Silberberg7e solution manual ch 04

Solution: The balanced equation is: 3CaCl2aq + 2Na3PO4aq  Ca3PO42s + 6NaClaq Finding the volume L of Na3PO4 needed to react with the CaCl2: = 0.1346 = 0.135 L Na 3 PO 4 4.10B Plan: We

CHAPTER 4 THREE MAJOR CLASSES OF

CHEMICAL REACTIONS

FOLLOW–UP PROBLEMS

4.1A Plan: Examine each compound to see what ions, and how many of each, result when the compound is dissolved

in water and match one compound’s ions to those in the beaker Use the total moles of particles and the molar ratio in the compound’s formula to find moles and then mass of compound

Solution:

a) LiBr(s) H O 2  Li+(aq) + Br–(aq)

Cs2CO3(s) H O 2  2Cs+(aq) + CO32–(aq)

BaCl2(s) H O 2  Ba2+(aq) + 2Cl–(aq)

Since the beaker contains +2 ions and twice as many –1 ions, the electrolyte is BaCl 2

4.1B Plan: Write the formula for sodium phosphate and then write a balanced equation showing the ions that result

when sodium phosphate is placed in water Use the balanced equation to determine the number of ions that result when 2 formula units of sodium phosphate are placed in water The molar ratio from the balanced equation gives the relationship between the moles of sodium phosphate and the moles of ions produced Use this molar ratio to calculate the moles of ions produced when 0.40 mol of sodium phosphate is placed in water

Solution:

a) The formula for sodium phosphate is Na3PO4 When the compound is placed in water, 4 ions are produced for each formula unit of sodium phosphate: three sodium ions, Na+, and 1 phosphate ion, PO43–

Na3PO4(s) H O 2  3Na+(aq) + PO43–(aq)

If two formula units of sodium phosphate are placed in water, twice as many ions should be produced:

2Na3PO4(s) H O 2  6Na+(aq) + 2PO43–(aq)

Any drawing should include 2 phosphate ions (each with a 3– charge) and 6 sodium ions (each with a 1+ charge)

b) Moles of ions = 0.40 mol Na3PO4

4 mol ions

1 mol Na3PO4 = 1.6 moles of ions

4.2A Plan: Write an equation showing the dissociation of one mole of compound into its ions Use the given

information to find the moles of compound; use the molar ratio between moles of compound and moles of ions in the dissociation equation to find moles of ions

Solution:

a) One mole of KClO4 dissociates to form one mole of potassium ions and one mole of perchlorate ions

KClO4(s) H O 2  K+(aq) + ClO4(aq)

Therefore, 2 moles of solid KClO4 produce 2 mol of K + ions and 2 mol of ClO 4 ions

c) (NH4)2CrO4(s) H O 2  2NH4(aq) + CrO42–(aq)

First convert formula units to moles

23

1 mol (NH ) CrO1.88 x 10 FU

a) One mole of Li2CO3 dissociates to form two moles of lithium ions and one mole of carbonate ions

Li2CO3(s) H O 2  2Li+(aq) + CO32–(aq)

Therefore, 4 moles of solid Li2CO3 produce 8 mol of Li + ions and 4 mol of CO 3 2– ions

b) Fe2(SO4)3(s) H O 2  2Fe3+(aq) + 3SO42–(aq)

First convert grams of Fe2(SO4)3 to moles of Fe2(SO4)3 and then use molar ratios to determine the moles of each ion produced

Moles of Fe2(SO4)3= (112 g Fe2(SO4)3) 1 mol Fe2(SO4)3

399.88 g Fe2(SO4)3 = 0.2801 = 0.280 mol Fe2(SO4)3 The dissolution of 0.280 mol Fe2(SO4)3(s) produces 0.560 mol Fe3+ (2 x 0.2801) and 0.840 mol SO 4 2–

(3 x 0.2801)

c) Al(NO3)3(s) H O 2  Al3+(aq) + 3NO3(aq)

First convert formula units of Al(NO3)3 to moles of Al(NO3)3 and then use molar ratios to determine the moles of each ion produced

Moles of Al(NO3)3= (8.09 x 1022 formula units Al(NO3)3) 1 mol Al(NO3)3

6.022 x 1023formula units Al(NO 3 )3

= 0.1343 mol Al(NO3)3 The dissolution of 0.134 mol Al(NO3)3 (s) produces 0.134 mol Al3+ and 0.403 mol NO 3 (3 x 0.1343)

4.3A Plan: Convert the volume from mL to liters Convert the mass to moles by dividing by the molar mass of KI

Divide the moles by the volume in liters to calculate molarity

4.3B Plan: Convert the volume from mL to liters Convert the mass to moles by first converting mg to g and then

dividing by the molar mass of NaNO3 Divide the moles by the volume in liters to calculate molarity

M = 0.00206 mol NaNO3

15.0 mL

1000 mL

1 L = 0.137 M

4.4A Plan: Divide the mass of sucrose by its molar mass to change the grams to moles Divide the moles of sucrose by

the molarity to obtain the volume of solution

4.5A Plan: Multiply the volume and molarity to calculate the number of moles of sodium phosphate in the solution

Write the formula of sodium phosphate Determine the number of each type of ion that is included in each

formula unit Use this information to determine the amount of each type of ion in the described solution

Solution:

Amount (mol) of Na3PO4 = 1.32 L 0.55 mol Na3PO4

1 L = 0.7260 = 0.73 mol Na3PO4

In each formula unit of Na3PO4, there are 3 Na+ ions and 1 PO43– ion

Amount (mol) of Na+ = 0.73 mol Na3PO4

4.5B Plan: Write the formula of aluminum sulfate Determine the number of aluminum ions in each formula unit

Calculate the number of aluminum ions in the sample of aluminum sulfate Convert the volume from mL to L Divide the number of moles of aluminum ion by the volume in L to calculate the molarity of the solution Solution:

In each formula unit of Al2(SO4)3, there are 2 Al3+ ions

Amount (mol) of Al3+ = 1.25 mol Al2(SO4)3

4.6B Plan: Determine the new concentration from the dilution equation (Mconc)(Vconc) = (Mdil)(Vdil) Convert the molarity

(mol/L) to g/mL in two steps (one step is moles to grams, and the other step is L to mL)

4.7B Plan: Count the number of particles in each solution per unit volume Determine the final volume of the solution

Use the dilution equation, (Nconc)(Vconc) = (Ndil)(Vdil), to determine the number of particles that will be present

when 300 mL of solvent is added to the 100 mL of solution represented in circle A (Because M is directly

proportional to the number of particles in a given solution, we can replace the molarity terms in the dilution equation with terms representing the number of particles.)

Solution:

There are 12 particles in circle A, 3 particles in circle B, 4 particles in circle C, and 6 particles in circle D

The concentrated solution (circle A) has a volume of 100 mL 300 mL of solvent are added, so the volume of the diluted solution is 400 mL

Ndil = NconcVconc

V dil = (12 particles)(100.0 mL)

(400 mL) = 3 particles

There are 3 particles in circle B, so circle B represents the diluted solution

4.8A Plan: Determine the ions present in each substance on the reactant side and write new cation-anion combinations

Use Table 4.1 to determine if either of the combinations of ions is not soluble If a precipitate forms there will be

a reaction and chemical equations may be written The molecular equation simply includes the formulas of the substances and balancing In the total ionic equation, all soluble substances are written as separate ions The net ionic equation comes from the total ionic equation by eliminating all substances appearing in identical form (spectator ions) on each side of the reaction arrow

Solution:

a) The resulting ion combinations that are possible are iron(III) phosphate and cesium chloride According to Table 4.1, iron(III) phosphate is insoluble, so a reaction occurs We see that cesium chloride is soluble

Total ionic equation:

Fe3+(aq) + 3Cl–(aq) + 3Cs+(aq) + PO43–(aq)  FePO4(s) + 3Cl–(aq) + 3Cs+(aq)

