79 Practice Exercises - 333546_SOLUTION MANUAL BRADY CHEMISTRY 6TH EDITION.pdf

5.1 (a) FeCl3(s) Fe³⁺(aq) + 3Cl^–(aq) (b) K3PO4(s) 3K⁺(aq) + PO43–

(aq) 5.2 (a) MgCl2(s) Mg²⁺(aq) + 2Cl^–(aq) (b) Al(NO3)3(s) Al³⁺(aq) + 3NO3

–(aq) (c) Na2CO3(s) 2Na⁺(aq) + CO3

2–(aq)

5.3 molecular: (NH4)2SO4(aq) + Ba(NO3)2(aq) BaSO4(s) + 2NH4NO3(aq) ionic: 2NH4+

(aq) + SO42–

(aq) + Ba²⁺(aq) + 2NO3–

(aq) BaSO4(s) + 2NH4+

(aq) + 2NO3–

(aq) net ionic: Ba²⁺(aq) + SO42–

(aq) BaSO4(s)

5.4 molecular: CdCl2(aq) + Na2S(aq) CdS(s) + 2NaCl(aq)

ionic: Cd²⁺(aq) + 2Cl^–(aq) + 2Na⁺(aq) + S^2–(aq) CdS(s) + 2Na⁺(aq) + 2Cl^–(aq) net ionic: Cd²⁺(aq) + S^2–(aq) CdS(s)

5.5 HCHO2(aq) + H2O H3O⁺(aq) + CHO2–

(aq)

5.6 Note that the drawing below is an anion. The negative charge has not been included in the drawing.

CH3CH2CH2CHOOH(l ) + H2O → CH3CH2CH2COO^-(aq) + H3O⁺(aq)

5.7 H3C6H5O7(s) + H2O H3O⁺(aq) + H2C6H5O7–

(aq) H2C6H5O7–

(aq) + H2O H3O⁺(aq) + HC6H5O72–

(aq) HC6H5O72–

(aq) + H2O H3O⁺(aq) + C6H5O73–

(aq)

5.8 (C2H5)3N(aq) + H2O (C2H5)3NH⁺(aq) + OH^–(aq)

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80

5.9 HONH2(aq) + H2O HONH3+

(aq) + OH^–(aq)

5.10 The drawing below is the cation formed and does not include the positive charge of the ion.

CH3CH2NH2(aq) + H2O i

_CH₃_CH₂_NH₃⁺_{(aq) + OH}^–_(aq)

5.11 CH3NH2(aq) + H2O

_CH₃_NH₃⁺_{(aq) + OH}^–_(aq)

5.12 HNO2(aq) + H2O

H3O⁺(aq) + NO2–

(aq)

5.13 Sodium arsenate

5.14 Calcium formate, calcium methanoate

5.15 HF: Hydrofluoric acid, sodium salt = sodium fluoride (NaF) HBr: Hydrobromic acid, sodium salt = sodium bromide (NaBr)

5.16 Iodic acid

5.17 NaHSO3, sodium hydrogen sulfite

5.18 H3PO4(aq) + NaOH(aq) NaH2PO4(aq) + H2O sodium dihydrogen phosphate NaH2PO4(aq) + NaOH(aq) Na2HPO4(aq) + H2O sodium hydrogen phosphate Na2HPO4(aq) + NaOH(aq) Na3PO4(aq) + H2O sodium phosphate

Chapter 5 net ionic: No reaction

5.20 (a) molecular: AgNO3(aq) + NH4Cl(aq) AgCl(s) + NH4NO3(aq)

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82

net ionic: 2HCHO2(aq) + Co(OH)2(s) 2CHO2–

(aq) + Co²⁺(aq) + 2H2O

5.25 (a) Formic acid, a weak acid will form.

molecular: KCHO2(aq) + HCl(aq) KCl(aq) + HCHO2(aq) ionic: K⁺(aq) + CHO2–

(aq) + H⁺(aq) + Cl^–(aq) K⁺(aq) + Cl^–(aq) + HCHO2(aq) net ionic: CHO2–

(aq) + H⁺(aq) HCHO2(aq)

(b) Carbonic acid will form and it will further dissociate to water and carbon dioxide:

CuCO3(s) + 2H⁺(aq) CO2(g) + H2O + Cu²⁺(aq)

molecular: CuCO3(s) + 2HC2H3O2(aq) CO2(g) + H2O + Cu(C2H3O2)2(aq) ionic: CuCO3(s) + 2HC2H3O2(aq) CO2(g) + H2O + Cu²⁺ + 2C2H3O2–

(aq) net ionic: CuCO3(s) + 2HC2H3O2(aq) CO2(g) + H2O + Cu²⁺ + 2C2H3O2

–(aq)

(d) Insoluble nickel hydroxide will precipitate.

