2MnO 4 6H + 5H 3AsO 4 + 2Mn 2+ + 3H 2O moles of MnO4- = 0.02377 L × 0.02144 M = 5.0963×10-4 mol To calculate the mass of As, use the mole ratios in the balanced equation in conjunction with the molar mass of As: 5 mol H 3 AsO3 1 mol As 74.922 g As 5.0963 104 mol MnO 4 2 mol MnO 4 1 mol H 3AsO3 1 mol As = 0.095456 g As 0.095456 g mass% As = 100% = 1.32% 7.25 g
EXERCISES Strong Electrolytes, Weak Electrolytes, and Nonelectrolytes 1.
(E) (a)
Because its formula begins with hydrogen, HC6 H 5O is an acid. It is not listed in Table 5-1, so it is a weak acid. A weak acid is a weak electrolyte.
(b)
Li 2SO 4 is an ionic compound, that is, a salt. A salt is a strong electrolyte.
(c)
MgI 2 also is a salt, a strong electrolyte.
(d)
CH3 CH 2 2 O is a covalent compound whose formula does not begin with H. Thus, it is neither an acid nor a salt. It also is not built around nitrogen, and thus it does not behave as a weak base. This is a nonelectrolyte.
(e)
Sr OH 2 is a strong electrolyte, one of the strong bases listed in Table 5-2. 169
Chapter 5: Introduction to Reactions in Aqueous Solutions
2.
(E) (a)
The best electrical conductor is the solution of the strong electrolyte: 0.10 M NaCl. In each liter of this solution, there are 0.10 mol Na + ions and 0.10 mol Cl ions.
(b) The poorest electrical conductor is the solution of the nonelectrolyte: 0.10 M CH 3CH 2OH . In this solution, the concentration of ions is negligible. 3.
(E) HCl is practically 100% dissociated into ions. The apparatus should light up brightly. A solution of both HCl and HC2H3O2 will yield similar results. In strongly acidic solutions, the weak acid HC2H3O2 is molecular and does not contribute to the conductivity of the solution. However, the strong acid HCl is practically dissociated into ions and is unaffected by the presence of the weak acid HC2H3O2. The apparatus should light up brightly.
4.
(E) NH3 (aq) is a weak base; HC2H3O2 (aq) is a weak acid. The reaction produces a solution of ammonium acetate, NH 4 C2 H 3 O 2 aq , a salt and a strong electrolyte.
NH 3 aq + HC2 H 3 O 2 aq NH 4 5.
6.
(E) (a) (c)
+
aq +
Barium bromide: strong electrolyte Ammonia: weak electrolyte
(E) sodium chloride(strong electrolyte) Na+ Cl-
Cl-
Na+
Na+
Cl-
(b)
aq
Propionic acid: weak electrolyte
hypochlorous acid (weak electrolyte)
Cl-
H O Cl
Na+
H O Cl
ammonium chloride (strong electrolyte) +
C2 H3 O2
+
+
O Cl
-
H O Cl H3O+ H O Cl
methanol (non electrolyte) H O CH3
H O CH3 H O CH3 H O CH3 H O CH3
H H H H N H Cl H N H Cl- H N H ClH H H
Ion Concentrations 7.
(E)
0.238 mol KNO 3 1 mol K + = 0.238 M K + 1 L soln 1 mol KNO 3
(a)
K+ =
(b)
NO3 =
0.167 mol Ca NO3 2 1 L soln
2 mol NO3 = 0.334 M NO3 1 mol Ca NO3 2
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Chapter 5: Introduction to Reactions in Aqueous Solutions
Al3+ =
(c)
Na
(d) 8.
+
0.083 mol Al2 SO 4 3 1 L soln
2 mol Al3+ = 0.166 M Al3+ 1 mol Al2 SO 4 3
0.209 mol Na 3 PO 4 3 mol Na + = = 0.627 M Na + 1 L soln 1 mol Na 3 PO 4
(E) Choice (d) is the solution with the greatest concentration of sulfate ions. 1 mol SO 4 2(a) 0.075 M H 2SO 4 0.075 M SO4 21 mol H 2SO 4 (b) 0.22 M MgSO 4
1 mol SO 4 2 0.22 M SO 4 21 mol MgSO 4
(c) 0.15 M Na 2SO 4
1 mol SO 4 2 0.15 M SO 4 21 mol Na 2SO 4
(d) 0.080 M Al2 (SO 4 )3 (e) 0.20 M CuSO 4
9.
3 mol SO 4 2 0.24 M SO 4 21 mol Al2 (SO 4 )3
1 mol SO 4 2 0.20 M SO 4 21 mol CuSO 4
(E) Conversion pathway approach:
OH =
0.132 g Ba OH 2 8H 2 O 1000 mL 1 mol Ba OH 2 8H 2 O 2 mol OH 275 mL soln 1 L 315.5 g Ba OH 2 8H 2 O 1 mol Ba OH 2 8 H 2 O
= 3.04 103 M OH Stepwise approach:
0.132 g Ba OH 2 8H 2 O 1000 mL = 0.480 g/L 275 mL soln 1 L
0.00152 mol Ba OH 2 ×8H 2O 0.480 g 1 mol Ba OH 2 8H 2 O = 315.5 g Ba OH 2 8H 2 O L L 0.00152 mol Ba OH 2 ×8H 2 O L 10.
2 mol OH 1 mol Ba OH 2 8 H 2 O
= 3.04 10-3 M OH -
(E)
0.126 mol KCl 1 mol K + = 0.126 M K+ K = 1 L soln 1 mol KCl 0.148 mol MgCl 2 1 mol Mg 2+ = 0.148 M Mg2+ Mg 2+ = 1 L soln 1 mol MgCl 2 +
Now determine the amount of Cl in 1.00 L of the solution.
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Chapter 5: Introduction to Reactions in Aqueous Solutions
0.126 mol KCl 1 mol Cl 0.148 mol MgCl2 2 mol Cl mol Cl = + 1 L soln 1 mol KCl 1 L soln 1 mol MgCl2 = 0.126 mol Cl + 0.296 mol Cl = 0.422 mol Cl 0.422 mol Cl = 0.422 M Cl Cl = 1 L soln
11.
(E) (a)
[Ca 2+ ] =
(c) 12.
+
100 mL solution
[Zn 2+ ] =
225 g Zn
2+
1 mL solution
1 g Ca 2+ 1000 mg Ca 1gK
2+
+
1000 mg K
1 g Zn
+
2+
1 mol Ca 2+ 40.078 g Ca
2+
3.54 104 M Ca 2+
1000 mL solution 1 L solution
2+
1 10 g Zn 6
1000 mL solution 1 L solution
1 mol K + 39.0983 g K
8.39 10 3 M K +
+
1 mol Zn 2+ 65.39 g Zn
2+
3.44 103 M Zn 2+
(E) [NaF] =
13
1 L solution 32.8 mg K
+
(b) [K ] =
14.2 mg Ca 2+
0.9 mg F 1 g 1 mol F 1 mol NaF = 4.7 105 M = 5 105 M NaF 1 L 1000 mg 19.00 g F 1 mol F
(E) In order to determine the solution with the largest concentration of K+, we begin by converting each concentration to a common concentration unit, namely, molarity of K+. 0.0850 M K 2 SO 4 2 mol K + 0.17 M K + 1 L solution 1 mol K 2 SO 4
1000 mL solution 1.25 g KBr 1 mol KBr 1 mol K + 0.105 M K + 100 mL solution 1 L solution 119.0023 g KBr 1 mol KBr 1000 mL solution 8.1 mg K + 1 g K+ 1 mol K + 0.207 M K + + + 1 mL solution 1 L solution 1000 mg K 39.0983 g K + Clearly, the solution containing 8.1 mg K per mL gives the largest K+ of the three solutions. 14.
(E) (c) NH 3 is a weak base and would have an exceedingly low H + ; the answer is not
1.00 M NH 3 . HC 2 H 3O 2 is a very weak acid; 0.011 M HC 2 H 3O 2 would have a low H + . H 2SO 4 has two ionizable protons per mole while HCl has but one. Thus, H 2SO 4 would have the highest H + in a 0.010 M aqueous solution.
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Chapter 5: Introduction to Reactions in Aqueous Solutions
15.
(M) Determine the amount of I in the solution as it now exists, and the amount of I in the solution of the desired concentration. The difference in these two amounts is the amount of I that must be added. Convert this amount to a mass of MgI 2 in grams. moles of I in final solution = 250.0 mL
1 L
0.1000 mol I
= 0.02500 mol I 1000 mL 1 L soln 1 L 0.0876 mol KI 1 mol I moles of I in KI solution = 250.0 mL = 0.0219 mol I 1000 mL 1 L soln 1 mol KI 1 mol MgI 278.11 g MgI 2 1000 mg 2 mass MgI 2 required = 0.02500 0.0219 mol I 2 mol I 1 mol MgI 2 1 g
= 4.3 102 mg MgI 2 16.
(M) The final volume is 975 mL. We can use dimensional analysis to obtain the [K+]. 12.0 mg K 2 SO 4 +
[K ] =
17.
18.
×1000 mL 1 g K 2 SO 4 1 mol K 2 SO 4 2 mol K + 1 mL × × × = 0.141 M K + 0.975 L solution 1000 mg K 2 SO 4 174.26 g K 2 SO 4 1 mol K 2 SO 4
(M) moles of chloride ion 0.625 mol KCl 1 mol Cl 0.385 mol MgCl 2 2 mol Cl = 0.225 L + 0.615 L 1 L soln 1 mol KCl 1 L soln 1 mol MgCl 2 0.615 mol Cl = 0.732 M = 0.141 mol Cl + 0.474 mol Cl = 0.615 mol Cl Cl = 0.225 L + 0.615 L
(M) amount of NO3 ion = 0.421 mol Mg NO 3 2 0.283 mol KNO 3 1 mol NO 3 2 mol NO 3 0.275 L + 0.328 L 1 L soln 1 mol KNO 3 1 L soln 1 mol Mg NO 3 2
= 0.0778 mol NO 3 + 0.276 mol NO 3
= 0.354 mol NO3-
0.354 mol NO3 NO3 = 0.275 L + 0.328 L + 0.784 L = 0.255 M
Predicting Precipitation Reactions 19.
(E) In each case, each available cation is paired with the available anions, one at a time, to determine if a compound is produced that is insoluble, based on the solubility rules of Chapter 5. Then a net ionic equation is written to summarize this information. (a) Pb 2+ aq + 2 Br aq PbBr2 s (b) No reaction occurs (all are spectator ions). (c) Fe3+ aq + 3 OH aq Fe OH 3 s 173
Chapter 5: Introduction to Reactions in Aqueous Solutions
20.
21.
(E) (a)
Ca2+(aq) + CO32-(aq) → CaCO3(s)
(b)
Ba2+(aq) + SO42-(aq) → BaSO4(s)
(c)
No precipitate forms. All ions stay in solution.
(E)
(a)
22.
23.
Mixture
Result (Net Ionic Equation)
HI a + Zn NO3 2 (aq):
No reaction occurs.