Net ionic equation:

Fe3+(aq) + PO43–(aq)  FePO4(s)

b) The resulting ion combinations that are possible are sodium nitrate (soluble) and cadmium hydroxide

(insoluble) A reaction occurs

Total ionic equation:

2Na+(aq) + 2OH–(aq) + Cd2+(aq) + 2NO3(aq)  Cd(OH)2(s) + 2Na+(aq) + 2NO3(aq)

Note: The coefficients for Na+ and OH– are necessary to balance the reaction and must be included

Net ionic equation:

Cd2+(aq) + 2OH–(aq)  Cd(OH)2(s)

c) The resulting ion combinations that are possible are magnesium acetate (soluble) and potassium bromide (soluble) No reaction occurs

4.8B Plan: Determine the ions present in each substance on the reactant side and write new cation-anion combinations

Use Table 4.1 to determine if either of the combinations of ions is not soluble If a precipitate forms there will be

a reaction and chemical equations may be written The molecular equation simply includes the formulas of the substances and balancing In the total ionic equation, all soluble substances are written as separate ions The net ionic equation comes from the total ionic equation by eliminating all substances appearing in identical form (spectator ions) on each side of the reaction arrow

Solution:

a) The resulting ion combinations that are possible are silver chloride (insoluble, an exception) and barium nitrate (soluble) A reaction occurs

Total ionic equation:

2Ag+(aq) + 2NO3(aq) + Ba2+(aq) + 2Cl–(aq)  2AgCl(s) + Ba2+(aq) + 2NO3(aq)

Net ionic equation:

Ag+(aq) + Cl–(aq)  AgCl(s)

b) The resulting ion combinations that are possible are ammonium sulfide (soluble) and potassium carbonate (soluble) No reaction occurs

c) The resulting ion combinations that are possible are lead(II) sulfate (insoluble, an exception) and nickel(II) nitrate (soluble) A reaction occurs

Total ionic equation:

Ni2+(aq) + SO42–(aq) + Pb2+(aq) + 2NO3(aq)  PbSO4(s) + Ni2+(aq) + 2NO3(aq)

Net ionic equation:

Pb2+(aq) + SO42–(aq)  PbSO4 (s)

4.9A Plan: Look at the ions (number and charge) produced when each of the given compounds dissolves in water and

find the match to the ions shown in the beaker Once the ions in each beaker are known, write new cation-anion combinations and use Table 4.1 to determine if any of the combination of ions is not soluble If a precipitate forms there will be a reaction and chemical equations may be written The molecular equation simply includes the formulas of the substances and must be balanced In the total ionic equation, all soluble substances are written as separate ions The net ionic equation comes from the total ionic equation by eliminating all substances appearing

in identical form (spectator ions) on each side of the reaction arrow

Solution:

a) Beaker A has four ions with a +2 charge and eight ions with a –1 charge The beaker contains dissolved

Zn(NO 3 ) 2 which dissolves to produce Zn2+ and NO3 ions in a 1:2 ratio The compound PbCl2 also has a +2 ion and –1 ion in a 1:2 ratio but PbCl2 is insoluble so ions would not result from this compound

b) Beaker B has three ions with a +2 charge and six ions with a –1 charge The beaker contains dissolved

Ba(OH) 2 which dissolves to produce Ba2+ and OH– ions in a 1:2 ratio Cd(OH)2 also has a +2 ion and a –1 ion in

a 1:2 ratio but Cd(OH)2 is insoluble so ions would not result from this compound

c) The resulting ion combinations that are possible are zinc hydroxide (insoluble) and barium nitrate (soluble)

The precipitate formed is Zn(OH) 2 The spectator ions are Ba 2+

and NO 3 Balanced molecular equation: Zn(NO3)2(aq) + Ba(OH)2(aq)  Zn(OH)2(s) + Ba(NO3)2(aq)

Total ionic equation:

Zn2+(aq) + 2NO3(aq) + Ba2+(aq) + 2OH– (aq)  Zn(OH)2(s) + Ba2+(aq) + 2NO3 (aq)

Net ionic equation:

Zn2+(aq) + 2OH–(aq)  Zn(OH)2(s)

d) Since there are only six OH– ions and four Zn2+ ions, the OH– is the limiting reactant

Mass of Zn(OH)2= (6 OH– ions) 0.050 mol OH

4.9B Plan: Look at the ions (number and charge) produced when each of the given compounds dissolves in water and

find the match to the ions shown in the beaker Once the ions in each beaker are known, write new cation-anion combinations and use Table 4.1 to determine if any of the combination of ions is not soluble If a precipitate forms there will be a reaction and chemical equations may be written The molecular equation simply includes the formulas of the substances and must be balanced In the total ionic equation, all soluble substances are written as separate ions The net ionic equation comes from the total ionic equation by eliminating all substances appearing

in identical form (spectator ions) on each side of the reaction arrow

Solution:

a) Beaker A has eight ions with a +1 charge and four ions with a –2 charge The beaker contains dissolved

Li 2 CO 3 which dissolves to produce Li+ and CO32–ions in a 2:1 ratio The compound Ag2SO4 also has a +1 ion and –2 ion in a 2:1 ratio but Ag2SO4 is insoluble so ions would not result from this compound

b) Beaker B has three ions with a +2 charge and six ions with a –1 charge The beaker contains dissolved CaCl 2

which dissolves to produce Ca2+ and Cl– ions in a 1:2 ratio Ni(OH)2 also has a +2 ion and a –1 ion in a 1:2 ratio but Cd(OH)2 is insoluble so ions would not result from this compound

c) The resulting ion combinations that are possible are calcium carbonate (insoluble) and lithium chloride

(soluble) The precipitate formed is CaCO 3 The spectator ions are Li +

and Cl – Balanced molecular equation: Li2CO3(aq) + CaCl2(aq)  CaCO3(s) + 2LiCl(aq)

Total ionic equation:

2Li+(aq) + CO32– (aq) + Ca2+(aq) + 2Cl– (aq)  CaCO3(s) + 2Li+(aq) + 2Cl– (aq)

Net ionic equation:

Ca2+(aq) + CO32–(aq)  CaCO3 (s)

d) Since there are only four CO32– ions and three Ca2+ ions, the Ca2+ is the limiting reactant

Mass of CaCO3= (3 Ca2+ ions) 0.20 mol Ca

1 mol CaCO3 = 60.0540 = 60 g CaCO 3

4.10A Plan: We are given the molarity and volume of calcium chloride solution, and we must find the volume of sodium

phosphate solution that will react with this amount of calcium chloride After writing the balanced equation, we find the amount (mol) of calcium chloride from its molarity and volume and use the molar ratio to find the amount (mol) of sodium phosphate required to react with the calcium chloride Finally, we use the molarity of the sodium phosphate solution to convert the amount (mol) of sodium phosphate to volume (L)

Solution:

The balanced equation is: 3CaCl2(aq) + 2Na3PO4(aq)  Ca3(PO4)2(s) + 6NaCl(aq)

Finding the volume (L) of Na3PO4 needed to react with the CaCl2:

= 0.1346 = 0.135 L Na 3 PO 4

4.10B Plan: We are given the mass of silver chloride produced in the reaction of silver nitrate and sodium chloride, and

we must find the molarity of the silver nitrate solution After writing the balanced equation, we find the amount

(mol) of silver nitrate that produces the precipitate by dividing the mass of silver chloride produced by its molar mass and then using the molar ratios from the balanced equation Then we calculate the molarity by dividing the moles of silver nitrate by the volume of the solution (in L)

The balanced equation is: AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)

