CHAPTER 4CH7 PROBLEM SOLVING CLASSR.D. A. BOLINAS

4.9 State whether each of the following substances is likely to be very soluble in water. Explain.(a) Lithium nitrate (b) Glycine, H2NCH2COOH(c) Pentane (d) Ethylene glycol, HOCH2CH2OH

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 Hg2+

• 4 . All common fluorides (F-) are soluble, except those of Pb2+ and Group 2A(2).

• 5. All common sulfates (SO42-) are soluble, except those of

Ca2+ , Sr2+ , Ba2+ , Ag+ , and Pb2+ .

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 Ca2+).

• 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 Group1A(1), Group 2A(2), and NH4+

• a) Lithium nitrate, an ionic compound, would be expected to be soluble in water, and the solubility rules confirm this.

• b) Glycine (H2NCH2COOH) is a covalent compound, but it contains polar N–H and O–H bonds. This would make the molecule interact well with polar water molecules, and make it likely that it would be soluble.c) Pentane (C5H12) has no bonds of significant polarity, so it would not be expected to be soluble in the polar solvent water.

• d) Ethylene glycol (HOCH2CH2OH) molecules contain polar O–H bonds, similar to water, so it would be expected to be soluble.

4.11 State whether an aqueous solution of each of the following substances conducts an electric current. Explain your reasoning.

(a) Potassium sulfate (b) Sucrose, C12H22O11

a) Yes; K2SO4 is an ionic compound that is soluble in water, producing K+ and SO42– ions.

b) No; sucrose is neither a salt, an acid, nor a base, so it would be a nonelectrolyte (even though it’s soluble in water).

4.13 How many total moles of ions are released when each of the following samples dissolves completely in water?

(a) 0.734 mol of Na2HPO4 (b) 3.86 g of CuSO4 5H⦁ 2O(c) 8.66x1020 formula units of NiCl2

4.15 How many moles and numbers of ions of each type are present in the following aqueous solutions?

(a) 88 mL of 1.75 M magnesium chloride(b) 321 mL of a solution containing 0.22 g aluminum sulfate/L(c) 1.65 L of a solution containing 8.83 1021 formula units of cesium nitrate per liter

4.17 How many moles of H+ ions are present in the following aqueous solutions?

(a) 1.4 mL of 0.75 M hydrobromic acid(b) 2.47 mL of 1.98 M hydriodic acid(c) 395 mL of 0.270 M nitric acid

• The acids in this problem are all strong acids, so you can assume that all acid molecules dissociate completely to yield H+ ions and associated anions. One mole of HBr, HI, and HNO3 each produce one mole of H+ upon dissociation, so moles H+ = moles acid. Molarity is expressed as moles/L instead of as M.a) Moles H+ = mol HBr = (1.4 mL) (10–3 L/1mL) (0.75 mol/L) = 1.05 x 10–3 = 1.0 x 10–3 mol H+

• b) Moles H+ = mol HI = (2.47 mL) (10–3L/1 mL) (1.98 mol/L) = 4.8906 x 10–3 = 4.89 x 10–3 mol H+ c) Moles H+ = mol HNO3 = (395 mL) (10–3 L/1 mL) (0.270 mol/L) = 0.10665 = 0.107 mol H+

4.19 Water “softeners” remove metal ions such as Ca2+ and Fe3+ by replacing them with enough Na+ ions to maintain the same number of positive charges in the solution. If 1.0 103 L of “hard” water is 0.015 M Ca2 and 0.0010 M Fe3 , how many moles of Na+ are needed to replace these ions?

• The moles of the calcium ions and the iron ions are needed. The moles of each of the ions to be replaced must be multiplied by the charge to get the total moles of charge. Since sodium has a +1 charge the total moles of charge equals the moles of sodium ions.

4.23 The beakers represent the aqueous reaction of AgNO3 and NaCl. Silver ions are gray. What colors are used to represent NO3

-, Na+, and Cl-? Write molecular, total ionic, and net ionicequations for the reaction.

+

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 Hg2+

• 4 . All common fluorides (F-) are soluble, except those of Pb2+ and Group 2A(2).

• 5. All common sulfates (SO42-) are soluble, except those of

Ca2+ , Sr2+ , Ba2+ , Ag+ , and Pb2+ .

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 Ca2+).