Ni²⁺(aq) + 2OH^–(aq) Ni(OH)2(s)

molecular: NiCl2(aq) + 2NaOH(aq) Ni(OH)2(s) + 2NaCl(aq)

ionic: Ni²⁺(aq) + 2Cl^–(aq) + 2Na⁺(aq) + 2OH^–(aq) Ni(OH)2(s) + 2Na⁺(aq) + 2Cl^–(aq) net ionic: Ni²⁺(aq) + 2OH^– (aq) Ni(OH)2(s)

5.26 CuO(s) + 2HNO3(aq) → Cu(NO3)2(aq) + H2O(l) Or

Cu(OH)2(s) + 2HNO3(aq) → Cu(NO3)2(aq) + 2H2O(l)

5.27 You want to use a metathesis reaction that produces CoS, which is insoluble, and a second product that is soluble. You may want the reactants to be soluble.

CoCl2(aq) + Na2S(aq) → CoS(s) + 2NaCl(aq)

5.28 mol HNO3 = (16.9 g HNO3) ³

1 mol HNO 63.02 g HNO

 

 

  = 0.268 mol HNO3

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83

M HNO3 = 0.268 mol HNO³ 1000 mL solution 175 ml solution 1 L solution

 

 

   = 1.53 M HNO3

The amount of HNO3 does not change as the solution is diluted.

5.29 mol NaCl in 1.223 g NaCl = (1.223 g NaCl) 1 mol NaCl

Total mol NaCl = 0.02093 mol NaCl + 0.02500 mol NaCl = 0.04593 mol NaCl

Molarity NaCl = 0.04593 mol NaCl 1000 mL solution 250.0 mL solution 1 L solution

  

  

 

  = 0.1837 M

5.30 mol HCl = 175 mL HCl solution 1 L solution 0.250 mol HCl 1000 mL solution 1 L solution

   

   

    = 0.0438 mol HCl

5.31 Determine the moles of HCl in 1.30 g. This is the number of moles of HCl in the 0.250 M solution so we need to determine what volume that is required to divide the moles by to get 0.250 M.

(1.30 g HCl)

1 mol HCl

^soln

5.32 If we were working with a full liter of this solution, it would contain 0.2 mol of Sr(NO3)2. The molar mass of the salt is 211.62 g mol^–1, so 0.2 mol is slightly more than 40 g. However, we are working with just 50 mL, so the amount of Sr(NO3)2 needed is slightly more than a twentieth of 40 g, or 2 g. The answer, 2.11 g, is close to this, so it makes sense.

g Sr(NO3)2 = 50 mL

( ) ( )

0.0125 mol AgNO 169.9 g AgNO 1 L sol’n

Chapter 5 1000 mL solution 1 L solution

   

   

    = 0.075 mol HCl

mL = (0.075 mol HCl) 1 L solution 1000 mL solution 0.10 mol HCl 1 L solution

   

   

    = 750 mL

To find the number of mL of water to add to the solution subtract the number of mL of the concentrated solution from the total volume:

750 mL solution – 150 mL = 600 mL Add 600 mL of water.

5.36 mol H3PO4 = (45.0 mL KOH) 1 L solution 0.100 mol KOH 1 mol H PO³ ⁴ 1000 mL solution 1 L solution 3 mol KOH

 

1 L solution 1000 mL solution 0.0475 mol H PO 1 L solution

  

Chapter 5

First determine the number of moles of Fe²⁺ present.

mol Fe²⁺ = (60.0 mL Fe²⁺)

0.250 mol FeCl 1 mol Fe 1000 mL solution 1 mol FeCl

 

 

  

  = 1.50 × 10^–2 mol Fe²⁺

Now, determine the amount of KOH needed to react with the Fe²⁺.

mL KOH = (1.50× 10^–2 mol Fe²⁺) 136.14 g CaSO 1 mol CaSO

  5.44 Balanced equation:

H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + 2H2O

Chapter 5

5.1 (a) Solvent – the medium into which something (a solute) is dissolved to make a solution