Amount (mol) of AgNO3 = 0.148 g AgCl 1 mol AgCl

1000 mL

1 L = 2.29 x 10 –2

M

4.11A Plan: Multiply the volume in liters of each solution by its molarity to obtain the moles of each reactant Write a

balanced equation Use molar ratios from the balanced equation to determine the moles of lead(II) chloride that may be produced from each reactant The reactant that generates the smaller number of moles is limiting Change the moles of lead(II) chloride from the limiting reactant to the grams of product using the molar mass of lead(II) chloride

Solution:

(a)The balanced equation is:

Pb(C2H3O2)2(aq) + 2NaCl(aq)  PbCl2(s) + 2NaC2H3O2(aq)

2

2

278.1 g PbCl0.221 mol PbCl

1 mol PbCl

  = 61.4601 = 61.5 g PbCl 2 (b) Ac is used to represent C2H3O2:

Amount (mol) Pb(Ac)2 + 2NaCl → PbCl2 + 2NaAc

Initial 0.402 0.442 0 0

Change – 0.221 – 0.442 +0.221 +0.442

Final 0.181 0 0.221 0.442

4.11B Plan: Write a balanced equation Multiply the volume in liters of each solution by its molarity to obtain the

moles of each reactant Use molar ratios from the balanced equation and the molar mass of iron(III) hydroxide to determine the mass (g) of iron(III) hydroxide that may be produced from each reactant The smaller mass is the amount of product actually made

Solution:

(a)The balanced equation is: FeCl3(aq) + 3NaOH(aq)  3NaCl(aq) + Fe(OH)3(s)

Mass(g) of Fe(OH)3 produced from FeCl3 =

1 mol mol Fe(OH)3 = 4.14 g Fe(OH)3

Mass(g) of Fe(OH)3 produced from NaOH =

1 mol mol Fe(OH)3 = 2.30 g Fe(OH)3

NaOH produces the smaller amount of product, so it is the limiting reagent and 2.30 g of Fe(OH)3 are produced (b) To complete the reaction table, we need the moles of FeCl3 and NaOH that react Remember that NaOH is the limiting reagent and will be consumed in this reaction

Amount (mol) of FeCl3 =

4.12A Plan: Convert the given volume from mL to L and multiply by the molarity (mol/L) to find moles of KOH Each

mole of the strong base KOH will produce one mole of hydroxide ions Finally, multiply the amount (mol) of hydroxide ions by Avogadro’s number to calculate the number of hydroxide ions produced

Solution:

KOH(s) H O 2  K+(aq) + OH–(aq)

No of OH– ions produced = 451 mL NaOH 1 L

4.12B Plan: Convert the given volume from mL to L and multiply by the molarity (mol/L) to find moles of HCl Each

mole of the strong acid HCl will produce one mole of hydrogen ions Finally, multiply the amount (mol) of

hydrogen ions by Avogadro’s number to calculate the number of hydrogen ions produced

4.13A Plan: According to Table 4.2, both reactants are strong and therefore completely dissociate in water Thus, the key

reaction is the formation of water The other product of the reaction is soluble

Solution:

Molecular equation: 2HNO3(aq) + Ca(OH)2(aq)  Ca(NO3)2(aq) + 2H2O(l)

Total ionic equation:

2H+(aq) + 2NO3(aq) + Ca2+(aq) + 2OH–(aq)  Ca2+(aq) + 2NO3(aq) + 2H2O(l)

Net ionic equation: 2H+(aq) + 2OH–(aq)  2H2O(l) which simplifies to H+

(aq) + OH–(aq)  H2O(l)

4.13B Plan: According to Table 4.2, both reactants are strong and therefore completely dissociate in water Thus, the key

reaction is the formation of water The other product of the reaction is soluble

Solution:

Molecular equation: HI(aq) + LiOH(aq)  LiI(aq) + H2O(l)

Total ionic equation: H+(aq) + I–(aq) + Li+(aq) + OH–(aq)  Li+(aq) + I–(aq) + H2O(l)

Net ionic equation: H +

(aq) + OH–(aq)  H2O(l)

4.14A Plan: The reactants are a weak acid and a strong base The acidic species is H+ and a proton is transferred to OH–

from the acid The only spectator ion is the cation of the base

Solution:

Molecular equation:

2HNO2(aq) + Sr(OH)2(aq)  Sr(NO2)2(aq) + 2H2O(l)

Ionic equation:

2HNO2(aq) + Sr2+(aq) + 2OH–(aq)  Sr2+(aq) + 2NO2 (aq) + 2H2O(l)

Net ionic equation:

2HNO2(aq) + 2OH–(aq)  2NO2 (aq) + 2H2O(l) or

HNO 2(aq) + OH–(aq)  NO2 (aq) + H2O(l)

The salt is Sr(NO 2 ) 2 , strontium nitrite, and the spectator ion is Sr 2+

4.14B Plan: The reactants are a strong acid and the salt of a weak base The acidic species is H3O+ and a proton is

transferred to the weak base HCO3 to form H2CO3, which then decomposes to form CO2 and water

Solution:

Molecular equation:

2HBr(aq) + Ca(HCO3)2(aq)  CaBr2(aq) + 2H2CO3(aq)

Ionic equation:

2H3O+(aq) + 2Br–(aq) + Ca2+(aq) + 2HCO3(aq)  2CO2(g) + 4H2O(l) + 2Br–(aq) + Ca2+(aq)

Net ionic equation:

2H3O+(aq) + 2HCO3(aq)  2CO2(g) + 4H2O(l) or H 3 O +(aq) + HCO3(aq)  CO2(g) + 2H2O(l)

The salt is CaBr 2 , calcium bromide

4.15A Plan: Write a balanced equation Determine the moles of HCl by multiplying its molarity by its volume, and,

through the balanced chemical equation and the molar mass of aluminum hydroxide, determine the mass of aluminum hydroxide required for the reaction

Solution:

The balanced equation is: Al(OH)3(s) + 3HCl(aq)  AlCl3(aq) + 3H2O(l)

Mass(g) of Al(OH)3 = 3.4 x 10–2 L HCl 0.10 mol HCl

4.15B Plan: Write a balanced equation Determine the moles of NaOH by multiplying its molarity by its volume, and,

through the balanced chemical equation and the molar mass of acetylsalicylic acid, determine the mass of

acetylsalicylic acid in the tablet

4.16A Plan: Write a balanced chemical equation Determine the moles of HCl by multiplying its molarity by its volume,

and, through the balanced chemical equation, determine the moles of Ba(OH)2 required for the reaction The amount of base in moles is divided by its molarity to find the volume

Solution:

The molarity of the HCl solution is 0.1016 M However, the molar ratio is not 1:1 as in the example problem

According to the balanced equation, the ratio is 2 moles of acid per 1 mole of base:

2HCl(aq) + Ba(OH)2(aq)  BaCl2(aq) + 2H2O(l)

4.16B Plan: Write a balanced chemical equation Determine the moles of H2SO4 by multiplying its molarity by its

volume, and, through the balanced chemical equation, determine the moles of KOH required for the reaction The amount of base in moles is divided by its volume (in L) to find the molarity

Solution:

The balanced equation is: 2KOH(aq) + H2SO4(aq)  K2SO4(aq) + 2H2O(l)

Amount (mol) of KOH = 20.00 mL H2SO4

4.17A Plan: Apply Table 4.4 to the compounds Do not forget that the sum of the O.N.’s (oxidation numbers) for a

compound must sum to zero, and for a polyatomic ion, the sum must equal the charge on the ion