(b)
CuSO 4 aq + Na 2 CO3 aq :
(c)
Cu NO3 2 aq +
Cu 2+ aq + CO3
aq CuCO3 s 3 Na 3 PO 4 aq : 3Cu 2+ aq + 2 PO 4 aq Cu 3 PO 4 2 s 2
(E)
Mixture
Result (Net Ionic Equation)
(a)
AgNO3 aq + CuCl 2 aq :
Ag + aq + Cl aq AgCl s
(b)
Na 2S aq + FeCl2 aq :
S2 aq + Fe 2+ aq FeS s
(c)
Na 2 CO3 aq + AgNO3 aq :
CO3
(E) (a)
Add K 2 SO 4 aq ; BaSO 4 s will form and MgSO4 will not precipitate.
2
aq + 2
Ag + aq Ag 2 CO3 s
BaCl2 s + K 2 SO 4 aq BaSO 4 s + 2 KCl aq
(b)
Add H 2 O l ; Na 2 CO3 s dissolves, but MgCO3 (s) will not dissolve (appreciably). water Na 2 CO3 s 2 Na + aq + CO3
(c)
24.
(M) (a)
2
aq
Add KCl(aq); AgCl(s) will form, while Cu(NO3)2 (s) will dissolve. AgNO3 s + KCl aq AgCl s + KNO3 aq Add H 2O. Cu NO3 2 s will dissolve, while PbSO4(s) will not dissolve (appreciably). water Cu NO3 2 s Cu 2+ aq + 2NO3
(b)
aq
Add HCl(aq). Mg(OH)2 (s) will dissolve, but BaSO4 (s) will not dissolve (appreciably). Mg OH 2 s + 2 HCl aq MgCl2 aq + 2 H 2 O(l) 174
Chapter 5: Introduction to Reactions in Aqueous Solutions
(c)
b g
Add HCl(aq). Both carbonates dissolve, but PbCl2(s) will precipitate while CaCl 2 aq remains dissolved. PbCO3 s + 2 HCl aq PbCl 2 s + H 2O(l) + CO 2 g CaCO3 s + 2 HCl aq CaCl 2 aq + H 2O(l) + CO 2 g
25.
(M) (a)
Mixture Sr NO3 2 aq + K 2SO 4 aq :
Net Ionic Equation 2 Sr 2+ aq + SO 4 aq SrSO 4 s
(b)
Mg NO3 2 aq + NaOH aq : Mg 2+ aq + 2 OH aq Mg OH 2 s
(c)
BaCl2 aq + K 2SO 4 aq :
2
Ba 2+ (aq) SO 4 (aq) BaSO4 (s) (upon filtering, KCl (aq) is obtained)
26. (M) (a)
Mixture BaCl2 aq + K 2SO 4 aq :
(b)
NaCl aq + AgNO3 aq :
AgCl s + Na + aq NO3 aq
alternatively BaCl2 aq + AgNO3 aq :
AgCl s + Ba 2+ (aq) + SO 4 2 aq
(c)
Net Ionic Equation 2 Ba 2+ aq + SO 4 aq BaSO 4 s
Sr NO3 2 aq + K 2SO 4 aq : Sr 2+ aq + SO 4
2
aq
SrSO 4 s
(upon filtering, KNO3 (aq) is obtained) Acid–Base Reactions 27.
(E) The type of reaction is given first, followed by the net ionic equation. (a) Neutralization: OH aq + HC2 H3 O2 aq H 2 O l + C2 H3 O 2
aq
(b) No reaction occurs. This is the physical mixing of two acids. (c) Gas evolution: FeS s + 2 H + aq H 2 S g + Fe 2+ aq
aq + H + aq "H 2 CO3 aq " H + aq Mg 2+ aq + H 2 g
(d) Gas evolution: HCO3 (e) Redox: Mg s + 2 28.
(E) (a)
NaHCO3 s + H + aq
H 2 O l + CO 2 g
Na + aq + H 2 O l + CO 2 g
175
Chapter 5: Introduction to Reactions in Aqueous Solutions
(b)
CaCO3 s + 2 H + aq
Ca 2+ aq + H 2 O l + CO 2 g
(c)
Mg OH 2 s + 2 H + aq
Mg 2+ aq + 2 H 2 O l
(d)
Mg OH 2 s + 2 H + aq
Mg 2+ aq + 2 H 2 O l
Al OH 3 s + 3 H + aq
(e) 29.
30.
31.
Al3+ aq + 3 H 2 O l
NaAl OH 2 CO3 s + 4 H + aq Al3+ aq + Na + aq +3 H 2 O l + CO 2 g
(M) As a salt:
NaHSO4 aq Na + aq + HSO 4
As an acid:
HSO 4
aq +
OH aq
aq 2 H 2 O l + SO 4 aq
(M) Because all three compounds contain an ammonium cation, all are formed by the reaction of an acid with aqueous ammonia. The identity of the anion determines which acid present. (a) 2 NH 3 aq + H 3PO 4 aq NH 4 2 HPO 4 aq
(b) NH3 aq + HNO3 aq
NH 4 NO3 aq
(c) 2 NH 3 aq + H 2SO 4 aq
NH 4 2 SO4 aq
(M) Use (b) NH3(aq): NH3 affords the OH- ions necessary to form Mg(OH)2(s). Applicable reactions: {NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)} 2
MgCl2(aq) Mg2+(aq) + 2 Cl-(aq) Mg2+(aq) + 2 OH-(aq) Mg(OH)2(s) 32.
(E) HCl(aq) reacts with KHSO3(aq) to give SO2(g) via the thermodynamically unstable intermediate sulfurous acid (H2SO3).
(b) KHSO3 reacts with HCl(aq) to form a gas according the net ionic equation below. Net ionic equation: H+(aq) + HSO3-(aq) “H2SO3(aq)” H2O(l) + SO2(g) (a), (c), and (d) do not form gaseous products. (a) H+(aq) + SO42-(aq) HSO4- (aq) (c) OH-(aq) + H+(aq) H2O(l) (d) CaCl2(aq) + HCl(aq) no reaction
176
Chapter 5: Introduction to Reactions in Aqueous Solutions
Oxidation–Reduction (Redox) Equations 33.
34.
(E) (a)
The O.S. of H is +1, that of O is 2 , that of C is +4 , and that of Mg is +2 on each side of this equation. This is not a redox equation.
(b)
The O.S. of Cl is 0 on the left and 1 on the right side of this equation. The O.S. of Br is 1 on the left and 0 on the right side of this equation. This is a redox reaction.
(c)
The O.S. of Ag is 0 on the left and +1 on the right side of this equation. The O.S. of N is +5 on the left and +4 on the right side of this equation. This is a redox reaction.
(d)
On both sides of the equation the O.S. of O is 2 , that of Ag is +1 , and that of Cr is +6 . Thus, this is not a redox equation.
(E) (a)
In this reaction, iron is reduced from Fe3+ (aq) to Fe2+ (aq) and manganese is reduced from a +7 O.S. in MnO 4 aq to a +2 O.S. in Mn2+ (aq). Thus, there are two reductions and no oxidation, which is impossible.
(b)
In this reaction, chlorine is oxidized from an O.S. of 0 in Cl2 (aq) to an O.S. of +1 in ClO aq and oxygen is oxidized from an O.S. of 1 in H 2 O 2 aq to an O.S. of 0 in O 2 g . Consequently there are two oxidation reactions and no reduction reactions, which is impossible.
35.
(E) (a)
aq + 6 H + aq + 4 e S2O32 aq + 3 H 2O(l) 2 NO3 aq +10 H + aq + 8 e N 2 O g + 5 H 2 O l Al s + 4 OH aq Al OH 4 aq + 3 e
Reduction:
2SO3
(b) Reduction: (c) 36.
(E) (a) (b) (c)
37.
Oxidation:
2
acidic H 2 C2 O 4 2 CO 2 + 2 H + + 2e-
Oxidation
+
2
2 Cr
-
MnO 2 + 4 OH
6 e + 14 H + Cr2 O7
2 H 2 O + 3 e + MnO4
acidic
basic
3+
+ 7 H2 O -
Reduction Re duction
(M) (a) Oxidation: { 2 I aq I 2 s + 2 e
Reduction: { MnO 4
aq + 8
H + aq + 5 e Mn 2+ aq + 4 H 2 O l
}5 }2
Net: 10 I aq + 2 MnO 4 aq +16 H + aq 5 I 2 s + 2 Mn 2+ aq + 8 H 2 O l
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Chapter 5: Introduction to Reactions in Aqueous Solutions
(b) Oxidation: { N 2 H 4 l N 2 g + 4 H + aq + 4 e
} 3
+ Reduction: { BrO3 aq + 6 H aq + 6 e Br aq + 3 H 2 O l
Net: (c)
3 N 2 H 4 l + 2 BrO3 aq 3 N 2 g + 2 Br aq + 6 H 2 O l
Oxidation: Fe 2+ aq Fe3+ aq + e Reduction: VO4 Net:
3
aq + 6
Fe2+ aq + VO 4
3
H + aq + e VO 2+ aq + 3 H 2 O l
aq + 6
H + aq Fe3+ aq + VO2+ aq + 3 H 2 O l
(d) Oxidation: { UO 2+ aq + H 2 O l UO2
Reduction: { NO3 aq + 4 H + aq + 3 e
Net: 38.
} 2
3 UO 2+ aq + 2 NO3
aq + 2
aq + 2 H+ aq + 2 e NO g + 2 H 2 O l
2
H + aq 3 UO 2
2
aq + 2
}3 }2
NO g + H 2 O l
(M) (a) Oxidation: { P4 s +16 H 2 O(l) 4 H 2 PO 4 aq + 24 H + aq + 20 e } 3
Reduction: { NO3 aq + 4 H + aq + 3 e NO g + 2 H 2 O(l) } 20 ____________________ ______ + 3 P4 s + 20 NO3 aq + 8 H 2 O + 8 H aq 12 H 2 PO 4 aq + 20 NO g Net:
aq + 5 H 2 O l 2 SO4 2 aq +10 H + aq + 8 e Reduction: { MnO 4 aq + 8 H + aq + 5 e Mn 2+ aq + 4 H 2 O(l)
(b) Oxidation: { S2 O3
Net: (c)
2
2
}8
2
5 S2 O3 aq + 8 MnO 4 aq +14 H + aq 10 SO 4 aq + 8 Mn 2+ aq + 7 H 2 O l
Oxidation: 2 HS aq + 3 H 2 O l S2 O3
2
aq + 8
H + aq + 8 e
Reduction: { 2 HSO3 aq + 4 H + aq + 4 e S2 O3
b g
Net:
b g
2 HS aq + 4 HSO 3 aq 3 S2 O 3
(d) Oxidation:
2
2
aq + 3
baqg + 3 H Oblg
H 2 O(l) } 2
2
2 NH 3 OH + aq N 2 O g + H 2 O l + 6 H + aq + 4 e
Reduction: { Fe3+ aq + e Fe 2+ aq Net:
}5
}4
4 Fe3+ aq + 2 NH 3OH + aq 4 Fe 2+ aq + N 2 O g + H 2 O l + 6 H + aq
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Chapter 5: Introduction to Reactions in Aqueous Solutions
39.