(b) Solute – Something dissolved in a solvent to make a solution

5.2 (a) Concentrated – a solution that has a large ratio of the amounts of solute to solvent (b) Dilute – a solution in which the ratio of the quantities of solute to solvent is small (c) Saturated – a solution that holds as much solute as it can at a given temperature

(d) Unsaturated – Any solution with a concentration less than that of a saturated solution of the same solute and solvent

(e) Supersaturated – a solution whose concentration of solute exceeds the equilibrium concentration (f) Solubility – the ratio of the quantity of solute to the quantity of solvent in a saturated solution 5.3 Chemical reactions are often carried out using solutions because this allows the reactants to move about

and come in contact with each other. Furthermore, solutions can be made with a high enough concentration to allow the reaction to proceed at a reasonable rate.

5.4 When a sugar crystal is added to

(a) a saturated sugar solution, the sugar crystal will not dissolve.

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87

(b) a supersaturated sugar solution, the sugar crystal will cause the extra sugar in solution to precipitate forming more sugar crystals.

5.5 Precipitate – a solid that separates from a solution usually as the result of a chemical reaction

For a precipitate to form spontaneously in a solution, the equilibrium must be disrupted. A supersaturated solution may form a precipitate spontaneously, or if the temperature changes in the direction that will cause a precipitate to form.

5.6 Electrolytes are soluble, ionic compounds. The following are likely to be electrolytes: CuBr2, iron(II) chloride, and (NH4)2SO4.

The following are molecular compounds and are not likely to be electrolytes: C12H22O11, and CH3OH.

5.7 nonelectrolyte does not have a charge, so it cannot allow ions to move. An ion is hydrated when it is surrounded by water molecules.

5.8 Dissociation – the dissolving of an ionic compound in water such that the individual ions that compose the ionic compound become separated from one another (via hydration), and move about freely in solution, acting more or less independently of one another.

5.9 (a) CaCl2(aq) Ca²⁺(aq) + 2Cl^–(aq) (b) (NH4)2SO4(aq) 2NH4+

(aq) + SO42–

(aq) (c) NaC2H3O2(aq) Na⁺(aq) + C2H3O2–

(aq) (d) Cu(ClO4)2(aq) Cu²⁺(aq) + 2ClO4

-(aq)

5.10 The spectator ions are Na⁺ and Cl^–. The net ionic equation is:

Co²⁺ + 2OH^– Co(OH)2(s)

5.11 There are no counter ions for the Al³⁺ or the OH^- so this must be a net ionic equation.

5.12 In a balanced ionic equation, both the mass and the electrical charge must be balanced. It must have the correct formulas of reactants and products. The product is not correct in the equation.

Co³⁺(aq) + HPO42–

(aq) CoPO4(s) + H⁺(aq)

5.13 Acid – sour taste, turns litmus red, corrode some metals, etc...

Base – bitter taste, turns litmus blue, soapy “feel”, etc...

5.14 If a solution is believed to be basic, red litmus paper should be used so that it would turn blue. The blue litmus paper may not change color if the solution is neutral.

5.15 According to the definition of Arrhenius, an acid gives H⁺ ions in water, and a base gives OH^– ions in water.

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88

5.16 (a) NaOH dissociation (b) HNO3 ionization (c) NH3 ionization (d) H2SO4 ionization 5.17 (a) P4O10 acidic solutions

(b) K2O basic solutions (c) SeO3 acidic solutions (d) Cl2O7 acidic solutions

5.18 Dynamic equilibrium is a condition in which two opposing processes are occurring at equal rates. Acetic acid is not a strong acid, so that it forms an equilibrium between the molecular form, HC2H3O2, and the ionized form, H⁺ and C2H3O2–

5.19 Double arrows are not used for the reaction of a strong acid with water because the reaction is not in equilibrium. These are not reversible reactions, i.e., the reverse reaction has practically no tendency to occur.

5.20 (a) HCN: weak (b) HNO3: strong (c) H2SO3: weak (d) HCl: strong (e) HCHO2: weak (f) HNO2: weak

5.21 (a) C5H5N: weak (b) Ba(OH)2: strong (c) KOH: strong (d) C6H5NH2: weak (e) Cs2O: strong (f) N2O5: acidic solution 5.22

N H H₃C

H H

Chapter 5

89

5.23 The student removed a hydrogen attached to the methyl group, CH3. Hydrogen attached to carbon atoms are not acidic protons and will not be removed in water. The correct structure of the ion is one where the hydrogen attached to the oxygen atom is removed. The structure should be:

5.24 The molecules is diethylamine, as base. In water, the molecule would add an H⁺ to the nitrogen atom. The structure of the resulting ion is:

The balanced chemical equation is given below.