Solution:

a) Sc = +3 O = –2 In most compounds, oxygen has a –2 O.N., so oxygen is often a good starting point If each

oxygen atom has a –2 O.N., then each scandium must have a +3 oxidation state so that the sum of O.N.’s equals zero: 2(+3) + 3(–2) = 0

b) Ga = +3 Cl = –1 In most compounds, chlorine has a –1 O.N., so chlorine is a good starting point If each

chlorine atom has a –1 O.N., then the gallium must have a +3 oxidation state so that the sum of O.N.’s equals zero: 1(+3) + 3(–1) = 0

c) H = +1 P = +5 O = –2 The hydrogen phosphate ion is HPO42– Again, oxygen has a –2 O.N Hydrogen has a +1 O.N because it is combined with nonmetals The sum of the O.N.’s must equal the ionic charge, so the following algebraic equation can be solved for P: 1(+1) + 1(P) + 4(–2) = –2; O.N for P = +5

d) I = +3 F = –1 The formula of iodine trifluoride is IF3 In all compounds, fluorine has a –1 O.N., so fluorine is often a good starting point If each fluorine atom has a –1 O.N., then the iodine must have a +3 oxidation state so that the sum of O.N.’s equal zero: 1(+3) + 3(–1) = 0

4.17B Plan: Apply Table 4.4 to the compounds Do not forget that the sum of the O.N.’s (oxidation numbers) for a

compound must sum to zero, and for a polyatomic ion, the sum must equal the charge on the ion

Solution:

a) K = +1 C = +4 O = –2 In all compounds, potassium has a +1 O.N, and in most compounds, oxygen has a –2

O.N If each potassium has a +1 O.N and each oxygen has a –2 O.N., carbon must have a +4 oxidation state so that the sum of the O.N.’s equals zero: 2(+1) + 1(+4) + 3(–2) = 0

b) N = –3 H = +1 When it is combined with a nonmetal, like N, hydrogen has a +1 O.N If hydrogen has a +1

O.N., nitrogen must have a –3 O.N so the sum of the O.N.’s equals +1, the charge on the polyatomic ion: 1(–3) + 4(+1) = +1

c) Ca = +2 P = –3 Calcium, a group 2A metal, has a +2 O.N in all compounds If calcium has a +2 O.N., the

phosphorus must have a –3 O.N so the sum of the O.N.’s equals zero: 3(+2) + 2(–3) = 0

d) S = +4 Cl = –1 Chlorine, a group 7A nonmetal, has a –1 O.N when it is in combination with any nonmetal

except O or other halogens lower in the group If chlorine has an O.N of –1, the sulfur must have an O.N of +4

so that the sum of the O.N.’s equals zero: 1(+4) + 4(–1) = 0

4.18A Plan: Apply Table 4.4 to determine the oxidation numbers for all the compounds in the reaction Do not forget

that the sum of the O.N.’s (oxidation numbers) for a compound must sum to zero, and for a polyatomic ion, the sum must equal the charge on the ion After determining oxidation numbers for all atoms in the reaction, identify the atoms for which the oxidation numbers have changed from the left hand side of the equation to the right hand side of the equation If the oxidation number of a particular atom increases, that atom has been oxidized, and the compound, element, or ion containing that atom on the reactant side of the equation is the reducing agent If the oxidation number of a particular atom decreases, that atom has been reduced, and the compound, element, or ion containing that atom on the reactant side of the equation is the oxidizing agent

Solution:

a) Oxidation numbers in NCl3: N = +3, Cl = –1

Oxidation numbers in H2O: H = +1, O = –2

Oxidation numbers in NH3: N = –3, H = +1

Oxidation numbers in HOCl: H = +1, O = –2, Cl = +1

The oxidation number of nitrogen decreases from +3 to –3, so N is reduced and NCl 3 is the oxidizing agent The oxidation number of chlorine increases from –1 to +1, so Cl is oxidized and NCl 3 is also the reducing agent

b) Oxidation numbers in AgNO3: N = +1, N = +5, O = –2

Oxidation numbers in NH4I: N = –3, H = +1, I = –1

Oxidation numbers in AgI: Ag = +1, I = –1

Oxidation numbers in NH4NO3: N (in NH4 ) = –3, H = +1, N (in NO3) = +5, O = –2

None of the oxidation numbers change, so this is not an oxidation-reduction reaction

4.18B Plan: Apply Table 4.4 to determine the oxidation numbers for all the compounds in the reaction Do not forget

that the sum of the O.N.’s (oxidation numbers) for a compound must sum to zero, and for a polyatomic ion, the sum must equal the charge on the ion After determining oxidation numbers for all atoms in the reaction, identify the atoms for which the oxidation numbers have changed from the left hand side of the equation to the right hand side of the equation If the oxidation number of a particular atom increases, that atom has been oxidized, and the compound, element, or ion containing that atom on the reactant side of the equation is the reducing agent If the oxidation number of a particular atom decreases, that atom has been reduced, and the compound, element, or ion containing that atom on the reactant side of the equation is the oxidizing agent

4.19A Plan: Write a balanced reaction Find moles of KMnO4 by multiplying its volume and molarity and determine

the moles of Ca2+ in the sample using the molar ratio from the reaction Divide moles of Ca2+ by the volume of the milk sample to obtain molarity In part (b), moles of Ca2+ will be converted to grams

Solution:

2KMnO4(aq) + 5CaC2O4(s) + H2SO4(aq) 

2MnSO4(aq) + K2SO4(aq) + 5CaSO4(s) + 10CO2(g) + 8H2O(l)

Moles of KMnO4 =   10 L3 4.56 x 10 mol KMnO3 4

This concentration is consistent with the typical value in milk

4.19B Plan: Find moles of K2Cr2O7 by multiplying its volume and molarity, and determine the moles of Fe2+ in the

sample using the molar ratio from the reaction provided in the problem statement (remember that the Fe2+ is part

of the FeSO4 compound and that there is one Fe2+ ion for every formula unit of FeSO4) Divide moles of Fe2+ by the volume of the FeSO4 solution to obtain molarity In part (b), convert the moles of Fe2+ to grams Then divide the mass of iron ion by the total mass of the ore and multiply by 100% to find the mass percent of iron in the ore Solution:

a) The balanced equation (provided in the problem statement) is:

6FeSO4(aq) + K2Cr2O7(aq) + 7H2SO4(aq)  3Fe2(SO4)3(aq) + Cr2(SO4)3(aq) + 7H2O(l) + K2SO4(aq)

2.58 g ore (100%) = 70.9302 = 70.9%

4.20A Plan: To classify a reaction, compare the number of reactants used versus the number of products formed Also

examine the changes, if any, in the oxidation numbers Recall the definitions of each type of reaction:

Sulfur changes from 0 to +4 oxidation state; it is oxidized and S 8 is the reducing agent

Fluorine changes from 0 to –1 oxidation state; it is reduced and F 2 is the oxidizing agent

b) Displacement; 2CsI(aq) + Cl2(aq)  2CsCl(aq) + I2(aq)

O.N.: Cs = +1 Cl = 0 Cs = +1 I = 0

I = –1 Cl = –1

Total ionic eqn: 2Cs+(aq) + 2I–(aq) + Cl2(aq)  2Cs+(aq) + 2Cl–(aq) + I2(aq)

Net ionic eqn: 2I–(aq) + Cl2(aq)  2Cl–(aq) + I2(aq)

Iodine changes from –1 to 0 oxidation state; it is oxidized and CsI is the reducing agent Chlorine changes from 0 to –1 oxidation state; it is reduced and Cl 2 is the oxidizing agent

c) Displacement; 3Ni(NO3)2(aq) + 2Cr(s)  2Cr(NO3)3(aq) + 3Ni(s)

O.N.: Ni = +2 Cr = 0 Cr = +3 Ni = 0

N = +5 N = +5

O = –2 O = –2

Total ionic eqn: 3Ni+2(aq) + 6NO3(aq) + 2Cr(s)  2Cr+3(aq) + 6NO3(aq) + 3Ni(s)

Net ionic eqn: 3Ni+2(aq) + 2Cr(s)  2Cr+3(aq) + 3Ni(s)

Nickel changes from +2 to 0 oxidation state; it is reduced and Ni(NO 3 ) 2 is the oxidizing agent