(M) (a) Oxidation: { MnO 2 s + 4 OH aq MnO 4 aq + 2 H 2 O(l) + 3 e } 2
Reduction: ClO3 aq + 3 H 2 O(l) + 6 e Cl aq + 6 OH aq
Net:
2 MnO 2 s + ClO3 aq + 2 OH aq 2MnO 4 aq + Cl aq + H 2 O(l)
(b) Oxidation: { Fe OH 3 s + 5 OH aq FeO 4
2
aq + 4
H 2 O(l) + 3 e
Reduction: { OCl aq + H 2 O(l) + 2 e Cl aq + 2OH - aq Net: 2 Fe OH 3 s + 3 OCl aq + 4 OH aq 2FeO 4
(c)
2
}2 }3
aq + 3
Cl aq + 5 H 2 O(l)
Oxidation: { ClO 2 (aq) + 2 OH aq ClO3 aq + H 2 O l + e } 5
Reduction: ClO 2 (aq) + 2 H 2 O l + 5 e Cl (aq) + 4 OH aq Net: (d)
40.
6 ClO2 (aq) + 6 OH aq 5ClO3 aq + Cl aq + 3 H 2 O (l)
Oxidation: (Ag (s) Ag+ (aq) + 1 e¯ ) 3 Reduction: 4 H2O(l) + CrO42- + 3 e¯ Cr(OH)3(s) + 5 OH Net: 3 Ag(s) + CrO42- + 4 H2O(l) 3 Ag+(aq) + Cr(OH)3(s) + 5 OH-
(M) (a) Oxidation: 8OH- + S2O42- 2SO42- + 6e- + 4H2O Reduction: {3 e- + 4H2O + CrO42- Cr(OH)3 + 5OH-} × 2 Net: 2CrO42- + 4H2O + S2O42- 2Cr(OH)3 + 2SO42- + 2OH(b)
Oxidation: N 2 H 4 l + 4 OH aq N 2 g + 4 H 2 O(l) + 4 e Reduction: { Fe CN 6 Net: 4 Fe CN 6
(c)
3
3
aq +
e Fe CN 6
4
aq
}4
aq + N 2 H 4 l + 4OH aq 4 Fe CN 6 aq + N 2 g + 4H 2 O l 4
Oxidation: { Fe OH 2 s + OH aq Fe OH 3 s + e
}4
Reduction: O 2 g + 2 H 2 O l + 4 e 4 OH aq
Net: (d)
4 Fe OH 2 s + O 2 g + 2 H 2 O l 4 Fe OH 3 s
Oxidation: { C 2 H 5 OH aq + 5 OH aq C 2 H 3O 2 aq + 4 H 2 O l + 4 e
}3
Reduction: { MnO 4 aq + 2 H 2 O l + 3 e MnO 2 s + 4 OH aq
}4
Net: 3 C 2 H 5 OH aq + 4 MnO 4 aq 3 C 2 H 3 O 2 aq + 4 MnO 2 s + OH aq + 4 H 2 O l
179
Chapter 5: Introduction to Reactions in Aqueous Solutions
41.
(M) (a) Oxidation: Cl2 g +12 OH aq 2 ClO3 aq + 6 H 2 O(l) +10 e
Reduction: { Cl 2 g + 2 e 2 Cl aq
(b)
42.
}5
Net:
6 Cl2 g +12 OH aq 10 Cl aq + 2 ClO3 aq + 6 H 2 O(l)
Or:
3 Cl2 g + 6 OH aq 5 Cl aq + ClO3 aq + 3 H 2 O(l)
aq + 2 H 2 O(l) 2 HSO3 aq + 2 H + aq + 2 e 2 2 Reduction: S2 O 4 aq + 2 H + aq + 2 e S2 O3 aq + H 2 O (l) 2 2 Net: 2 S2 O 4 aq + H 2 O(l) 2 HSO3 aq + S2 O3 aq Oxidation: S2 O 4
2
(M) 2 (a) Oxidation: { MnO 4 aq MnO4 aq + e
aq + 2 H 2 O(l) + 2 e MnO2 s + 4 OH aq 2 MnO 4 aq + 2 H 2 O(l) 2 MnO4 aq + MnO 2 s + 4 OH aq
Reduction: MnO 4 Net: (b)
(c)
3
}2
2
Oxidation: { P4 s + 8 OH aq 4 H 2 PO 2
aq + 4 e Reduction: P4 s +12 H 2 O(l) +12 e 4 PH 3 g +12 OH aq Net: 4 P4 s +12 OH aq +12 H 2 O(l) 12 H 2 PO 2 aq + 4 Oxidation: S8 s + 24 OH aq 4 S2 O3
2
aq +12
}3 PH 3 g
H 2 O l + 16 e
S8 s +16 e 8 S2 aq
Reduction: 2 2 S8 s + 24 OH aq 8 S2 aq + 4 S2 O3 aq +12 H 2 O l Net: (d) Oxidation: As 2S3 s + 40 OH aq 2 AsO 4
3
aq + 3 SO4 2 aq + 20 H 2O + 28 e
Reduction: { H 2 O 2 aq +2 e- 2 OH - aq Net: As 2S3 s +12 OH aq +14 H 2 O 2 aq 2 AsO 4 43.
} 14 3
aq + 3 SO4 2 aq + 20 H 2O(l)
(M) (a) Oxidation: { NO 2 aq + H 2 O l NO3 aq + 2 H + aq + 2 e + 2+ Reduction: { MnO 4 aq + 8 H aq + 5 e Mn aq + 4 H 2 O l
}5 }2
Net: 5 NO2 aq + 2 MnO 4 aq + 6 H + aq 5 NO3 aq + 2 Mn 2+ aq + 3 H 2 O l
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Chapter 5: Introduction to Reactions in Aqueous Solutions
(b)
(c)
Oxidation: { Mn2+ (aq) + 4 OH- (aq) MnO2 (s) + 2 H2O (l) + 2 e- } 3 Reduction: { MnO4- (aq) + 2 H2O (l) + 3 e- MnO2 (s) + 4 OH- (aq) } 2 Net: 3 Mn2+ (aq) + 2 MnO4- (aq) + 4 OH- (aq) 5 MnO2 (s) + 2 H2O (l) Oxidation: { C2 H 5 OH CH 3 CHO + 2 H + aq + 2 e
} 3
Reduction: Cr2 O 7 2 aq +14 H + aq + 6 e 2 Cr 3+ aq + 7 H 2 O l Net: Cr2 O 7 2 aq + 8 H + aq + 3 C2 H 5 OH 2 Cr 3+ aq + 7 H 2 O l + 3 CH 3 CHO 44.
(M) (a) Oxidation: [ Al s Al3+ aq + 3e
} ×2
Reduction: { 2 HI aq + 2 e 2 I aq + H 2 g }×3
2 Al(s) + 6 HI(aq) 2 AlI3(aq) + 3 H2(g) (b)
Oxidation: Zn s Zn 2+ aq + 2 e Reduction: { VO 2+ aq + 2 H + aq + e V3+ aq + H 2 O l } 2 Net:
(c)
Zn(s) + 2 VO 2+ aq + 4 H + aq Zn 2+ (aq) 2 V 3+ (aq) + 2 H 2 O(l)
Oxidation: H 2 O + CH 3 OH CO 2 + 6 H + + 6 e
Reduction:{ ClO3 aq + 2 H + aq + e ClO 2 aq + H 2 O l } 6 Net: 45.
CH 3 OH + 6 ClO3 aq + 6 H + 6 ClO 2 aq + 5 H 2 O l CO 2
(D) For the purpose of balancing its redox equation, each of the reactions is treated as if it takes place in acidic aqueous solution. (a)
2 H2O(g) + CH4(g) CO2(g) + 8 H+(g) + 8 e{2 e- + 2 H+(g) + NO(g) ½ N2(g) + H2O(g) }4 CH4(g) + 4 NO(g) 2 N2(g) + CO2(g) + 2 H2O(g)
(b)
{H2S(g) 1/8 S8(s)+ 2 H+(g) + 2 e}2 + 4 e + 4 H (g) + SO2(g) 1/8 S8(s) + 2 H2O(g) 2 H2S(g) + SO2(g) 3/8 S8(s) + 2 H2O(g) or 16 H2S(g) + 8 SO2(g) 3 S8(s) + 16 H2O(g)
(c)
{Cl2O(g) + 2 NH4+(aq) + 2 H+(aq) + 4 e- 2 NH4Cl(s) + H2O(l) {2 NH3(g) N2(g) + 6 e- + 6 H+(aq) 6 NH3(g) + 6 H+(aq) 6 NH4+(aq) 10 NH3(g) +3 Cl2O(g) 6 NH4Cl(s) + 2 N2(g) + 3 H2O(l)
181
}3 } 2
Chapter 5: Introduction to Reactions in Aqueous Solutions
46.
(D) For the purpose of balancing its redox equation, each of the reactions is treated as if it takes place in acidic aqueous solution. (a) CH4(g) + NH3(g) HCN(g) + 6 e- + 6 H+ } 3 { 2 e- + 2 H+(g) + ½ O2(g) H2O(g) CH4(g) + NH3(g) + 3/2 O2(g) HCN(g) + 3 H2O(g) (b) { H 2 g 2 H + aq + 2 e }5 2 NO g +10 H + aq +10 e 2 NH 3 g + 2 H 2 O(l) 5 H 2 g + 2 NO g 2 NH 3 g + 2 H 2 O g
(c)
{Fe(s) Fe3+(aq) + 3 e} 4 } 3 {4 e- + 2 H2O(l) + O2(g) 4 OH-(aq) 4 Fe(s) + 6 H2O(l) + 3 O2(g) 4 Fe(OH)3(s)
Oxidizing and Reducing Agents 47.
(E) The oxidizing agents experience a decrease in the oxidation state of one of their elements, while the reducing agents experience an increase in the oxidation state of one of their elements.
aq is the reducing agent; the O.S. of S = +4 in SO32 and = +6 in SO 4 2 . MnO 4 aq is the oxidizing agent; the O.S. of Mn = +7 in MnO 4 and + 2 in Mn 2+ . 2
(a)
SO3
(b)
H 2 g is the reducing agent; the O.S. of H = 0 in H 2 g and = +1 in H 2 O g . NO 2 g is the oxidizing agent; the O.S. of N = +4 in NO 2 g and 3 in NH 3 (g).
(c) Fe CN 6
4
aq is the reducing agent; the O.S. of
Fe = +2 in Fe CN 6
4
3
and = +3 in Fe CN 6 . H 2O 2 aq is the oxidizing agent; the O.S. of O = 1 in H 2 O 2 and = 2 in H 2 O .
48.
(E) 2 2 (a) 2 S2 O3 aq + I 2 s S4 O6 aq + 2 I aq
aq + 4 2 S2 O3 aq + 4
(b) S2 O3 (c)
2
Cl2 g + 5 H 2 O 1 2 HSO4 OCl aq + 2 OH aq 2
182
aq + 8 Cl aq + 8 H + aq 2 SO4 aq + 4 Cl aq + H 2 O 1
Chapter 5: Introduction to Reactions in Aqueous Solutions
Neutralization and Acid–Base Titrations 49.
(E) The problem is most easily solved with amounts in millimoles.
VNaOH
0.128 mmol HCl 1 mmol H + 1 mmol OH = 10.00 mL HCl aq 1 mL HCl aq 1 mmol HCl 1 mmol H +
50.