3 2 2 2 3 2 2 2

(CH CH ) NH( ) + H O( ) l l → (CH CH ) NH (

⁺

aq ) + OH (

⁻

aq )

5.25 (a) hydrogen selenide (b) hydroselenic acid 5.26 (a) periodic acid

(b) iodic acid (c) iodous acid (d) hypoiodous acid (e) hydroiodic acid 5.27 (a) IO4–

IO3–

IO2–

IO^– I^–

(b) periodate iodate iodite hypoiodite iodide

5.28 (a) H2CrO4 (b) H2CO3

Chapter 5

90

5.29 (a) sodium bicarbonate or sodium hydrogen carbonate (b) potassium dihydrogen phosphate

5.30 NaH2PO4 Na2HPO4 Na3PO4

5.31 (a) hypochlorous acid sodium hypochlorite NaOCl

(b) iodous acid sodium iodite NaIO2

(d) perchloric acid sodium perchlorate NaClO4

5.32 H3AsO3

5.33 sodium butyrate 5.34 propionic acid

5.35 Formation of a weak electrolyte, water, a gas, or an insoluble solid.

5.36 A metathesis reaction is also called a double replacement reaction.

5.37 Since AgBr is insoluble, the concentrations of Ag⁺ and Br^– in a saturated solution of AgBr are very small.

When solutions of AgNO3 and NaBr are mixed, the concentrations of Ag⁺ and Br^– are momentarily larger than those in a saturated AgBr solution. Since this solution is immediately supersaturated in the moment of mixing, a precipitate of AgBr forms spontaneously.

5.38 3Ca²⁺(aq) + 2PO43–

(aq) Ca3(PO4)2(s) 3Mg²⁺(aq) + 2PO43–

(aq) Mg3(PO4)2(s)

5.39 The substance is an electrolyte that dissolves readily in water:

Na2CO3·10H2O(s) 2Na⁺(aq) + CO32–

(aq) + 10H2O

The carbonate anion then serves to cause the precipitation of calcium cations:

Ca²⁺(aq) + CO3

2–(aq) CaCO3(s)

5.40 HCHO2 will react with the following:

(a) KOH HCHO2 + KOH KCHO2 + H2O (b) MgO 2HCHO2 + MgO Mg(CHO2)2 + 2H2O (c) NH3 HCHO2 + NH3 NH4+

+ CHO2–

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91

5.41 Any solution containing ammonium ion will react with a strong base to yield ammonia. The presence of ammonia is easily detected by its odor.

5.42 (a) HCl(aq) + NaHCO3(aq) NaCl(aq) + H2O + CO2(g) (b) 2HCl(aq) +Na2S(aq) 2NaCl(aq) + H2S(g)

5.43 Molarity is the number of moles of solute per liter of solution, also known as molar concentration.

mmol 1 mol 1000 mL 1000 mol mol

mL 1000 mmol 1 L 1000 L L

       

= =

       

       

5.44 The two conversion factors are:

0.25 mol HCl

5.46 The number of moles of HNO3 in the solution has not changed because none of the original sample was removed. Instead, the concentration has decreased since more water was added.

5.47 The number of moles of CaCl2 is the same in both solutions, but A is 0.10 M CaCl2 and B is 0.20 M CaCl2. The volume of solution A is 50 mL, therefore the volume of B is 25 mL:

mol CaCl2 present = ²

5.48 Qualitative analysis is the use of experimental procedures to determine what elements are present in a substance.

Quantitative analysis determines the percentage composition of a compound or the percentage of a component in a mixture.

Qualitative analysis answers the question, "what is in the sample?" Quantitative analysis answers the question, "how much is in the sample?"

5.49 (a) Buret – a long glass tube fitted with a stopcock, graduated in mL, and used for the controlled, measured addition of a volume of a solution to a receiving flask.

(b) Titration – a procedure for obtaining quantitative information about a reactant by a controlled addition of one substance to another until a signal (usually a color change of an indicator) shows that equivalent quantities have reacted.