• 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 Group1A(1), Group 2A(2), and NH4+

• Plan: Use Table 4.1 to predict the products of this reaction. Ions not involved in the precipitate are spectatorions and are not included in the net ionic equation.

•Solution: Assuming that the left beaker is AgNO3 (because it has gray Ag+ ion) and the right must be NaCl, then the NO3– is blue, the Na+ is brown, and the Cl– is green. (Cl– must be green since it is present with Ag+ in the precipitate in the beaker on the right.)

• Molecular equation: AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

Total ionic equation: Ag+(aq) + NO3–(aq) + Na+(aq) + Cl–(aq) → AgCl(s) + Na+(aq) + NO3–(aq) • Net ionic equation: Ag+(aq) + Cl–(aq) → AgCl(s)

4.27 When each of the following pairs of aqueous solutions is mixed, does a precipitation reaction occur? If so, write balanced molecular, total ionic, and net ionic equations:

(a) Potassium carbonate + barium hydroxide

(b) Aluminum nitrate + sodium phosphate

• a) Barium carbonate (BaCO3) precipitates; the potassium hydroxide is a strong base. • Molecular: K2CO3(aq) + Ba(OH)2(aq) → BaCO3(s) + 2 KOH(aq) • Total ionic: 2 K+(aq) + CO32–(aq) + Ba2+(aq) + 2 OH–(aq) →

BaCO3(s) + 2 K+(aq) + 2 OH–(aq) • Net ionic: Ba2+(aq) + CO32–(aq) → BaCO3(s)

• b) Aluminum phosphate (AlPO4) precipitates; the sodium nitrate is soluble. • Molecular: Al(NO3)3(aq) + Na3PO4(aq) → AlPO4(s) + 3

NaNO3(aq)Total ionic: Al3+(aq) + 3 NO3–(aq) + 3 Na+(aq) + PO43–(aq) → AlPO4(s) + 3 NO3–(aq) + 3 Na+(aq)

4.33 Aluminum sulfate, known as cake alum, has a wide range of uses, from dyeing leather and cloth to purifying sewage. In aqueous solution, it reacts with base to form a white precipitate.

(a) Write balanced total and net ionic equations for its reaction with aqueous NaOH.

(b) What mass of precipitate forms when185.5 mL of 0.533 M NaOH is added to 627 mL of a solutionthat contains 15.8 g of aluminum sulfate per liter?

4.39 Complete the following acid-base reactions with balanced molecular, total ionic, and net ionic equations:

(a) Cesium hydroxide(aq) + nitric acid(aq)(b) Calcium hydroxide(aq) + acetic acid(aq)

• a) Molecular: CsOH(aq) + HNO3(aq) → CsNO3(aq) + H2O(l)

• Total ionic: Cs+(aq) + OH–(aq) + H+(aq) + NO3–(aq) → Cs+(aq) + NO3–(aq) + H2O(l)

• Net ionic: OH–(aq) + H+(aq) → H2O(l)Spectator ions are Cs+ and NO3–.

• b) Molecular: Ca(OH)2(aq) + 2 HC2H3O2(aq) → Ca(C2H3O2)2(aq) + 2 H2O(l)

• Total ionic: Ca2+(aq) + 2 OH–(aq) + 2 HC2H3O2 (aq) → Ca2+(aq) + 2 C2H3O2–(aq) + 2 H2O(l)

• Net ionic: OH–(aq) + HC2H3O2(aq) → C2H3O2–(aq) + H2O(l)Spectator ion is Ca2+.

4.42 If 25.98 mL of a standard 0.1180 M KOH solution reacts with52.50 mL of CH3COOH solution,what is the molarity of the acid solution?

• Plan: Write a balanced equation and use the molar ratios to convert the amount of KOH to the amount of CH3COOH.

• The reaction is: KOH(aq) + CH3COOH(aq) → KCH3COO(aq) + H2O(l)

4.43 If 26.25 mL of a standard 0.1850 M NaOH solution is requiredto neutralize 25.00 mL of H2SO4, what is the molarity of the acid solution?

• The reaction is: 2 NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2 H2O(l)

4.44 An auto mechanic spills 88 mL of 2.6M H2SO4 solution from a rebuilt auto battery. How many milliliters of 1.6 M NaHCO3 must be poured on the spill to react completely with the sulfuric acid?

[Hint: H2O and CO2 are among the products.]