1 mL NaOH aq 1 mmol NaOH = 13.3 mL NaOH aq soln 0.0962 mmol NaOH 1 mmol OH
(M)
10.00 mL acid [NaOH] = 51.
0.1012 mmol H 2SO 4 2 mmol NaOH 1 mL acid 1 mmol H 2SO 4 = 0.08683 M 23.31 mL base
(E) The net reaction is OH aq + HC3 H5 O2 aq H 2 O(l) + C3 H5 O2 aq .
Conversion pathway approach: Vbase = 25.00 mL acid
0.3057 mmol HC3 H 5 O 2 1 mL acid
1 mmol KOH 1 mmol HC3 H 5 O 2
1 mL base 2.155 mmol KOH
= 3.546 mL KOH solution
Stepwise approach: 25.00 mL acid
0.3057 mmol HC3 H 5 O 2
= 7.643 mmol HC3 H 5 O 2
1 mL acid 1 mmol KOH 7.643 mmol HC3H 5O 2 = 7.643 mmol KOH 1 mmol HC3 H 5 O 2
7.643 mmol KOH 52.
1 mL base 2.155 mmol KOH
= 3.546 mL KOH solution
(E) Titration reaction: Ba OH 2 aq + 2 HNO3 aq Ba NO3 2 aq + 2 H 2 O 1
Vbase = 50.00 mL acid
0.0526 mol HNO3 1 mmol Ba OH 2 1 mL base 1 mL acid 2 mmol HNO3 0.0844 mmol Ba OH 2
= 15.6 mL Ba OH 2 solution 53.
(E) NaOH aq + HCl aq NaCl aq + H 2 O(l) is the titration reaction. 0.02834 L [NaOH] =
0.1085 mol HCl 1mol NaOH 1L soln 1 mol HCl = 0.1230 M NaOH 0.02500 L sample
183
Chapter 5: Introduction to Reactions in Aqueous Solutions
54.
(E)
28.72 mL acid NH 3 = 55.
1.021mmol HCl 1mmol H + 1mmol NH 3 1 mL acid 1mmol HCl 1mmol H + = 5.86 M NH 3 5.00 mL sample
(M) The mass of acetylsalicylic acid is converted to the amount of NaOH, in millimoles, that will react with it. 0.32 g HC9 H 7 O 4 1 mol HC9 H 7 O 4 1 mol NaOH 1000 mmol NaOH NaOH = 23 mL NaOH aq 180.2 g HC9 H 7 O 4 1 mol HC9 H 7 O 4 1 mol NaOH = 0.077 M NaOH
56.
(M) (a)
(b)
0.10 mol HCl 36.5 g HCl 100 g conc soln' 1 mL 1 L soln 1 mol HCl 38 g HCl 1.19 g conc soln' 2 = 1.6 10 mL conc. acid
vol conc. acid = 20.0 L
The titration reaction is HCl aq + NaOH aq NaCl aq + H 2 O 1 20.93mL base HCl =
(c)
57.
0.1186 mmol NaOH 1 mmol HCl 1 mL base 1mmol NaOH = 0.09929 M HCl 25.00 mL acid
First of all, the volume of the dilute solution (20 L) is known at best to a precision of two significant figures. Secondly, HCl is somewhat volatile (we can smell its odor above the solution) and some will likely be lost during the process of preparing the solution.
(M) The equation for the reaction is HNO3 aq + KOH aq KNO3 aq + H 2 O 1 .
This equation shows that equal numbers of moles are needed for a complete reaction. We compute the amount of each reactant. mmol HNO 3 = 25.00 mL acid
0.132 mmol HNO 3 = 3.30 mmol HNO 3 1 mL acid
0.318 mmol KOH = 3.18 mmol KOH 1 mL base There is more acid present than base. Thus, the resulting solution is acidic. mmol KOH = 10.00 mL acid
58.
(M) Here we compute the amount of acetic acid in the vinegar and the amount of acetic acid needed to react with the sodium carbonate. If there is more than enough acid to react with the solid, the solution will remain acidic.
acetic acid in vinegar = 125 mL
0.762 mmol HC 2 H 3O 2 = 95.3 mmol HC2 H3O2 1 mL vinegar
184
Chapter 5: Introduction to Reactions in Aqueous Solutions
Na 2 CO3 s + 2 HC2 H 3O 2 aq 2 NaC 2 H 3O 2 aq + H 2 O(l) + CO 2 g
acetic acid required for solid: = 7.55 g Na 2 CO3
1000 mmol Na 2 CO3 2 mmol HC2 H 3 O 2 = 143 mmol HC2 H 3 O 2 106.0 g Na 2 CO3 1 mmol Na 2 CO3
Clearly there is not enough acetic acid present to react with all of the sodium carbonate. The resulting solution will not be acidic. In fact, the solution will contain only a trace amount of acetic acid (HC2H3O2). 59.
(M) Vbase = 5.00 mL vinegar
60.
1.01 g vinegar 4.0 g HC2 H 3O 2 1 mol HC2 H 3O 2 1 mL 100.0 g vinegar 60.0 g HC2 H 3O 2
1 mol NaOH 1 L base 1000 mL = 34 mL base 1 mol HC 2 H 3O 2 0.1000 mol NaOH 1 L
(M) The titration reaction is . 2 NaOH aq + H 2SO 4 aq Na 2SO 4 aq + 2 H 2 O(l)
It is most convenient to consider molarity as millimoles per milliliter when solving this problem. 0.935mmol NaOH 1mmol H 2SO 4 1mL base 2 mmol NaOH H 2SO 4 = = 4.65 M H2SO4 5.00 mL battery acid Thus, the battery acid is not sufficiently concentrated. 49.74 mL base
61.
(E) Answer is (d): 120 % of necessary titrant added in titration of NH3
5 NH3 + 5 HCl + 1 HCl 62. (E) (a) (b)
required for equivalence point
5 NH4+ + 6 Cl- + H3O+ (depicted in question's drawing )
20 % excess
H2O(l) + K+(aq) + Cl-(aq) CH3COOH(aq) + CH3COO-(aq) + H2O(l) + Na+(aq)
185
Chapter 5: Introduction to Reactions in Aqueous Solutions
Stoichiometry of Oxidation–Reduction Reactions 63.
(M) Conversion pathway approach:
0.1078 g As 2 O 3
[ MnO 4 ]=
1mol As 2 O 3 197.84 g As 2 O 3 22.15 mL
4 mol MnO 4
5 mol As 2 O3 1L
1mol KMnO 4 1mol MnO 4
= 0.01968 M KMnO 4
1000 mL
Stepwise approach: mol KMnO 4 L solution 1mol As 2 O3 0.1078 g As 2 O3 = 5.449 10-4 mol As 2 O3 197.84 g As 2 O3
[ KMnO 4 ]=
5.449 10-4 mol As 2 O3 4.359 10-4 mol MnO 4 22.15 mL
1L 1000 mL
[ KMnO 4 ]=
64.
1mol KMnO 4 1mol MnO 4
= 4.359 10-4 mol KMnO 4
0.02215 L solution
mol KMnO 4 L solution
=
4.359 10-4 mol KMnO 4 0.02215 L solution
= 1.968 10-2 M
(E) The balanced equation for the titration is: 5 SO32 aq + 2 MnO 4 aq + 6 H + aq 5 SO 4 2 aq + 2 Mn 2+ aq + 3 H 2O l 31.46 mL
SO3 2 = 65.
4 mol MnO 4 = 4.359 10-4 mol MnO 4 5 mol As 2 O3
0.02237 mmol KMnO 4 1mL soln
1mmol MnO 4 1 mmol KMnO 4
25.00 mL SO 3
2
5 mmolSO3 2 2 mmol MnO 4
soln
= 0.07038 M SO3 2
(M) First, we will determine the mass of Fe, then the percentage of iron in the ore. 2 1 L 0.05051 mol Cr2 O7 6 mol Fe 2+ 55.85 g Fe mass Fe = 28.72 mL 2 1000 mL 1 L soln 1 mol Fe 2+ 1 mol Cr2 O7
mass Fe = 0.4861 g Fe % Fe =
0.4861g Fe 100% 53.23% Fe 0.9132 g ore
186
Chapter 5: Introduction to Reactions in Aqueous Solutions
66.
(M) First, balance the titration equation.
Oxidation: { Mn 2 (aq) 4 OH (aq) MnO 2 (s) 2 H 2O(l) 2e
}3
Reduction: { MnO 4 (aq) 2 H 2 O (l) 3e MnO 2 (s) 4 OH (aq) } 2 Net:
3Mn 2 (aq) 2 MnO 4 (aq) 4 OH (aq) 5MnO 2 (s) 2 H 2 O(l)
Mn 2+
67.
0.04162 mmol MnO 4 3mmol Mn 2 37.21mL titrant 1mL titrant 2 mmol MnO 4 = = 0.09292 M Mn 2+ 25.00 mL soln
(M) First, balance the titration equation:
}5 aq 2 CO2 g + 2 e Reduction: { MnO 4 aq + 8 H + aq + 5 e Mn 2+ aq + 4 H 2 O l } 2 2 Net: 5 C2 O 4 aq + 2 MnO 4 aq +16 H + aq 10 CO 2 g + 2 Mn 2+ aq + 8 H 2 O l Oxidation: { C2 O 4
2
mass Na 2C2O4 =1.00 L satd soln Na 2 C2 O 4
1000 mL 25.8 mL satd soln KMnO 4 0.02140 mol KMnO 4 1L 5.00 mL satd soln Na 2 C2 O 4 1000 mL KMnO4
1 mol MnO 4 5 mol C2 O4 2 1 mol Na 2 C2 O4 134.0 g Na 2 C2 O 4 1 mol KMnO4 2 mol MnO4 1 mol Na 2 C2 O 4 1 mol C2 O 4 2
mass Na 2C2O4 = 37.0 g Na 2 C2 O 4
68.