(d) End point – that point during a titration when the indicator changes color, the titration is stopped, and the total added volume of the titrant is recorded.

Chapter 5

92

5.50 The indicator provides a visible signal that the solution has changed from an acid to a base.

(a) Phenolphthalein is colorless in acid solution.

(b) Phenolphthalein is pink in base solution.

Review Problems

5.51 (a) ionic: 2NH4+

(aq) + CO32–

(aq) + Mg²⁺(aq) + 2Cl^–(aq) 2NH4+

(aq) + 2Cl^–(aq) + MgCO3(s) net: Mg²⁺(aq) + CO32–

(aq) MgCO3(s)

(b) ionic: Cu²⁺(aq) + 2Cl^–(aq) + 2Na⁺(aq) + 2OH^–(aq) Cu(OH)2(s) + 2Na⁺(aq) + 2Cl^–(aq) net: Cu²⁺(aq) + 2OH^–(aq) Cu(OH)2(s)

(aq) + 6Na⁺(aq) + 2PO43–

(aq) Fe3(PO4)2(s) + 6Na⁺(aq) + 3SO42–

(aq) Net: 3Fe²⁺(aq) + 2PO43–

(aq) Fe3(PO4)2(s) (d) ionic: 2Ag⁺(aq) + 2C2H3O2

–(aq) + Ni²⁺(aq) + 2Cl^–(aq) 2AgCl(s) + Ni²⁺(aq) + 2C2H3O^–(aq) Net: 2Ag⁺(aq) + 2Cl^–(aq) 2AgCl(s)

5.52 (a) ionic: Cu²⁺(aq) + SO42–

(aq) + Ba²⁺(aq) + 2Cl^–(aq) BaSO4(s) + Cu²⁺(aq) + 2Cl^–(aq) net: Ba²⁺(aq) + SO42–

(aq) BaSO4(s) (b) Fe³⁺(aq) + 3NO3

–(aq) + 3Li⁺(aq) + 3OH^–(aq) Fe(OH)3(s) + 3Li⁺(aq) + 3NO3 –(aq) net: Fe³⁺(aq) + 3OH^–(aq) Fe(OH)3(s)

(aq) + 3Ca²⁺(aq) + 6Cl^–(aq) Ca3(PO4)2(s) + 6Na⁺(aq) + 6Cl^–(aq) net: 3Ca²⁺(aq) + 2PO43–

(aq) Ca3(PO4)2(s) (d) 2Na⁺(aq) + S^2–(aq) + 2Ag⁺(aq) + 2C2H3O2

–(aq) 2Na⁺(aq) + 2C2H3O2

–(aq) + Ag2S(s) net: 2Ag⁺(aq) + S^2–(aq) Ag2S(s)

5.53 This is an ionization reaction: HClO4(l) + H2O H3O⁺(aq) + ClO4–

(aq) 5.54 HBr(l) + H2O H3O⁺(aq) + Br^–(aq)

5.55 N2H4(aq) + H2O

_N₂_H₅⁺_{(aq) + OH}^–_(aq)

5.56 C5H5N(aq) + H2O

_C₅_H₅_NH⁺_{(aq) + OH}^–_(aq)

5.57 HNO2(aq) + H2O

_H₃_O⁺_{(aq) + NO}₂^–_(aq)

5.58 HC5H9O2(aq) + H2O

H3O⁺(aq) + C5H9O2–

(aq)

5.59 H2CO3(aq) + H2O

_H₃_O⁺_{(aq) + HCO}₃^–_(aq)

Chapter 5

5.69 The electrical conductivity would decrease as the solution is neutralized because there are half the amount of ions as products of this reaction than there were to start with.

Cu(OH)2(s) + 2H⁺(aq) 2H2O(l) + Cu²⁺(aq)

5.70 The electrical conductivity would increase since HC2C3O2 is a weak acid and is only partially dissociated, and as the NH3 is added two ions are formed, increasing the concentration of ions in solution.

NH3(aq) + HC2H3O2(aq) NH4+

(aq) + C2H3O2–

(aq)

Once the point of neutralization has been reached, adding more NH3 will not significantly change the amount of electrolytes in solution since it is a weak base.