Chromium changes from 0 to +3 oxidation state; it is oxidized and Cr is the reducing agent

4.20B Plan: To classify a reaction, compare the number of reactants used versus the number of products formed Also

examine the changes, if any, in the oxidation numbers Recall the definitions of each type of reaction:

Co(s) + 2H+(aq) + 2Cl–(aq)  Co2+(aq) + 2Cl–(aq) + H2(g) (total ionic equation)

Co(s) + 2H+(aq)  Co2+(aq) + H2(g) (net ionic equation)

Cobalt changes from 0 to +2 oxidation state; it is oxidized and Co is the reducing agent

Hydrogen changes from +1 to 0 oxidation state; it is reduced and HCl is the oxidizing agent

b) Combination; 2CO (g) + O2(g)  2CO2(g)

Water is polar because the distribution of its bonding electrons is unequal, resulting in polar bonds, and the shape

of the molecule (bent) is unsymmetrical

4.2 Plan: Review the discussion on water soluble compounds

Solution:

Ionic and polar covalent compounds are most likely to be soluble in water Because water is polar, the partial charges in its molecules are able to interact with the charges, either ionic or dipole-induced, in other substances 4.3 Plan: Solutions that conduct an electric current contain electrolytes

Solution:

Ions must be present in an aqueous solution for it to conduct an electric current Ions come from ionic compounds

or from other electrolytes such as acids and bases

4.4 Plan: Review the discussion on ionic compounds in water

Solution:

The ions on the surface of the solid attract the water molecules (cations attract the “negative” ends and anions attract the

“positive” ends of the water molecules) The interaction of the solvent with the ions overcomes the attraction of the oppositely charged ions for one another, and they are released into the solution

4.5 Plan: Recall that ionic compounds dissociate into their ions when dissolved in water Examine the charges of the

ions in each scene and the ratio of cations to anions

Solution:

a) CaCl2 dissociates to produce one Ca2+ ion for every two Cl– ions Scene B contains four 2+ ions

and twice that number of 1– ions

b) Li2SO4 dissociates to produce two Li+ ions for every one SO42– ion Scene C contains eight 1+ ions

and half as many 2– ions

c) NH4Br dissociates to produce one NH4+ ion for every one Br– ion Scene A contains equal numbers of 1+

solution Only Scene B has twice as many nitrate ions (red circles) as magnesium ions (blue circles)

4.7 Plan: Review the discussion of ionic compounds in water

The interaction with water depends on the structure of the molecule If the interaction is strong, the substance will

be soluble; otherwise, the substance will not be very soluble Covalent compounds that contain polar

groups interact well with the polar solvent water and therefore dissolve in water Covalent compounds that do not contain polar bonds are not soluble in water

4.9 Plan: Review the discussion of covalent compounds in water

Solution:

Some covalent compounds that contain the hydrogen atom dissociate into ions when dissolved in water These

compounds form acidic solutions in water; three examples are HCl, HNO 3 , and HBr

4.10 Plan: Count the total number of spheres in each box The number in box A divided by the volume change in each

part will give the number we are looking for and allow us to match boxes

Solution:

The number in each box is: A = 12, B = 6, C = 4, and D = 3

a) When the volume is tripled, there should be 12/3 = 4 spheres in a box This is box C

b) When the volume is doubled, there should be 12/2 = 6 spheres in a box This is box B

c) When the volume is quadrupled, there should be 12/4 = 3 spheres in a box This is box D

4.11 Plan: Recall that molarity = moles of solute/volume (L) of solution

Solution:

a) M dil = molarity of the diluted solution Mconc = molarity of the concentrated solution

V dil = volume of the diluted solution Vconc = volume of the concentrated solution

The equation works because the quantity (moles) of solute remains the same when a solution

is diluted; only the amount of solvent changes

M x V = moles; Mdil x Vdil = Mconc x Vconc

molesdil = molesconc

b) Molarity = moles solute

a) Solution B has the highest molarity as it has the largest number of particles, 12, in a volume of 50 mL

b) Solutions A and F both have 8 particles in a volume of 50 mL and thus the same molarity Solutions C, D, and E all have 4 particles in a volume of 50 mL and thus have the same molarity

c) Mixing Solutions A and C results in 8 + 4 = 12 particles in a volume of 100 mL That is a lower molarity than

that of Solution B which has 12 particles in a volume of 50 mL or 24 particles in a volume of 100 mL

d) Adding 50 mL to Solution D would result in 4 particles in a total volume of 100 mL; adding 75 mL to Solution

F would result in 4 particles in a volume of 100 mL The molarity of each solution would be the same

e) Solution A has 8 particles in a volume of 50 mL while Solution E has the equivalent of 4 particles in a volume

of 50 mL The molarity of Solution E is half that of Solution A Therefore half of the volume, 12.5 mL, of

Solution E must be evaporated When 12.5 mL of solvent is evaporated from Solution E, the result will be 2 particles in 12.5 mL or 8 particles in 50 mL as in Solution A

4.13 Plan: Remember that molarity is moles of solute/volume of solution

Solution:

Volumes may not be additive when two different solutions are mixed, so the final volume may be slightly

different from 1000.0 mL The correct method would state, “Take 100.0 mL of the 10.0 M solution and add water

until the total volume is 1000 mL.”

4.14 Plan: Compounds that are soluble in water tend to be ionic compounds or covalent compounds that have polar

bonds Many ionic compounds are soluble in water because the attractive force between the oppositely charged ions in an ionic compound are replaced with an attractive force between the polar water molecule and the ions when the compound is dissolved in water Covalent compounds with polar bonds are often soluble in water since the polar bonds of the covalent compound interact with those in water

Solution:

a) Benzene, a covalent compound, is likely to be insoluble in water because it is nonpolar and water is polar b) Sodium hydroxide (NaOH) is an ionic compound and is therefore likely to be soluble in water

c) Ethanol (CH3CH2OH) will likely be soluble in water because it contains a polar –OH bond like water

d) Potassium acetate (KC2H3O2) is an ionic compound and will likely be soluble in water

4.15 Plan: Compounds that are soluble in water tend to be ionic compounds or covalent compounds that have polar

bonds Many ionic compounds are soluble in water because the attractive force between the oppositely charged ions in an ionic compound are replaced with an attractive force between the polar water molecule and the ions when the compound is dissolved in water Covalent compounds with polar bonds are often soluble in water since the polar bonds of the covalent compound interact with those in water

Solution:

a) Lithium nitrate is an ionic compound and is expected to be soluble in water

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 be expected to be insoluble 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

4.16 Plan: Substances whose aqueous solutions conduct an electric current are electrolytes such as ionic compounds,

acids, and bases

Solution:

a) Cesium bromide, CsBr, is a soluble ionic compound, and a solution of this salt in water contains Cs+ and Br–

ions Its solution conducts an electric current

b) HI is a strong acid that dissociates completely in water Its aqueous solution contains H+ and I– ions, so it

conducts an electric current

4.17 Plan: Substances whose aqueous solutions conduct an electric current are electrolytes such as ionic compounds,

acids, and bases

Solution:

a) Potassium sulfate, K2SO4, is an ionic compound that is soluble in water, producing K+ and SO42– ions Its

solution conducts an electric current

b) Sucrose is neither an ionic compound, an acid, nor a base, so it would be a nonelectrolyte (even though it’s

soluble in water) Its solution does not conduct an electric current

4.18 Plan: To determine the total moles of ions released, write an equation that shows the compound dissociating into

ions with the correct molar ratios Convert mass and formula units to moles of compound and use the molar ratio

to convert moles of compound to moles of ions

1 mol NH Cl

  = 0.64 mol of ions b) Each mole of Ba(OH)2•8H2O forms 1 mole of Ba2+ ions and 2 moles of OH– ions, or a total of 3 moles of ions: Ba(OH)2•8H2O(s) → Ba2+(aq) + 2OH–(aq) The waters of hydration become part of the larger bulk of water