(M) Balanced equation: 2 2 2 3 S2 O 4 aq + 2 CrO 4 aq + 4 H 2 O l 6 SO3 aq + 2 Cr OH 3 s + 2 H + aq (a)
mass Cr OH 3 = 100. L soln
0.0126 mol CrO 4 2 1 L soln
2 mol Cr OH 3 2 mol CrO 4
2
103.0 g Cr OH 3 1 mol Cr OH 3
= 130 g Cr(OH)3 (b)
mass Na 2 S2 O 4 = 100. L soln
0.0126 mol CrO 4 2 1 L soln
3 mol S2 O 4 2 2 mol CrO 4
2
174.1 g Na 2S2 O 4 = 329 g Na 2S2 O 4 1 mol Na 2S2 O 4
Integrative and Advanced Exercises 69. (M) (a) 2 Na(s) + 2 H2O(l)
2NaOH(aq) + H2(g)
(b) Fe3+(aq) + 3OH-(aq) Fe(OH)3(s)
187
1 mol Na 2 S2 O 4 1 mol S2 O 4 2
Chapter 5: Introduction to Reactions in Aqueous Solutions
(c) Fe(OH)3(s) + 3 H3O+(aq) Fe3+(aq) + 6 H2O(l)
or Fe(OH)3(s) + 3 H+(aq) Fe3+(aq) + 3 H2O(l) 70. (M)
FeCl 2 (aq) H 2 S(g) (a) 2 HCl(aq) FeS(s) Reduction :
2 Cl (aq) Cl 2 (aq) 2 e Mn 2 (aq) 2 H 2 O (l) MnO 2 (s) 4 H (aq) 2 e
Net:
2 Cl (aq) MnO 2 (s) 4 H (aq) Cl2 (g) Mn 2 (aq) 2 H 2 O(l)
(b) Oxidation:
Spectator Ions: 4 HCl(aq) MnO 2 (s) Cl2 (g) MnCl2 (aq) 2 H 2 O(l) (c) Because NH3 (aq) is a weak base, the reaction takes place in alkaline solution. Oxidation :
2 NH 3 (aq) N 2 (g) 6 H 6 e
Reduction :
{Br2 2 e 2 Br (aq)}
Net :
2 NH 3 (aq) 3 Br2 N 2 (g) 6 H 6 Br (aq)
3
The spectator ion is NH 4 (aq); first, add 6 NH 3 (aq) on each side, to “neutralize” H . N 2 (g) 6 H 6 Br (aq) 6 NH 3 (aq) 2 NH 3 (aq) 3 Br2 6 NH 3 (aq) Then, recognize that NH 3 (aq) H (aq) NH 4 (aq) , and NH 4 Br(aq) is really
NH 4 (aq) Br (aq) . 8 NH 3 (aq) 3 Br2 N 2 (g) 6 NH 4 Br(aq) 2 HClO 2 (aq) BaSO 4 (s) (d) Ba(ClO 2 ) 2 (s) H 2 SO 4 (aq) 71. (M) A possible product, based on solubility rules, is Ca 3 (PO 4 ) 2 . We determine the % Ca in this compound.
3 40.078 g Ca 2 30.974 g P 8 15.999 g O 120.23 g Ca 61.948 g P 127.99 g O 310.17 g 120.23 g Ca % Ca 100% 38.763% 310.17 g Ca 3 (PO 4 ) 2 molar mass
Thus, Ca 3 (PO 4 ) 2 is the predicted product. The net ionic equation follows. 2
3 Ca 2 (aq) 2 HPO 4 (aq) Ca 3 (PO 4 ) 2 (s) 2 H (aq)
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72. (M) We can calculate the initial concentration of OH-.
0.0250 mol Ba(OH) 2 2 mol OH [OH ] 0.0500 M 1 L soln 1 L soln
We can determine the ratio of the dilute (volumetric flask) to the concentrated (pipet) solutions. Vc C c Vd C d Vc 0.0500 M Vd 0.0100 M
Vd 0.0500 M 5.00 Vc 0.0100 M
If we pipet 0.0250 M Ba(OH)2 with a 50.00-mL pipet into a 250.0-mL flask, and fill this flask, with mixing, to the mark with distilled water, the resulting solution will be 0.0100 M OH . 73. (M) (a) A small amount of Na2CO3(s) mixed in with NaOH(s) will have a very small effect on the pH of the final solution which for most practical cases negligile. Lets assume that the contamination is ~0.2% by mass relative to NaOH, and 1 L of a 0.1000 M solution is made.
To make a 0.1000 M NaOH solution, the amount of NaOH needed is as follows: 0.1000 mol NaOH
39.98 g NaOH 3.998 g NaOH 1 mol NaOH
Since 0.2% of this by weight is Na2CO3, the total mass of Na2CO3 is 0.007996 g. HCl reacts with NaOH and Na2CO3 as follows: NaOH(s) HCl(aq) NaCl(aq) H 2 O (l) AND Na 2 CO 3 (s) 2 HCl(aq) 2 NaCl(aq) CO 2 (g) H 2 O (l) The amount of HCl reacted with each is therefore: 1 mol NaOH 1 mol HCl 0.09980 mol HCl reacted 39.98 g NaOH 1 mol NaOH 1 mol Na 2 CO3 2 mol HCl 1.509 104 mol HCl reacted 0.007996 g Na 2CO3 105.98 g Na 2CO3 1 mol NaOH
3.990 g NaOH
The amount of HCl needed to react with 0.2% of Na2CO3 is 0.15% of the total HCl reacted, which is a very small amount except very precise measurements. (b) As the proportion of Na2CO3(s) grows, the error it introduces becomes more significant and makes an unstandardized solution unusable for precise work. For example, trying to make the same 0.1000 M NaOH, having a 2% contamination affects the results as follows:
3.998 g NaOH × 0.02 = 0.07996 g Na2CO3, making the actual NaOH mass 3.918 g.
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1 mol NaOH 1 mol HCl 0.09800 mol HCl reacted 39.98 g NaOH 1 mol NaOH 1 mol Na 2 CO3 2 mol HCl 1.509 103 mol HCl reacted 0.07996 g Na 2 CO3 105.98 g Na 2CO3 1 mol NaOH
3.918 g NaOH
The amount of HCl needed to react with 2% of Na2CO3 contamination is 1.5% of the total HCl reacted, which is significant enough to make a standardized solution not trustworthy. 74. (D) Let us first determine the mass of Mg in the sample analyzed.
Conversion pathway approach: 1 mol Mg 2 P 2 O7 2 mol Mg 24.305 g 0.0120 g Mg 222.55 g Mg 2 P 2 O7 1 mol Mg 2 P2 O 7 1 mol Mg 0.0120 g Mg ppm Mg 106 g sample 108 ppm Mg 110.520 g sample mass Mg 0.0549 g Mg 2 P2 O7
Stepwise approach: 0.0549 g Mg 2 P2 O 7
1 mol Mg 2 P 2 O7 = 2.47 10-4 mol Mg 2 P 2 O7 222.55 g Mg 2 P 2 O 7
2.47 10-4 mol Mg 2 P 2 O7 4.93 10-4 mol Mg
2 mol Mg 4.93 10-4 mol Mg 1 mol Mg 2 P2 O 7
24.305 g 0.0120 g Mg 1 mol Mg
ppm Mg 106 g sample
0.0120 g Mg 108 ppm Mg 110.520 g sample
75. (M) Let V represent the volume of added 0.248 M CaCl2 that must be added.
We know that [Cl ] = 0.250 M, but also, 0.335 L [Cl ]
0.248 mol CaCl 2 0.186 mol KCl 1 mol Cl 2 mol Cl V 1 L soln 1 mol KCl 1 L soln 1 mol CaCl 2 0.335 L V
0.250 (0.335 V ) 0.0838 0.250 V 0.0623 0.496 V
190
V
0.0838 0.0623 0.0874 L 0.496 0.250
Chapter 5: Introduction to Reactions in Aqueous Solutions
76. (D) (a) Cu2+ would produce a colored solid, while for NH4+/Na+, no solids are expected (these cations form very soluble salts). Therefore, Cu2+, NH4+, and Na+ are not present. Thus the possible cations are Ba2+ and Mg2+ (both give colorless solutions). (b) Gas evolution when the solid reacts with HCl(aq) suggests the presence of carbonate (CO32-).
2 H+(aq) + CO32-(aq) “H2CO3(aq)” H2O(l) + CO2(g) If the solid is indeed a mixture of carbonates, the net ionic equation for the reaction of the solid with HCl(aq) would be: BaCO3(s) + MgCO3(s) + 4 H+(aq) Ba2+(aq) + Mg2+(aq) + 2 H2O(l) + 2 CO2(g) The solution above contains Ba2+ + Mg2+ ions, so the addition of (NH4)2SO4(aq) should result in the formation of the insoluble sulfates of these two metal ions. This is consistent with the observations. Net ionic equation: Ba2+(aq) + Mg2+(aq) + 2 SO42- BaSO4(s) + MgSO4(s). If the solution above the solid should contain small quantities of all of the ions present in the solid, then it should contain Ba2+(aq), Mg2+(aq), and CO32-(aq). Addition of KOH should result in the formation of the hydroxide of the two metal ions and potassium carbonate. Note: K2CO3 and Ba(OH)2 are both relatively soluble species, hence, neither of these species should precipitate out of solution. However, Mg(OH)2, is relatively insoluble. A precipitate (Mg(OH)2(s) which is white) is expected to form. This is consistent with the observations provided. We can conclude that the solid is likely a mixture of BaCO3(s) and MgCO3(s). Without additional data, this conclusion would indeed explain all of the observations provided. 77. (M) (a) Oxidation: {
IBr(aq) 3 H 2 O(l) IO3 (aq) Br (aq) 6 H (aq) 4 e }
Reduction: { BrO3 6 H (aq) 6 e Br (aq) 3 H 2 O
}
3 2
Net: 3 IBr(aq) 3 H 2 O(l) 2 BrO3 (aq) 3 IO3 (aq) 5 Br (aq) 6 H (aq) (b) Oxidation : {Sn(s) Sn 2 (aq) 2 e }
3
Reduction: C 2 H 5 NO3 (aq) 6 H (aq) 6 e C 2 H 5 OH(aq) NH 2 OH(aq) H 2 O(l) Net: 3 Sn(s) C2 H 5 NO3 (aq) 6 H (aq) 3 Sn 2 (aq) C2 H 5 OH(aq) NH 2 OH(aq) H 2 O(l) (c) Oxidation: {As 2 S3 (s) 8 H 2 O(l) 2 H3 AsO 4 (aq) 3 S(s) 10 H (aq) 10 e }
Reduction : {NO 3 (aq) 4 H (aq) 3 e NO(g) 2 H 2 O }
3
10
Net: 3 As 2 S3 (s) 4 H 2 O(l) 10 NO3 (aq) 10 H (aq) 6 H3 AsO 4 (aq) 9 S(s) 10 NO(g)
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Chapter 5: Introduction to Reactions in Aqueous Solutions
(d) Oxidation:I 2 (aq) 6 H 2 O(l) 2 IO3 (aq) 12 H (aq) 10 e
1
Reduction : {H 5 IO 6 (aq) H (aq) 2 e IO 3 (aq) 3 H 2 O }
5
Net : I 2 (aq) 5 H 5 IO 6 (aq) 7 IO 3 (aq) 9 H 2 O(l) 7 H (aq)
(e) Oxidation: {2 S2 F2 (g) 6 H 2 O(l) H 2 S4 O6 (aq) 4 HF(aq) 6 H (aq) 6 e } 4
Reduction: {4 S2 F2 (g) 8 H (aq) 8 e- S8 (s) 8 HF(aq) }
3
Net: 20 S2 F2 (g) 24 H 2 O(l) 4 H 2 S4 O6 (aq) 3 S8 (s) 40 HF(aq)
78. (M) (a) Oxidation : {Fe 2 S3 (s) 6 OH (aq) 2 Fe(OH)3 (s) 3 S(s) 6 e } 2
Reduction : {O 2 (g) 2 H 2 O(l) 4 e 4 OH (aq)
}
3
Net : 2 Fe 2 S 3 (s) 3 O 2 (g) 6 H 2 O(l) 4 Fe(OH) 3 (s) 6 S(s) (b) Oxidation: {4 OH (aq) O 2 (g) 2 H 2O(l) 4 e }
3
Reduction : {O 2 (aq) 2 H 2 O(l) 3 e 4 OH (aq) }
4
3 O 2 (g) 4 OH (aq) Net : 4 O 2 (aq) 2 H 2 O(l) 2
(c) Oxidation: {CrI3 (s) 32 OH (aq) CrO 4 (aq) 3 IO 4 (aq) 16 H 2 O 27 e }
Reduction: {H 2 O2 (aq) 2 e 2 OH (aq)}
27 2
2
Net : 2 CrI 3 (s) 10 OH (aq) 27 H 2 O 2 (aq) 2 CrO 4 (aq) 6 IO 4 (aq) 32 H 2 O
(d) Oxidation : {Ag(s) 2 CN (aq) [Ag(CN) 2 ] (aq) e } 4
Reduction: 2H 2 O(l) O 2 (g) 4 e 4 OH (aq) Net : 4 Ag(s) 8 CN (aq) 2 H 2 O O 2 (g) 4[Ag(CN) 2 ] (aq) 4 OH (aq) (e) Oxidation: B2 Cl 4 (aq) 8 OH (aq) 2 BO 2 (aq) 4 Cl (aq) 4 H 2 O 2 e
Reduction: 2 H 2 O(l) 2 e H 2 (g) 2 OH (aq) Net: B2 Cl4 (aq) 6 OH (aq) 2 BO2 (aq) 4 Cl (aq) 2 H 2 O(l) H 2 (g)
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79. (M)
Oxidation: {P4 (s) 16 H 2 O(l) 4 H 3 PO 4 (aq) 20 H (aq) 20 e } Reduction : {P4 (s) 12 H (aq) 12 e 4 PH 3 (g)
3
5
}
Net: 8 P4 (s) 48 H 2 O(l) 12 H3 PO4 (aq) 20 PH3 (g) Or: 2 P4 (s) 12 H 2 O(l) 3 H3 PO 4 (aq) 5 PH 3 (g)
80. (D) (a) [FeS2 + 8 H2O → Fe3+ + 2 SO42− + 16 H+ + 15 e−] × 4 [O2 + 4 H+ + 4 e− → 2 H2O] × 15
overall: 4 FeS2(s) + 15 O2(g) + 2 H2O(l) → 4 Fe3+(aq) + 8 SO42−(aq) + 4 H+(aq) (b) One kilogram of tailings contains 0.03 kg (30 g) of S. We have
moles of FeS2 = 30 g S
1 mol S 1 mol FeS2 0.468 mol FeS2 32.07 g S 2 mol S
moles of H+ = 0.468 mol FeS2
4 mol H + 0.467 mol H + 4 mol FeS2
moles of CaCO3 = 0.467 mol H +
1 mol CaCO3 0.234 mol CaCO3 2 mol H +
mass of CaCO3 = 0.234 mol CaCO3
100.09 g CaCO3 23.4 g CaCO3 1 mol CaCO3
81. (D) The neutralization reaction is H 2 SO 4 (aq) Ba(OH) 2 (aq) BaSO 4 (s) 2 H 2 O(l) .