5.71 (a) 2H⁺(aq) + CO32–

5.73 These reactions have the following "driving forces":

(a) formation of insoluble Cr(OH)3

(b) formation of water, a weak electrolyte 5.74 These reactions have the following "driving forces":

(a) formation of a gas, CO2

Chapter 5

5.79 There are numerous possible answers. One of many possible sets of answers would be:

(a) NaHCO3(aq) + HCl(aq) NaCl(aq) + CO2(g) + H2O (b) FeCl2(aq) + 2NaOH(aq) Fe(OH)2(s) + 2NaCl(aq) (c) Ba(NO3)2(aq) + K2SO3(aq) BaSO3(s) + 2KNO3(aq) (d) 2AgNO3(aq) + Na2S(aq) Ag2S(s) + 2NaNO3(aq) (e) ZnO(s) + 2HCl(aq) ZnCl2(aq) + H2O

5.80 We need to choose a set of reactants that are both soluble and that react to yield only one solid product.

Choose (b). It can form CuCO3, depending on the concentration of (NH4)2CO3. The solution needs to be kept acidic enough to prevent the formation of Cu(OH)2. Choices (a), (c), and (e) all have insoluble reactants, and for (d), the K2CO3 is basic enough to form Cu(OH)2.

M NaOH solution = 0.125 mol NaOH 1000 mL NaOH 250.0 mL NaOH 1 L NaOH

0.126 mol CaCl 1000 mL CaCl 200.0 mL CaCl 1 L CaCl

   

   

    = 0.631 M CaCl2

Chapter 5

0.250 mol FeCl 162.2 g FeCl 1 L

Chapter 5

0.084 mol HNO 1000 mL HNO 450 mL HNO 1 L HNO

Chapter 5

1000 mL solution 2 mol Cl 1 mole NiCl

−

Chapter 5

1 mol NaHCO 84.01 g NaHCO 0.052 mol HCl

1 L AgNO 0.0625 mol AgNO 3 mol AgCl 143.32 AgCl 1000 mL AgNO 1 L AgNO 3 mol AgNO 1 mol AgCl

Chapter 5

1 mol AlCl 1000 mL AlCl 0.250 moles AlCl 1000 mL NaOH soln 1 mol NaOH

 − 

Therefore, the mass of Fe2O3 that remains unreacted is:

(4.72 g – 0.333 g) = 4.39 g Next determine the number of moles of Mg(OH)2 present:

mol Mg(OH)2 = (3.50 g Mg(OH)2) ²

will react with 0.015 mol H2SO4. This produces 0.015 mol of MgSO4(aq) in 30.0 mL of solution and leaves 0.0600 – 0.015 = 0.045 mol Mg(OH)2 unreacted. The concentration of Mg²⁺ is: 0.0300 L soln 1 mol MgSO

 + 

 

 

  

   = 0.50 M Mg²⁺

The number of grams of Mg(OH)2 not dissolved is:

g Mg(OH)2 = (0.045 mol Mg(OH)2) ²

5.111 First, calculate the number of moles HCl based on the titration according to the following equation:

Chapter 5

103

NaOH(aq) + HCl(aq) NaCl(aq) + H2O

mol HCl = (23.25 mL NaOH) 0.105 mol NaOH 1 mol HCl 1000 mL NaOH 1 mol NaOH

  

  

   = 2.44 × 10^–3 mol HCl

Next, determine the concentration of the HCl solution:

2.44 × 10^–3 mol ÷ 0.01975 L = 0.124 M HCl 5.112 (a) The balanced equation for the titration is:

NaOH(aq) + HC2H3O2(aq) NaC2H3O2(aq) + H2O

mol HC2H3O2 = (28.28 mL NaOH) 0.368 mol NaOH 1 mol HC H O² ³ ² 1000 mL NaOH 1 mol NaOH

 

(b) First convert the density of vinegar to a value appropriate for one liter of solution:

1.01 g/mL × 1000 mL/L = 1010 g/L

We know that one liter of this vinegar contains 0.833 mol of acetic acid so we can determine the mass of acetic acid that is present in one liter of this vinegar:

( )

² ³ ²

The % by weight of acetic acid in vinegar solution is then given by the following:

(50.0 g HC2H3O2/L ÷ 1010 g/L) × 100 = 4.95 % acetic acid

This is the mass of acetic acid in one L of solution divided by the total mass of one L of solution, multiplied by 100%.