Convert mass to moles using the molar mass

c) Each mole of LiCl produces 2 moles of ions (1 mole of Li+ ions and 1 mole of Cl– ions):

LiCl(s) → Li+(aq) + Cl–(aq) Recall that a mole contains 6.022x1023 entities, so a mole of LiCl contains

6.022x1023 units of LiCl, more easily expressed as formula units

4.19 Plan: To determine the total moles of ions released, write an equation that shows the compound dissociating into

ions with the correct molar ratios Convert mass and formula units to moles of compound and use the molar ratio

to convert moles of compound to moles of ions

Solution:

a) Each mole of Rb2SO4 dissolves in water to form 2 moles of Rb+ ions and 1 mole of SO42– ions, or a total of

3 moles of ions: Rb2SO4(s) → 2Rb+(aq) + SO42–(aq)

Moles of ions =  2 4

3 mol ions0.805 mol Rb SO

1 mol Rb SO

  = 2.415 = 2.42 mol of ions b) Each mole of Ca(NO3)2 forms 1 mole of Ca2+ ions and 2 moles of NO3 ions, or a total of

3 moles of ions: Ca(NO3)2(s) → Ca2+(aq) + 2NO3(aq) Convert mass to moles using molar mass

c) Each mole of Sr(HCO3)2 produces 3 moles of ions (1 mole of Sr2+ ions and 2 moles of HCO3– ions):

Sr(HCO3)2(s) → Sr2+(aq) + 2HCO3(aq) Recall that a mole contains 6.022x1023 entities, so a mole of Sr(HCO3)2contains 6.022x1023 units of Sr(HCO3)2, more easily expressed as formula units

4.20 Plan: To determine the total moles of ions released, write an equation that shows the compound dissociating into

ions with the correct molar ratios Convert mass and formula units to moles of compound and use the molar ratio

to convert moles of compound to moles of ions

Convert mass to moles using the molar mass

c) Each mole of FeCl3 forms 1mole of Fe3+ ions and 3 moles of Cl– ions, or a total of 4 moles of ions:

FeCl3(s)  Fe3+(aq) + 3Cl–(aq) Recall that a mole contains 6.022x1023 entities, so a mole of FeCl3 contains 6.022x1023 units of FeCl3, more easily expressed as formula units

3 3

2.23x10 FU FeCl

1 mol FeCl6.022x10 FU FeCl

4.21 Plan: To determine the total moles of ions released, write an equation that shows the compound dissociating into

ions with the correct molar ratios Convert mass and formula units to moles of compound and use the molar ratio

to convert moles of compound to moles of ions

Solution:

a) Each mole of Na2HPO4 forms 2 moles of Na+ ions and 1 mole of HPO42– ions, or a total of 3 moles of ions:

Na2HPO4(s)  2Na+(aq) + HPO42–(aq)

3 mol ions0.734 mol Na HPO

1 mol Na HPO

  = 2.202 = 2.20 mol of ions b) Each mole of CuSO4•5H2O forms 1 mole of Cu2+ ions and 1 mole of SO42– ions, or a total of 2 moles of ions: CuSO4•5H2O(s)  Cu+2(aq) + SO42–(aq) The waters of hydration become part of the larger bulk of water

Convert mass to moles using the molar mass

c) Each mole of NiCl2 forms 1mole of Ni2+ ions and 2 moles of Cl– ions, or a total of 3 moles of ions:

NiCl2(s)  Ni2+(aq) + 2Cl–(aq) Recall that a mole contains 6.022x1023 entities, so a mole of NiCl2 contains 6.022x1023 units of NiCl2, more easily expressed as formula units

2 2

8.66x10 FU NiCl

1 mol NiCl6.022x10 FU NiCl

Solution:

a) Calculating moles of solute in solution:

Moles of Ca(C2H3O2)2 =  10 L3 0.267 mol Ca(C H O )2 3 2 2

1 mol Ca(C H O )

mmoles/mL

Solution:

a) Converting mass of solute to moles:

Moles of KOH = 8.42 g KOH 1 mol KOH

2

2

6.022 x10 Cu ions119.6 mol Cu ions

4.24 Plan: In all cases, use the known quantities and the definition of molarity moles solute

multiply the volume by the molarity to obtain moles of solute, and convert moles to mass in grams (b) The simplest way will be to convert the milligrams to millimoles Molarity may not only be expressed as moles/L, but also as mmoles/mL (c) Convert the milliliters to liters and find the moles of solute and moles of ions by

multiplying the volume and molarity Use Avogadro’s number to determine the number of ions present

1 mol K SO

  = 4.6519 = 4.65 g K 2 SO 4 b) Calculating mmoles of solute:

169.9 g AgNO

Calculating the molarity:

Molarity of AgNO3 = 0.2707574 mol AgNO3

0.385 mol MnSO

  = 1.08368 = 1.08 L MnSO 4 solution c) Volume (mL) of ATP solution = 1.68 mmol ATP 1mL2

4.26 Plan: To determine the total moles of ions released, write an equation that shows the compound dissociating into

ions with the correct molar ratios Convert the information given to moles of compound and use the molar ratio

to convert moles of compound to moles of ions Avogadro’s number is used to convert moles of ions to numbers

of ions

Solution:

a) Each mole of AlCl3 forms 1mole of Al3+ ions and 3 moles of Cl– ions: AlCl3(s)  Al3+(aq) + 3Cl–(aq)

Molarity and volume must be converted to moles of AlCl3

Moles of AlCl3 =   10 L3 0.45 mol AlCl3

4.27 Plan: To determine the total moles of ions released, write an equation that shows the compound dissociating into

ions with the correct molar ratios Convert the information given to moles of compound and use the molar ratio

to convert moles of compound to moles of ions Avogadro’s number is used to convert moles of ions to numbers

b) Each mole of Al2(SO4)3 forms 2 moles of Al3+ ions and 3 moles of SO42– ions:

Al2(SO4)3(s)  2Al3+(aq) + 3SO42–(aq)

Number of Al3+ ions =   23 3

3

6.022x10 Al4.12812x10 mol Al

3

6.022x10 NO0.024194 mol NO

4.28 Plan: These are dilution problems Dilution problems can be solved by converting to moles and using the new

volume; however, it is much easier to use M1V1 = M2V2 The dilution equation does not require a volume in liters; it only requires that the volume units match In part c), it is necessary to find the moles of sodium ions in each separate solution, add these two mole amounts, and divide by the total volume of the two solutions

4.29 Plan: These are dilution problems Dilution problems can be solved by converting to moles and using the new

volume; however, it is much easier to use M1V1 = M2V2 The dilution equation does not require a volume in liters; it only requires that the volume units match

M

M = 39.8571 = 39.9 mL

4.30 Plan: Use the density of the solution to find the mass of 1 L of solution Volume in liters must be converted to

volume in mL The 70.0% by mass translates to 70.0 g solute/100 g solution and is used to find the mass of HNO3 in 1 L of solution Convert mass of HNO3 to moles to obtain moles/L, molarity

4.31 Plan: Use the molarity of the solution to find the moles of H2SO4 in 1 mL Convert moles of H2SO4 to mass of

H2SO4, divide that mass by the mass of 1 mL of solution, and multiply by 100 for mass percent Use the density

of the solution to find the mass of 1 mL of solution

1 mol H SO

  = 1.79486 g H2SO4Mass of 1 mL of solution = 1 mL 1.84 g

1001.84 g solution = 97.5467 = 97.5% H 2 SO 4 by mass

4.32 Plan: Recall the definition of molarity moles solute

L of solution

M  Convert mass of solute to moles and volume from

mL to liters Divide the moles by the volume

4.33 Plan: The first part of the problem is a simple dilution problem (M1V1 = M2V2) The volume in units of gallons can

be used In part b), convert mass of HCl to moles and use the molarity to find the volume that contains that number of moles

container Add slowly and with mixing 0.90 gal of 11.7 M HCl into the water Dilute to 3.0 gal with water

b) Converting from mass of HCl to moles of HCl:

Volume (mL) of solution = 0.264948 mol HCl 1 L 1 mL3

11.7 mol HCl 10 L

= 22.64513 = 22.6 mL muriatic acid solution

4.34 Plan: Convert the mass of the seawater in kg to g and use the density to convert the mass of the seawater to

volume in L Convert mass of each compound to moles of compound and then use the molar ratio in the

dissociation of the compound to find the moles of each ion The molarity of each ion is the moles of ion divided

by the volume of the seawater To find the total molarity of the alkali metal ions [Group 1A(1)], add the moles of the alkali metal ions and divide by the volume of the seawater Perform the same calculation to find the total molarity of the alkaline earth metal ions [Group 2A(2)] and the anions (the negatively charged ions)

Each mole of NaCl forms 1 mole of Na+ ions and 1 mole of Cl– ions: NaCl(s)  Na+(aq) + Cl–(aq)

Moles of NaCl = 26.5 g NaCl 1 mol NaCl

58.44 g NaCl

  = 0.4534565 mol NaCl Moles of Na+ = 0.4534565 mol NaCl 1 mol Na

Moles of Cl– = 0.4534565 mol NaCl 1 mol Cl

84.01 g NaHCO

Moles of Na+ = 3

3

1 mol Na0.00374955 mol NaHCO

Each mole of NaBrforms 1 mole of Na+ ions and 1 mole of Br– ions: NaBr(s)  Na+(aq) + Br–(aq)

Moles of NaBr = 0.098 g NaBr 1 mol NaBr

102.89 g NaBr

  = 0.0009524735 mol NaBr Moles of Na+ =0.0009524735 mol NaBr 1 mol Na

Br–: 0.0009524735 mol Br–Dividing each of the numbers of moles by the volume (0.97560976 L) and rounding to the proper number of significant figures gives the molarities

M = molL



= 0.054361 = 0.0544 M Mg2+

M SO42– =

2 4

0.0278308 mol SO0.97560976 L

b) The alkali metal cations are Na+ and K+ Add the molarities of the individual ions

0.469612 M Na+ + 0.014437 M K+ = 0.484049 = 0.484 M total for alkali metal cations

c) The alkaline earth metal cations are Mg2+ and Ca2+ Add the molarities of the individual ions

0.054361 M Mg2+ + 0.011083 M Ca2+ = 0.065444 = 0.0654 M total for alkaline earth cations

d) The anions are Cl–, SO42–, HCO3, and Br– Add the molarities of the individual ions

0.55307 M Cl– + 0.028526 M SO42– + 0.003843 M HCO3 + 0.0009763 M Br–

= 0.5864153 = 0.586 M total for anions

4.35 Plan: Use the molarity and volume of the ions to find the moles of each ion Multiply the moles of each ion by

that ion’s charge to find the total moles of charge Since sodium ions have a +1 charge, the total moles of charge equals the moles of sodium ions

Solution:

3 0.015 mol Ca1.0 x10 L

Total moles of charge = 30 mol + 3.0 mol = 33 mol charge

Moles Na+ = 33 mol charge 1 mol Na

4.37 Plan: Write the total ionic and net ionic equations for the reaction given The total ionic equation shows all

soluble ionic substances dissociated into ions The net ionic equation eliminates the spectator ions New

equations may be written by replacing the spectator ions in the given equation by other spectator ions

Solution:

The reaction given has the following total ionic and net ionic equations:

Total ionic equation: Ba2+(aq) + 2NO3(aq) + 2Na+(aq) + CO32–(aq)  BaCO3(s) + 2Na+(aq) + 2NO3(aq)

The spectator ions are underlined and are omitted:

Net ionic equation: Ba2+(aq) + CO32–(aq)  BaCO3(s)

New equations will contain a soluble barium compound and a soluble carbonate compound

The “new” equations are:

Molecular: BaCl2(aq) + K2CO3(aq)  BaCO3(s) + 2KCl(aq)

Total ionic: Ba2+(aq) + 2Cl–(aq) + 2K+(aq) + CO32–(aq)  BaCO3(s) + 2K+(aq) + 2Cl–(aq)

Molecular: BaBr2(aq) + (NH4)2CO3(aq)  BaCO3(s) + 2NH4Br(aq)

Total ionic: Ba2+(aq) + 2Br–(aq) + 2NH4(aq) + CO32–(aq)  BaCO3(s) + 2NH4 (aq) + 2Br–(aq)

4.38 If the electrostatic attraction between the ions is greater than the attraction of the ions for water molecules, the ions

will form a precipitate This is the basis for the solubility rules

4.39 Plan: Write the new cation-anion combinations as the products of the reaction and use the solubility rules to

determine if any of the new combinations are insoluble The spectator ions are the ions that are present in the soluble ionic compound

Solution:

a) Ca(NO3)2(aq) + 2NaCl(aq) → CaCl2(aq) + 2NaNO3(aq)

Since the possible products (CaCl2 and NaNO3) are both soluble, no reaction would take place

b) 2KCl(aq) + Pb(NO3)2(aq) → 2KNO3(aq) + PbCl2(s)

According to the solubility rules, KNO3 is soluble but PbCl2 is insoluble so a precipitation reaction takes place The K+ and NO3 would be spectator ions, because their salt is soluble

4.40 Plan: Use the solubility rules to predict the products of this reaction Ions not involved in the precipitate are

spectator ions and are not included in the net ionic equation

Solution:

Assuming that the left beaker is AgNO3 (because it has gray Ag+ ions) 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)

4.41 Plan: Write the new cation-anion combinations as the products of the reaction and use the solubility rules to

determine if any of the new combinations are insoluble The total ionic equation shows all

soluble ionic substances dissociated into ions The spectator ions are the ions that are present in

the soluble ionic compound The spectator ions are omitted from the net ionic equation

Solution:

a) Molecular: Hg2(NO3)2(aq) + 2KI(aq)  Hg2I2(s) + 2KNO3(aq)

Total ionic: Hg22+(aq) + 2NO3(aq) + 2K+(aq) + 2I–(aq)  Hg2I2(s) + 2K+(aq) + 2NO3(aq)

Net ionic: Hg22+(aq) + 2I–(aq)  Hg2I2(s)

Spectator ions are K+ and NO3

b) Molecular: FeSO4(aq) + Sr(OH)2(aq)  Fe(OH)2(s) + SrSO4(s)

Total ionic: Fe2+(aq) + SO42–(aq) + Sr2+(aq) + 2OH–(aq)  Fe(OH)2(s) + SrSO4(s)

Net ionic: This is the same as the total ionic equation because there are no spectator ions

4.42 Plan: Write the new cation-anion combinations as the products of the reaction and use the solubility rules to

determine if any of the new combinations are insoluble The total ionic equation shows all

soluble ionic substances dissociated into ions The spectator ions are the ions that are present in

the soluble ionic compound The spectator ions are omitted from the net ionic equation

Solution:

a) Molecular: 3CaCl2(aq) + 2Cs3PO4(aq)  Ca3(PO4)2(s) + 6CsCl(aq)

Total ionic: 3Ca2+(aq) + 6Cl–(aq) + 6Cs+(aq) + 2PO43–(aq)  Ca3(PO4)2(s) + 6Cs+(aq) + 6Cl–(aq)

Net ionic: 3Ca2+(aq) + 2PO43–(aq)  Ca3(PO4)2(s)

Spectator ions are Cs+ and Cl–

b) Molecular: Na2S(aq) + ZnSO4(aq)  ZnS(s) + Na2SO4(aq)