• The reaction is: 2 NaHCO3(aq) + H2SO4(aq) → Na2SO4(aq) + 2 H2O(l) + 2 CO2(g)

4.51 Give the oxidation number of sulfur in the following:

(a) SOCl2 (b) H2S2 (c) H2SO3 (d) Na2S

1. IN GROUP 1A(1), O.N. = +1 in all compounds

2. IN GROUP 2A(2) O.N. = +2 in all cpds3. For HYDROGEN, O.N. = +1 with non-

metals, O.N.= -1 with metals and Boron.

4. FLUORINE: O.N. = -1 in all compounds

5. OXYGEN: O.N. = -1 in peroxides (-O-O-) , O.N. = -2 in all other compounds

6. FOR GROUP 7A(17): O.N. = -1 with metals, nonmetals except O, and other halogens lower in the group.

RULES FOR ASSIGNING OXIDATION NUMBERS:• 1. Elemental form = 0 O.N. (e.g. O2, Na, Fe…)• 2. monoatomic ion, O.N. = CHARGE (Na+ = 1)• 3. in a compound, SUM OF O.N. = 0, in polyatomic ion, SUM = CHARGE.

1. IN GROUP 1A(1), O.N. = +1 in all compounds2. IN GROUP 2A(2) O.N. = +2 in all cpds3. For HYDROGEN, O.N. = +1 with non-metals, O.N.= -1 with metals and Boron.4. FLUORINE: O.N. = -1 in all compounds5. OXYGEN: O.N. = -1 in peroxides (-O-O-) , O.N. = -2 in all other compounds6. FOR GROUP 7A(17): O.N. = -1 with metals, nonmetals except O, and other

halogens lower in the group.

• Consult Table 4.3 for the rules for assigning oxidation numbers.

• a) +4 • b) -1 • c) +4 • d) -2

4.57 Identify the oxidizing and reducing agents in the following:

(a) Sn(s) + 2H+(aq) Sn2+(aq) + H2(g)

(b) 2H+(aq) + H2O2 (aq) + 2Fe2+ (aq) 2Fe3+ (aq) + 2H2O(l)

• LEORA vs. GEROA• Loses Electrons, OXIDIZED, Reducing Agent• Gains Electrons, REDUCED, Oxidizing Agent

• a) Oxidizing agent = H+ Reducing agent = Sn • b) Oxidizing agent = H2O2 Reducing agent = Fe2+

4.59 Identify the oxidizing and reducing agents in the following:

(a) 8H+ (aq) +Cr2O72-(aq) + 3SO3

2-(aq) 2Cr3+(aq) + 3SO4

2-(aq) + 4H2O(l) (b) NO3

-(aq) + 4Zn(s) + 7OH- (aq) + 6H2O(l) 4Zn(OH)4

2-(aq) + NH3(aq)

• a) Oxidizing agent = Cr2O72– Reducing agent = SO32– b) Oxidizing agent = NO3– Reducing agent = Zn

4.66 Balance each of the following redox reactions and classify it as a combination, decomposition, or displacement reaction:

(a) Sb(s) + Cl2(g) SbCl3(s)(b) AsH3(g) As(s) + H2(g) (c) Zn(s) + Fe(NO3)2(aq)

Zn(NO3)2(aq) +Fe(s)

• Combination: X + Y → Z;• decomposition: Z → X + Y• Single displacement: X + YZ → XZ + Y• Double displacement: WX + YZ → WZ + YX

• a) 2 Sb(s) + 3 Cl2(g) → 2 SbCl3(s) COMBINATION• b)2AsH3(g)→2As(s)+3H2(g) DECOMPOSITION

c) Zn(s) + Fe(NO3)2(aq) → Zn(NO3)2(aq) + Fe(s) DISPLACEMENT

4.85 For the following aqueous reactions, complete and balance the molecular equation and write a net ionic equation:

(a) Manganese(II) sulfide + hydrobromic acid(b) Potassium carbonate + strontium nitrate (c) Potassium nitrite + hydrochloric acid(d) Calcium hydroxide + nitric acid(e) Barium acetate + iron(II) sulfate(f) Barium hydroxide + hydrocyanic acid(g) Copper(II) nitrate + hydrosulfuric acid(h) Magnesium hydroxide + chloric acid(i) Potassium chloride + ammonium phosphate