First, we determine the molarity of H2SO4 in the 10.00 mL of diluted acid. 32.44 mL base
0.00498 mmol Ba(OH) 2
molarity of H 2SO 4
1 mL base 10.00 mL acid
1 mmol H 2SO 4 1 mmol Ba(OH) 2
0.0162 M H 2SO 4
Next, we determine the molarity of H2SO4 in the concentrated solution, using Vc × Cc = Vd × Cd. 1.00 mL Cc 250.0 mL 0.0162 M %H 2 SO 4
Cc
250.0 mL 0.0162 M 4.05 M H 2 SO 4 1.00 mL
4.05 mol H 2 SO 4 98.08 g H 2 SO 4 1L 1 mL 100% 32.1% H 2 SO 4 1 L soln 1 mol H 2 SO 4 1000 mL 1.239 g
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Chapter 5: Introduction to Reactions in Aqueous Solutions
82. (D) The titration reaction is: CaCO3 (s) 2 H (aq) Ca 2 (aq) H 2 O(l) CO 2 (g) .
Now we calculate the mass of the marble, through several steps, as follows. 2.52 mol HCl 5.04 mol HCl 1 L soln NaOH(aq) HCl(aq) NaCl(aq) H 2 O is the titration reaction.
initial moles HCl 2.00 L
final molarity of HCl
0.02487 L
0.9987 mol NaOH 1 mol HCl 1L 1 mol NaOH 2.484 M 0.01000 L
2.484 mol HCl 4.968 mol HCl 1 L soln 1 mol CaCO3 100.1 g CaCO3 mass CaCO3 (5.04 4.968) mol HCl 3.6 g CaCO3 ~ 4g CaCO 3 2 mol HCl 1 mol CaCO3 There are two reasons why the final result can be determined to but one significant figure. The first is the limited precision of the data given, as in the three-significant-figure limitation of the initial concentration and the total volume of the solution. The second is that the initial solution is quite concentrated for the job it must do. If it were one-tenth as concentrated, the final result could be determined more precisely because the concentration of the solution would have decreased to a greater extent. final amount HCl 2.00 L
83. (M)
Oxidation : {2 Cl (aq) Cl 2 (g) 2 e }
3
2
Reduction : Cr2 O 7 (aq) 14 H (aq) 6 e 2 Cr 3 (aq) 7 H 2 O 2
Net : 6 Cl (aq) Cr2 O 7 (aq) 14 H (aq) 2 Cr 3 (aq) 7 H 2 O 3 Cl 2 (g)
We need to determine the amount of Cl2(g) produced from each of the reactants. The limiting reactant is the one that produces the lesser amount of Cl2.. 1.15 g 30.1 g HCl 1 mol HCl 1 mol Cl 3 mol Cl 2 1 mL 100. g soln 36.46 g HCl 1 mol HCl 6 mol Cl 1.54 mol Cl 2
amount Cl 2 325 mL
2
98.5 g K 2 Cr2 O 7 1 mol K 2 Cr2 O 7 1 mol Cr2 O 7 3 mol Cl 2 amount Cl 2 62.6 g 100. g sample 294.2 g K 2 Cr2 O 7 1 mol K 2 Cr2 O 7 1 mol Cr2 O 7 2 0.629 mol Cl 2 , the amount produced from the limiting reactant
Then we determine the mass of Cl2(g) produced. = 0.629 mol Cl2 ×
194
70.91 g Cl2 = 44.6 g Cl2 1 mol Cl2
Chapter 5: Introduction to Reactions in Aqueous Solutions
84. (M)
Oxidation: {As 2 O3 (s) 5 H 2 O(l) 2H 3 AsO 4 (aq) 4 H (aq) 4 e } 5 Reduction: {MnO 4 (aq) 8 H (aq) 5 e Mn 2 (aq) 4 H 2 O(l)}
4
2+ Net: 5 As2O3(s) + 9 H2O(l) + 4 MnO4–(aq) + 12 H+(aq) 10 H3AsO4(aq) + 4 Mn (aq)
KMnO4 molarity = 0.02140 M, as in Example 5-10. 99.96 g As 2 O3 1 mol As 2 O3 4 mol MnO 4 1 mol KMnO 4 soln. volume 0.1304 g 100.00 g sample 197.84 g As 2 O3 5 mol As 2 O3 1 mol MnO 4
1 L soln 1000 mL 24.63 mL soln 0.02140 mol KMnO 4 1L
85. (M)
Cl 2 (g) NaClO 2 (aq) NaCl(aq) ClO 2 (g)
(not balanced)
2 NaCl(aq) 2 ClO 2 (g) Cl 2 (g) 2 NaClO 2 (aq) amount ClO 2 1 gal
2 mol ClO2 67.45 g ClO 2 3.785 L 2.0 mol NaClO 2 1 gal 1 L soln 2 mol NaClO 2 1 mol ClO2
97 g ClO 2 produced 5.0 102 g ClO 2 (g) 100 g ClO 2 calculated
86. (M) CaCO3(s) + 2 HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq) nHCl(initial) = 0.05000 L × 0.5000 M = 0.02500 mol HCl nOH- = 0.04020 L × 0.2184 M = 0.00878 mol OHnHCl(excess) = 0.00878 mol OH-
1 mol HCl = 0.00878 mol HCl 1 mol OH -
nHCl(reacted) = nHCl(initial) - nHCl(excess) = 0.02500 mol HCl - 0.00878 mol HCl= 0.01622 mol HCl 2+ 2+ 1000 mg Ca 2+ mass Ca2+ = 0.01622 mol HCl 1 mol CaCO3 1 mol Ca 40.078 g Ca = 325 mg Ca 2+ 2+ 2+
2 mol HCl
195
1 mol CaCO3
1 mol Ca
1 g Ca
Chapter 5: Introduction to Reactions in Aqueous Solutions
87.
(M) Let XKOH = mass of KOH in grams and XLiOH be the mass of LiOH in grams.
(Note: Molar masses: KOH = 56.1056 g mol-1 and LiOH = 23.9483 g mol-1) moles of HCl = CV = 0.3520 M 0.02828 L = 0.009956 mol HCl We can set up two equations for the two unknowns: XKOH + XLiOH = 0.4324 g and since moles of HCl = moles of OH- (Stoichiometry is 1:1) 0.009956 mol OH- =
X KOH X LiOH 56.1056 23.9483
Make the substitution that XKOH = 0.4324 g - XLiOH 0.009956 mol OH- =
(0.4324 X LiOH ) X X LiOH X 0.4324 + LiOH = + LiOH 56.1056 23.9483 56.1056 56.1056 23.9483
Collect terms: 0.009956 mol OH- = 0.007707 mol OH- + 0.02393XLiOH mol OH0.009956 mol OH- - 0.007707 mol OH- = 0.02393XLiOH mol OH- = 0.002249 mol OH-
XLiOH = 0.09397 g LiOH hence, XKOH = 0.4324 g - 0.09397 g = 0.3384 g Mass % LiOH =
0.09397 g 0.3384 g 100% 21.73% Mass % KOH = 100% 78.26% 0.4324 g 0.4324 g
88. (D) (a) First, balance the redox equations needed for the calculation.
Oxidation: {HSO3- (aq)+ H2O(l) SO42- (aq) + 3 H+ (aq) + 2 e- }
×
3
Reduction: {IO3- (aq) + 6 H+ (aq) + 6 e- I-(aq) + 3 H2O(l) }
×
1
Net: 3 HSO3- (aq) + IO3- (aq) 3 SO42- (aq) + 3 H+ (aq) + I- (aq) The solution volume of 5.00 L contains 29.0 g NaIO3. This represents 29.0 g/197.9g/mol NaIO3 = 0.147 mol NaIO3. (b) From the above equation, we need 3 times that molar amount of NaHSO3, which is 3(0.147 mol) = 0.441 mol NaHSO3; the molar mass of NaHSO3 is 104.06 g/mol.
The required mass then is 0.441(104.06) = 45.9 g.
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For the second process: Oxidation: {2 I-(aq) I2(aq) + 2 e- }
× 5
Reduction: {2 IO3- (aq) + 12 H+ (aq) + 10 e- I2(aq) + 6 H2O(l) }
× 1
Net:
5 I-(aq) + IO3- (aq) + 6 H+ (aq) 3 I2(aq) + 3 H2O(l)
In Step 1, we produced 1 mol of I- for every mole of IO3- reactant; therefore we had 0.147 mol I-. In step 2, we require 1/5 mol IO3- for every mol of I-. We require only 1.00 L of the solution in the question instead of the 5.00 L in the first step. 89.