5.113 Since lactic acid is monoprotic, it reacts with sodium hydroxide on a one to one mole basis:

(a) mol HC3H5O3 = (17.25 mL NaOH) 0.155 mol NaOH 1 mol HC H O³ ⁵ ³ 1000 mL NaOH 1 mol NaOH

 

5.114 Note that ascorbic acid is diprotic.

g H2C6H6O6 = (21.46 mL NaOH) ² ⁶ ⁶ ⁶ ^{2 6} ⁶ ⁶

Chapter 5

The percentage of Pb in the sample can be calculated as 0.7386 g Pb

5.117 The equation for the reaction indicates that the two materials react in equimolar amounts, i.e. the stoichiometry is 1 to 1:

AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)

(a) Because this reaction is 1:1, we can see by inspection that the AgNO3 is the limiting reagent. We know this because the concentration of the AgNO3 is lower than the NaCl. Since we start with equal volumes, there are fewer moles of the AgNO3.

mol AgCl = (25.0 mL AgNO3 soln) ³

3 3

0.375 mol AgNO 1 mol AgCl 1000 mL AgNO soln 1 mol AgNO

  

  

  

= 9.38 × 10^–3 mol AgCl

(b) Assuming that AgCl is essentially insoluble, the concentration of silver ion can be said to be zero since all of the AgNO3 reacted. The number of moles of chloride ion would be reduced by the precipitation of 9.38 × 10^–3 mol AgCl, such that the final number of moles of chloride ion would be:

0.0250 L × 0.460 mol/L – 9.38 × 10^–3 mol = 2.12 × 10^–3 mol Cl^– The final concentration of Cl^– is, therefore:

2.12 × 10^–3 mol ÷ 0.0500 L = 0.0424 M Cl^– All of the original number of moles of NO3–

and of Na⁺ would still be present in solution, and their concentrations would be:

Chapter 5

1000 mL NaCl soln 1 mol NaCl

M Na = = 0.230 M Na

First, determine the initial number of moles of Ca²⁺ ion that are present:

mol Ca²⁺ = (38.0 mL Ca(NO3)2)

Next, determine the initial number of moles of phosphate ion that are present:

mol PO4

Now determine the number of moles of calcium ion that are required to react with this much phosphate ion, and compare the result to the amount of calcium ion that is available:

mol Ca²⁺ = (6.48 × 10^–3mol PO43–

Since there is not this much Ca²⁺ available according to the above calculation, then we can conclude that Ca²⁺ must be the limiting reagent, and that subsequent calculations should be based on the number of moles of it that are present:

g Ca3(PO4)2 = (5.32 × 10^–3 mol Ca²⁺) ³ ^{4 2} ³ ^{4 2}

(b) If we assume that the Ca3(PO4)2 is completely insoluble, then its concentration may be said to be essentially zero. The concentrations of the other ions are determined as follows:

For nitrate:

Chapter 5

For phosphate, we determine the number of moles that react with calcium:

mol PO43– and subtract from the original number of moles that were present:

mol PO43–

= 6.48 × 10^–3 mol PO43–

– 3.55 × 10^–3 mol PO43–

= 2.93 × 10^–3 mol PO43–

This allows a calculation of the final phosphate concentration:

M (b) nonelectrolyte (f) nonelectrolyte (c) strong electrolyte (g) strong electrolyte (d) nonelectrolyte (h) weak electrolyte

5.120 (a) molecular: CaCO3(s) + 2HNO3(aq) Ca(NO3)2(aq) + H2O + CO2(g)

Chapter 5

107

Percentage by weight = 0.1589 g HC H O⁹ ⁷ ⁴

0.250 g sample = 63.56% aspirin in the sample

5.122 (a) 3Ba²⁺(aq) + 2Al³⁺(aq) + 6OH^–(aq) + 3SO42–

(aq) 3BaSO4(s) + 2Al(OH)3(s)

(b) Because we know the amounts of both starting materials this is a limiting reactant problem. So start by assuming that the barium hydroxide is the limiting reactant.

2 2 1 mol BaSO 233.39 g BaSO

2.84 g BaSO 1 mol BaSO 233.39 g BaSO

6.47 g BaSO

Therefore the barium hydroxide is the limiting reactant. Now we can calculate the mass of aluminum hydroxide that is produced.

3 2 2

(c) All of the barium ion and hydroxide ion are reacted so the concentration of each is 0. We started with the following:

Chapter 5

108

Similarly for SO42–

, the concentration of SO42–

remaining

2– represents the amount precipitated as BaSO4.