(D)
Mg(OH)2(aq) + 2 HCl(aq) → MgCl2(aq) + 2 H2O(l)
(1)
Al(OH)3(aq) + 3 HCl(aq) → AlCl3(aq) + 3 H2O(l)
(2)
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
(3)
initial moles of HCl =
0.0500 L
0.500 mol = 0.0250 mol 1L
moles of HCl that reacted with NaOH = moles of HCl left over from reaction with active ingredients = 0.0165 L
0.377 mol NaOH 1 mol HCl = 6.22 10-3 mol 1L 1 mol NaOH
moles of HCl that react with active ingredients = 0.0250 mol - 6.22 10-3 mol = 0.0188 mol # moles HCl that # moles HCl that react with Mg(OH) + react with Al(OH) = total moles of HCl reacted/used 3 2
# moles HCl that react with Mg(OH)2 = 1 mol Mg(OH) 2 2 mol HCl X grams Mg(OH) 2 58.32 g Mg(OH) 2 1 mol Mg(OH) 2
# moles HCl that react with Al(OH)3 = 1 mol Al(OH)3 3 mol HCl 0.500-X grams Al(OH)3 78.00 g Al(OH)3 1 mol Al(OH)3
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2X 3(0.500 X) 0.0188 58.32 78.00 X = 0.108, therefore the mass of Mg(OH)2 in the sample is 0.108 grams. % Mg(OH)2 = (0.108/0.500) × 100 = 21.6
%Al = 100 − %Mg(OH)2 = 78.4
90. (D) The first step is to balance the chemical equation. Using the method outlined in the text,
we obtain the following result: 8 H+(aq) + MnO4(aq) + 5 Fe2+(aq) Mn2+(aq) + 5 Fe3+(aq) + 4 H2O(l)
0.04217 L
0.01621 mol KMnO 4 1 mol MnO 4 5 mol Fe 2+ = 0.003418 mol Fe 2+ 1L 1 mol KMnO 4 1 mol MnO 4
55.847 g = 0.1909 g Fe 1 mol 0.2729 g sample - 0.1909 g Fe = 0.0820 g oxygen 0.003418 mol Fe in sample
1 mol = 0.005125 mol O 0.003418 = 1.5 16.00 g 0.003418 mol Fe 0.003418 = 1 (FeO1.5 ) 2 = Fe 2 O3 0.0820 g O
91. (M) 0.1386 g AgI
1 mol AgI 1 mol CHI3 1 mol C19 H16 O 4 308.33 g C19 H16 O 4 234.77 g AgI 3 mol AgI 1 mol CHI3 1 mol C19 H16 O 4
0.06068 g C19 H16 O 4
% C19 H16 O 4 =
0.06068 g 100 0.4346 % 13.96 g
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92. (M) The first step is to balance the chemical equation. By using the method described in the
text, we obtain the following result: 3 CuS(s) + 8 NO3−(aq) + 11 H+(aq) → 3 Cu2+(aq) + 8 NO(g) + 3 HSO4−(aq) + 4 H2O(l) The next step is to calculate the volume of solution required: 1 kg CuS
1000 g 1 mol CuS 8 mol HNO3 63.02 g HNO3 = 1757.7 g HNO3 1 kg 95.61 g CuS 3 mol CuS 1 mol HNO3
1757.7 g HNO3
100 g soln 1 mL soln = 1793.6 mL = 2000 mL (to 1 sig fig) 70 g HNO3 1.40 g soln
93. (M) (a)
CaO(s) + H2O(l) → Ca2+(aq) + 2 OH−(aq) H2PO4−(aq) + 2 OH−(aq) → PO43−(aq) + 2 H2O(l) HPO4−(aq) + OH−(aq) → PO43−(aq) + H2O(l) 5 Ca2+(aq) + 3 PO43−(aq) + OH−(aq) → Ca5(PO4)3OH(s) 10.0 103 g P 1 mol P 1 mol PO34 5 mol Ca 2 1.00 10 L L 30.97 g P 1 mol P 3 mol PO34 4
(b)
1 mol CaO 56.08 g CaO = 301.80 g CaO = 302 g = 0.302 kg 1 mol Ca 2 1 mol CaO
FEATURE PROBLEMS 94.
(D) From the volume of titrant, we can calculate both the amount in moles of NaC5 H 5 and (through its molar mass of 88.08 g/mol) the mass of NaC5 H 5 in a sample. The remaining mass in a sample is that of C 4 H 8O (72.11 g/mol), whose amount in moles we calculate. The ratio of the molar amount of C 4 H 8O in the sample to the molar amount of NaC5 H 5 is the value of x.
Conversion pathway approach: 0.1001 mol HCl 1 mol NaOH 1 mol NaC5 H 5 1 L soln 1 mol HCl 1 mol NaOH = 0.001493 mol NaC5 H 5
moles of NaC5 H 5 = 0.01492 L
88.08 g NaC5 H 5 mass of C4 H8O = 0.242 g sample 0.001493 mol NaC5 H 5 1 mol NaC5 H 5 = 0.111 g C4 H8O
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1mol C4 H8O 72.11g C4 H8O = 1.03 0.001493 mol NaC5 H 5
0.110 g C4 H8O
x=
Stepwise approach: 0.1001 mol HCl 0.01492 L = 1.493 10-3 mol HCl 1 L soln 1 mol NaOH 1.493 10-3 mol HCl = 1.493 10-3 mol NaOH 1 mol HCl 1 mol NaC5 H 5 1.493 10-3 mol NaOH 1.493 10-3 mol NaC5 H 5 1 mol NaOH 88.08 g NaC5 H 5 1.493 10-3 mol NaC5 H5 0.1315 g NaC5 H 5 1 mol NaC5 H 5 mass of C4 H8O = 0.242 g sample 0.1315 g NaC5 H5 = 0.111 g C4 H8O 0.111g C4 H8O
1mol C4 H8O = 1.54 10-3 mol C4 H8O 72.11g C4 H8O
1.54 10-3 mol C4 H8O = 1.03 0.001493 mol NaC5 H 5 For the second sample, parallel calculations give 0.001200 mol NaC5 H 5 , 0.093 g C 4 H 8 , x = 1.1. There is rounding error in this second calculation because it is limited to two significant figures. The best answer is from the first run x ~1.03 or 1. The formula is NaC5H5(THF)1. 95.
(D) First, we balance the two equations. Oxidation: H 2 C2 O 4 aq 2 CO 2 g + 2 H + aq + 2 e
Reduction: MnO 2 s + 4 H + aq + 2 e Mn 2+ aq + 2 H 2 O l Net:
H 2 C 2 O 4 aq + MnO 2 s + 2 H + aq 2 CO 2 g + Mn 2+ aq + 2 H 2O l
Oxidation: { H 2 C2 O 4 aq 2 CO 2 g + 2 H + aq + 2 e
}5
Reduction: { MnO 4 aq + 8 H + aq + 5 e Mn 2+ aq + 4 H 2 O l
}2
Net: 5 H 2 C2 O 4 aq + 2 MnO 4 aq + 6 H + aq 10 CO 2 g + 2 Mn 2+ aq + 8 H 2 O l
Next, we determine the mass of the excess oxalic acid. 0.1000 mol KMnO 4 1mol MnO 4 5 mol H 2 C2 O 4 mass H 2C2 O 4 2H 2 O 0.03006 L 1L 1mol KMnO 4 2 mol MnO 4
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1 mol H 2 C 2 O 4 2H 2 O 126.07 g H 2 C 2 O 4 2H 2 O = 0.9474 g H 2C2O4 2H 2O 1 mol H 2 C 2 O 4 1 mol H 2 C 2 O 4 2H 2 O
The mass of H 2 C 2 O 4 2 H 2 O that reacted with MnO2 = 1.651 g 0.9474 g = 0.704 g H 2C2O4 2H 2O mass MnO 2 = 0.704 g H 2 C 2 O 4 2H 2 O
1 mol H 2 C 2 O 4
1 mol MnO 2
126.07 g H 2 C 2 O 4 2H 2 O 1 mol H 2 C 2 O 4
86.9 g MnO 2 1 mol MnO 2
= 0.485 g MnO 2
% MnO 2 96.
0.485g MnO 2 100% 91.0% MnO 2 0.533g sample
(D) Reactions: 2 NH3 + H2SO4 → (NH4)2SO4 and H2SO4 + 2NaOH → 2 H2O + Na2SO4 n OH- (used to find moles H 2SO 4 in excess) 0.03224 L 0.4498 M = 0.01450 mol OH 1 mol H 2SO 4 0.00725108 mol H 2SO 4 2 mol OH n OH- (used to find moles H 2SO 4 in separate unreacted sample) 0.02224 L 0.4498 M n H2SO4 ( in excess) = 0.01450 mol OH -
= 0.0100035 mol OH 1 mol H 2SO 4 0.005002 mol H 2SO 4 2 mol OH NOTE: this was in a 25.00 mL sample: we need to scale up to 50.00 mL.
n H2SO4 (initial) = 0.0100035 mol OH -
Hence, n H2SO4 (initial) = 2 0.005002 mol H 2SO 4 = 0.0100035 mol H 2SO 4 n H2SO4 (reacted) = n H2SO4 (initial) n H2SO4 (excess) = 0.0100035 mol H 2SO 4 0.00725108 mol H 2SO 4 = 0.00275 mol H 2SO 4 n NH3 0.00275 mol H 2SO 4
2 mol NH 3 0.005505 mol NH 3 1 mol H 2SO 4
mass NH3 0.005505 mol NH 3
1 mol N 14.0067 g N 0.0771 g N in sample 1 mol NH 3 1 mol N
100 g protein = 0.482 g protein in sample 16 g N 0.482 g protein in sample percent protein in sample = 100% = 38.6 % protein 1.250 g sample
mass protein in sample = 0.0771 g N in sample
97.
(D)
The molecular formula for CH3CH2OH is C2H6O and for CH3COOH is C2H4O2. The first step is to balance the oxidation–reduction reaction. Oxidation: [C2H6O + H2O → C2H4O2 + 4 H+ + 4 e−] × 3 Reduction: [Cr2O72− + 14 H+ + 6e− → 2 Cr3+ + 7 H2O] × 2 Overall: 3 C2H6O + 2 Cr2O72− + 16 H+ → 3 C2H4O2 + 4 Cr3+ + 11 H2O 201
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Before the breath test: 0.75 mg K 2 Cr2 O7 1g 1 mol 1000 mL = 8.498 10-4 M 3 mL 1000 mg 294.19 g 1L = 8×10-4 M (to 1 sig fig) For the breath sample: BrAC =
0.05 g C2 H 6 O 2.38×107 g C2 H 6 O 1 mL blood = mL breath 100 mL blood 2100 mL breath
mass C2H6O =
2.38×107 g C2 H 6 O × 500. mL breath = 1.19 × 10−4 g C2H6O mL breath
Calculate the amount of K2Cr2O7 that reacts: –
1 mol C2 H 6 O 2 mol Cr2 O72 1 mol K 2 Cr2 O7 1.19 10 g C2 H 6 O – 46.068 g C2 H 6 O 3 mol C2 H 6 O 1 mol Cr2 O72 4
= 1.72 106 mol K 2 Cr2 O7 # mol K2Cr2O7 remaining = moles K2Cr2O7 before – moles K2Cr2O7 that reacts moles K2Cr2O7 before = 0.75 mg K 2 Cr2 O7
1g 1 mol = 2.5 10-6 mol 1000 mg 294.19 g
# mol K2Cr2O7 remaining = 2.5 × 10−6 mol − 1.72 × 10−6 mol = 0.78 × 10−6 mol concentration of K2Cr2O7 after the breath test = 0.78 × 10−6 mol/0.003 L = 2.6 × 10−4 mol/L = 3 × 10−4 mol/L (to 1 sig fig) 98. (D) (a) Step 1: Assign oxidation states to each element in the reaction and identify the species
being oxidized and reduced. The oxidation state of Cr is +6 in Cr2O72− and +3 in Cr3+. The oxidation state of Cl is −1 in Cl− and 0 in Cl2. Each Cr gains three electrons and each Cl loses one electron.
Step 2: Write separate, unbalanced equations for the oxidation and reduction half-reactions. Oxidation: Cl → Cl2 Reduction: Cr2 O72 → Cr3+
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Step 3: Balance the separate half-equations, in this order: First, with respect to the element being oxidized or reduced Oxidation: 2 Cl− → Cl2 Reduction: Cr2O72− → 2 Cr3+
Then, by adding electrons to one side or the other to account for the number of electrons produced (oxidation) or consumed (reduction) Keep in mind that each Cl loses one electron and each Cr gains three electrons. Oxidation: 2 Cl− → Cl2 + 2 e− Reduction: Cr2O72− + 6 e− → 2 Cr3+
Step 4: Combine the half-equations algebraically so that the total number of electrons cancels out. Oxidation: [2 Cl− → Cl2 + 2 e−] × 3 Reduction: [Cr2O72− + 6 e− → 2 Cr3+] × 1 Overall: 6 Cl− + Cr2O72− → 3 Cl2 + 2 Cr3+
Step 5: Balance the net charge by adding either H+ (for acidic solutions) or OH− (for basic solutions). The reaction occurs in acidic solution, so we use H+ to balance charge. The total charge on the left side of the overall equation is 6(−1) + (−2) = −8. The total charge on the right side is 3(0) + 2(+3) = +6. To balance charge, add 14 H+ to the left side: 6 Cl− + Cr2O72− + 14 H+ → 3 Cl2 + 2 Cr3+
Step 6: Balance H and O by adding H2O. Add 7 H2O to the right side to balance hydrogen. This also balances oxygen. The balanced equation is given below: 6 Cl− + Cr2O72− + 14 H+ → 3 Cl2 + 2 Cr3+ + 7 H2O (b) Step 1: Assign oxidation states to each element in the reaction and identify the species
being oxidized and reduced. The oxidation state of C is +3 in C2O42− (treating the C’s as equivalent) and +4 in CO32-. The oxidation state of Mn is +7 in MnO4− and +4 in MnO2. Each Mn gains three electrons and each C loses one electron.
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Step 2: Write separate, unbalanced equations for the oxidation and reduction half-reactions. Oxidation: C2O42− → CO32− Reduction: MnO4− → MnO2
Step 3: Balance the separate half-equations, in this order: First with respect to the element being oxidized or reduced Oxidation: C2O42− → 2 CO32− Reduction: MnO4− → MnO2
Then, by adding electrons to one side or the other to account for the number of electrons produced (oxidation) or consumed (reduction) Keep in mind that each C loses one electron and each Mn gains three electrons. Oxidation: C2O42− → 2 CO32− + 2 e− Reduction: MnO4− + 3 e− → MnO2
Step 4: Combine the half-equations algebraically so that the total number of electrons cancels out. Oxidation: [C2O42− → 2 CO32− + 2 e−] × 3 Reduction: [MnO4− + 3e− → MnO2] × 2 Overall: 3 C2O42− + 2 MnO4− → 6 CO32− + 2 MnO2
Step 5: Balance the net charge by adding either H+ (for acidic solutions) or OH− (for basic solutions). The reaction occurs in basic solution, so we use OH− to balance charge. The total charge on the left side of the overall equation is 3(−2) + 2(−1) = −8. The total charge on the right side is 6(−2) + 2(0) = −12. To balance charge, we must add 4 OH− to the left side: 3 C2O42− + 2 MnO4− + 4 OH− → 6 CO32− + 2 MnO2
Step 6: Balance H and O by adding H2O. Add 2 H2O to the right-hand side to balance hydrogen. This also balances oxygen. The balanced equation is given below: 3 C2O42− + 2 MnO4− + 4 OH− → 6 CO32− + 2 MnO2 + 2 H2O
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SELF-ASSESSMENT EXERCISES 99.
(E) (a) : The process/reaction is reversible (b) [] : A shorthand notation to indicate molar concentration (c) A spectator ion does not participate in a reaction and remains unchanged. (d) A weak acid is one that does not fully dissociate in water. A solution of that acid contains a large amount of the undissociated acid in molecular form.
100. (E) (a) Half-equation method of balancing redox reactions: A method of balancing equations in which the major oxidation and reduction half-reactions are balanced separately (b) Disproportionation reaction : A reaction in which a substance is both oxidized and reduced (c) Titration: A method of determining the concentration of an unknown by reacting it with a reactant of known volume and concentration (d) Standardization of a solution: The process of determining the exact concentration of a solution 101. (E) (a) Strong electrolyte vs. strong acid: A strong electrolyte dissociates fully in water to its constituent ions. A strong acid dissociates fully in water to give a proton and an anion. A strong acid is always a strong electrolyte, but a strong electrolyte need not be an acid (b) Oxidizing vs. reducing agent: An oxidizing agent removes electrons from a reactant (it gets reduced itself), while a reducing agent gives electrons to a reactant (it gets oxidized itself). (c) Precipitation vs. neutralization reaction: In a precipitation reaction, an insoluble product is formed and is removed from the solution phase. In a neutralization reaction, an acid and a base react with each other to form salt and water. (d) Half-reaction vs. overall reaction: A half-reaction represents either the oxidation or reduction portion of a reaction. The overall reaction is the summation of the oxidation and reduction half-reactions. 102. (E) The answer is (b). Conversion pathway approach: 0.0050 mol Ba(OH) 2 2 mol OH 0.300 L = 0.0030 mol 1L 1mol Ba(OH) 2
Stepwise approach: 0.0050 mol Ba(OH) 2 0.300 L = 1.5 10-3 mol Ba(OH) 2 1L 2 mol OH -3 1.5 10 mol Ba(OH) 2 = 0.0030 mol 1mol Ba(OH) 2 205
Chapter 5: Introduction to Reactions in Aqueous Solutions
103. (E) The answer is (d), because H2SO4 is a strong diprotic acid and theoretically yields 0.20 mol of H+ for every 0.10 mol of H2SO4. 104. (E) The answer is (c). Based on the solubility guidelines in Table 5-1, carbonates (CO32-) are insoluble. 105. (M) The answer is (a). Reaction with ZnO gives ZnCl2 (soluble) and H2O. There is no reaction with NaBr and Na2SO4, since all species are aqueous. By the process of elimination, (a) is the answer. 106. (E) Balanced equation: 2 KI + Pb(NO 3 ) 2 2KNO3 + PbI 2 Net ionic equation: 2I- + Pb 2 PbI 2 (s)
107. (E) Balanced equation: Na 2 CO3 + 2HCl 2NaCl + H 2 O + CO 2 Net ionic equation: CO 23 + 2H H 2O (l) + CO 2 (g)
108. (M) Balanced equation: 2 Na 3PO 4 + 3 Zn(NO 3 ) 2 6NaNO3 + Zn 3 (PO 4 ) 2 (a) Net ionic equation: 3 Zn 2+ + PO34 Zn 3 (PO 4 ) 2 (s) (b) (c)
Balanced equation: 2 NaOH + Cu(NO 3 ) 2 Cu(OH) 2 + 2 NaNO3 Net ionic equation: Cu 2+ + 2 OH Cu(OH) 2 (s) Balanced equation: NiCl2 + Na 2 CO 3 NiCO3 + 2 NaCl Net ionic equation: Ni 2+ + CO32 NiCO3 (s)
109. (M) (a) Species oxidized: N in NO (b) Species reduced: O2 (c) Oxidizing agent: O2 (d) Reducing agent: NO (e) Gains electrons: O2 (f) Loses electrons: NO 110. (M) The answer is (b). The charges need to be balanced on both sides. Using a coefficient of 4, the charges on both sides of the reaction becomes +12.
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111. (D) The answer is (d), 5 ClO- to 1 I2. The work to balance the half-reactions is shown below: Reduction: 5ClO + 2H + + 2e Cl + H 2 O Oxidation: I 2 + 6H 2 O 2IO3 + 10e + 12H To combine the above reactions, the oxidation reaction should be multiplied by 5. The combined equation is: Combined: 5ClO + I 2 + H 2 O 5Cl + 2IO3 + 2H
112. (M) The answer is (a). The balanced half-reaction is as follows: NpO 2 + 4 H + + e- Np 4 + 2H 2 O 113. (M) (a) False. Based on solubility rules, BaCl2 dissolves well in water. Therefore, it is a strong electrolyte. (b) True. Since H- is a base, H2O is by necessity an acid. It also reduces H- (-1) to H2 (0). (c) False. The product of such a reaction would be NaCl and H2CO3, neither of which precipitates out. (d) False. HF is among the strongest of weak acids. It is not a strong acid, because it doesn’t completely dissociate. (e) True. For every mole of Mg(NO3)2, there are 3 moles of ions, in contrast to 2 moles of ions for NaNO3. 114. (M) (a) No. Oxidation states of C, H or O do not change throughout the reaction. (b) Yes. Li is oxidized to Li+ and H in H2O is reduced from +1 to 0 in H2. (c) Yes. Ag is oxidized and Pt is reduced. (d) No. Oxidation states of Cl, Ca, H, and O remain unchanged. 115. (E) The dissociation of acetic acid in water can be expressed as follows: HAc + H2O → Ac- + H3O+ If HAc dissociates by 5% and assuming an initial value of 100 HAc molecules, 5 molecules of HAc dissociate to give 5 H+ and 5 Ac- ions, while 95 HAc molecules remain. In other words, for every 19 HAc molecules, there is one H+ and one Ac-. 116. (M) To construct a concept map, one must first start with the most general concepts. These concepts are defined by or in terms of other simpler concepts discussed in those sections. In this case, the overarching concept is Oxidation-Reduction reactions. This concept can be further explained by defining electron transfer. Oxidation involves the loss of electrons, reduction the gain. Subtopics that fall under the rubric of oxidation-reduction reactions are oxidizing agents (which get reduced) and reducing agents. The concept of half-reactions is another subtopic, and balancing half reaction falls under half-reactions. Balancing the overall reaction falls under the subtopic of balancing half reactions. Take a look at the subsection headings and problems for more refining of the general and specific concepts.
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