5.123 Since the number of moles in the final solution must be equal to the number of moles contributed by both solutions, the equation MfVf = MiVi may be used, and the volumes of the final solution must equal the volumes of the two solution combined.

Vf = V1 + V2

Vf = 30.0 mL + V2

(

) (

₂

) ( )

0.25 mol 0.10 mol 0.45 mol

V V 30 mL

1000 mL 1000 mL 1000 mL

     

= +

     

     

(

₂

) (

₂

) ( )

0.25 mol 0.10 mol 0.45 mol

30 mL + V V 30 mL

1000 mL 1000 mL 1000 mL

     

= +

     

     

( ) (

₂

) (

₂

) ( )

0.25 mol 0.25 mol 0.10 mol 0.45 mol

30 mL + V V 30 mL

1000 mL 1000 mL 1000 mL 1000 mL

       

= +

       

       

( ) ( ) (

₂

) (

₂

)

0.45 mol 0.25 mol 0.25 mol 0.10 mol

30 mL - 30 mL V V

1000 mL 1000 mL 1000 mL 1000 mL

       

= −

       

       

multiply through by 1000 mL

(0.45 mol)(30 mL) – (0.25 mol)(30 mL) = (0.25 mol)(V2) – (0.10 mol)(V2)

Chapter 6

109 Practice Exercises

6.1 2Na(s) + O2(g) Na2O2(s)

Oxygen is reduced since it gains electrons.

Sodium is oxidized since it loses electrons.

6.2 2Al(s) + 3Cl2(g) 2AlCl3(aq)

Aluminum is oxidized and is, therefore, the reducing agent.

Chlorine is reduced and is, therefore, the oxidizing agent.

6.3 Fe2O3(s) + 2Al(s) 2Fe(s) + Al2O3(s)

Fe2O3 is reduced and is, therefore, the oxiding agent.

Al is oxidized and is, therefore, the reducing agent.

6.4 ClO2–

: O –2 Cl +3 6.5 (a) Ni +2; Cl –1

(b) Mg +2; Ti +4; O –2 (c) K +1; Cr +6; O –2 (d) H +1; P +5, O –2 (e) V +3; C 0; H +1; O –2 (f) N -3; H +1

6.6 There is a total charge of +8, divided over three atoms, so the average charge is +8/3.

6.7 (a) Mo +3; Cl -1 (b) Mo +4; S -2 (c) Mo +6; O -2, Cl -1 (d) Mo +6; P -3

6.8 First the oxidation numbers of all atoms must be found.

N2O5 + 3NaHCO3 2NaNO3 + 2CO2 + H2O Reactants: Products:

N = +5 N = +5 O = –2 O = –2 Na = +1 Na = +1 C = +4

H = +1 O = –2 C = +4

O = –2

H = +1 O = –2

None of the oxidation numbers change, therefore it is not an oxidation reaction.

KClO3 + 3HNO2 KCl + 3HNO3

Reactants: Products:

K = +1 K = +1

Chapter 6

110 Cl = +5 Cl = –1

O = –2 O = -2 H = +1 H = +1 N = +3 N = +5 O = –2 O = –2

The oxidation numbers for K and Na do not change. However, the oxidation numbers for the chlorine atom decreases. The oxidation numbers for nitrogen increase.

Therefore, KClO3 is reduced and HNO2 is oxidized.

This means KClO3 is the oxidizing agent and HNO2 is the reducing agent.

This reaction is the redox reaction. In the other reaction, the oxidation numbers of the atoms do not change.

6.9 First the oxidation numbers of all atoms must be found.

Cl2 + 2NaClO2 2ClO2 + 2NaCl Reactants: Products:

Cl = 0 Cl = +4 O = –2 Na = +1 Na = +1 Cl = +3 Cl = –1 O = –2

The oxidation numbers for O and Na do not change. However, the oxidation numbers for all chlorine atoms change. There is no simple way to tell which chlorines are reduced and which are oxidized in this reaction.

One analysis would have the Cl in Cl2 end up as the Cl in NaCl, while the Cl in NaClO2 ends up as the Cl in ClO2. In this case Cl2 is reduced and is the oxidizing agent, while NaClO2 is oxidized and is the reducing agent.

6.10 If H2O2 acts as an oxidizing agent, it gets reduced itself in the process. Examining the oxidation numbers:

In document 333546_SOLUTION MANUAL BRADY CHEMISTRY 6TH EDITION.pdf (Page 79-200)