Total ionic: 2Na+(aq) + S2–(aq) + Zn2+(aq) + SO42–(aq)  ZnS(s) + 2Na+(aq) + SO42–(aq)

Net ionic: Zn2+(aq) + S2–(aq)  ZnS(s)

Spectator ions are Na+ and SO42–

4.43 Plan: A precipitate forms if reactant ions can form combinations that are insoluble, as determined by the solubility

rules in Table 4.1 Create cation-anion combinations other than the original reactants and determine if they are insoluble Any ions not involved in a precipitate are spectator ions and are omitted from the net ionic equation Solution:

a) NaNO3(aq) + CuSO4(aq)  Na2SO4(aq) + Cu(NO3)2(aq)

No precipitate will form The ions Na+ and SO42– will not form an insoluble salt according to the first solubility rule which states that all common compounds of Group 1A ions are soluble The ions Cu2+ and NO3 will not form

an insoluble salt according to the solubility rule #2: All common nitrates are soluble There is no reaction b) A precipitate will form because silver ions, Ag+, and bromide ions, Br–, will combine to form a solid salt, silver bromide, AgBr The ammonium and nitrate ions do not form a precipitate

Molecular: NH4Br(aq) + AgNO3(aq)  AgBr(s) + NH4NO3(aq)

Total ionic: NH4(aq) + Br–(aq) + Ag+(aq) + NO3(aq)  AgBr(s) + NH4 (aq) + NO3(aq)

Net ionic: Ag+(aq) + Br–(aq)  AgBr(s)

4.44 Plan: A precipitate forms if reactant ions can form combinations that are insoluble, as determined by the solubility

rules in Table 4.1 Create cation-anion combinations other than the original reactants and determine if they are insoluble Any ions not involved in a precipitate are spectator ions and are omitted from the net ionic equation Solution:

a) Barium carbonate (BaCO3) precipitates since the solubility rules state that all common carbonates are insoluble Molecular: K2CO3(aq) + Ba(OH)2(aq)  BaCO3(s) + 2KOH(aq)

Total ionic: 2K+(aq) + CO32–(aq) + Ba2+(aq) + 2OH–(aq)  BaCO3(s) + 2K+(aq) + 2OH–(aq)

Net ionic: Ba2+(aq) + CO32–(aq)  BaCO3(s)

b) Aluminum phosphate (AlPO4) precipitates since most common phosphates are insoluble; the sodium nitrate is soluble

Molecular: Al(NO3)3(aq) + Na3PO4(aq)  AlPO4(s) + 3NaNO3(aq)

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

Net ionic: Al3+(aq) + PO43–(aq)  AlPO4(s)

4.45 Plan: A precipitate forms if reactant ions can form combinations that are insoluble, as determined by the solubility

rules in Table 4.1 Create cation-anion combinations other than the original reactants and determine if they are insoluble Any ions not involved in a precipitate are spectator ions and are omitted from the net ionic equation Solution:

a) New cation-anion combinations are potassium nitrate (KNO3) and iron(III) chloride (FeCl3) The solubility rules state that all common nitrates and chlorides (with some exceptions) are soluble, so no precipitate forms

3KCl(aq) + Fe(NO3)3(aq)  3KNO3(aq) + FeCl3(aq)

b) New cation-anion combinations are ammonium chloride and barium sulfate The solubility rules state that most chlorides are soluble; however, another rule states that sulfate compounds containing barium are insoluble Barium sulfate is a precipitate and its formula is BaSO4

Molecular: (NH4)2SO4(aq) + BaCl2(aq)  BaSO4(s) + 2NH4Cl(aq)

Total ionic: 2NH4(aq) + SO42–(aq) + Ba2+(aq) + 2Cl–(aq)  BaSO4(s) + 2NH4(aq) + 2Cl–(aq)

Net ionic: Ba2+(aq) + SO42–(aq)  BaSO4(s)

4.46 Plan: A precipitate forms if reactant ions can form combinations that are insoluble, as determined by the solubility

rules in Table 4.1 Create cation-anion combinations other than the original reactants and determine if they are insoluble Any ions not involved in a precipitate are spectator ions and are omitted from the net ionic equation Solution:

a) New cation-anion combinations are nickel(II) sulfide (NiS) and sodium sulfate (Na2SO4) The solubility rules state that all compounds of Group 1A(1) like sodium are soluble; another rule states that common sulfide

compounds are insoluble Nickel(II) sulfide is a precipitate

Molecular: Na2S(aq) + NiSO4(aq)  NiS(s) + Na2SO4(aq)

Total ionic: 2Na+(aq) + S–2(aq) + Ni2+(aq) + SO42–(aq)  NiS(s) + 2Na+(aq) + SO42–(aq)

Net ionic: Ni2+(aq) + S2–(aq)  NiS(s)

b) New cation-anion combinations are lead(II) bromide (PbBr2) and potassium nitrate (KNO3) The solubility rules state that all common nitrate compounds are soluble; another rule states that while most common bromide compounds are soluble, lead(II) bromide is insoluble Lead(II) bromide is a precipitate

Molecular: Pb(NO3)2(aq) + 2KBr(aq)  PbBr2(s) + 2KNO3(aq)

Total ionic: Pb2+(aq) + 2NO3(aq) + 2K+(aq) + 2Br–(aq)  PbBr2(s) + 2K+(aq) + 2NO3(aq)

Net ionic: Pb2+(aq) + 2Br–(aq)  PbBr2(s)

4.47 Plan: Write a balanced equation for the chemical reaction described in the problem By applying the solubility

rules to the two possible products (NaNO3 and PbI2), determine that PbI2 is the precipitate By using molar relationships, determine how many moles of Pb(NO3)2 are required to produce 0.628 g of PbI2 The molarity is calculated by dividing moles of Pb(NO3)2 by its volume in liters

Solution:

The reaction is: Pb(NO3)2(aq) + 2NaI(aq)  PbI2(s) + 2NaNO3(aq)

461.0 g PbI 1 mol PbI

Moles of Pb2+ = moles of Pb(NO3)2 = 0.001362256 mol Pb2+

Molarity of Pb2+ = moles Pb2 2 = 0.001362256 mol 1 mL3

4.48 Plan: Write a balanced equation for the chemical reaction described in the problem By applying the solubility

rules to the two possible products (KNO3 and AgCl), determine that AgClis the precipitate By using molar relationships, determine how many moles of AgNO3 are required to produce 0.842 g of AgCl The molarity is calculated by dividing moles of AgNO3 by its volume in liters

Solution:

The reaction is AgNO3(aq) + KCl(aq)  AgCl(s) + KNO3(aq)

Moles of AgNO3 =   1 mol AgCl 1 mol AgNO3

Moles of Ag+ = moles of AgNO3 = 0.0058717 mol Ag+

Molarity of Ag+ = moles Ag = 0.0058717 mol 1 mL3

4.49 Plan: The first step is to write and balance the chemical equation for the reaction Multiply the molarity and

volume of each of the reactants to determine the moles of each To determine which reactant is limiting, calculate the amount of barium sulfate formed from each reactant, assuming an excess of the other reactant The reactant that produces less product is the limiting reagent Use the limiting reagent and the mole ratio from the balanced chemical equation to determine the mass of barium sulfate formed

Solution:

The balanced chemical equation is:

BaCl2(aq) + Na2SO4(aq)  BaSO4(s) + 2NaCl(aq)

Moles of BaCl2 =   10 L3 0.160 mol BaCl2

Finding the moles of BaSO4 from the moles of BaCl2 (if Na2SO4 is in excess):

2

2

1 mol BaSO0.00560 moL BaCl

1 mol BaCl

  = 0.00560 mol BaSO4 Moles of Na2SO4 =   10 L3 0.065 mol Na SO2 4

Finding the moles of BaSO4 from the moles of Na2SO4 (if BaCl2 is in excess):

2 4

2 4

1 mol BaSO0.00377 moL Na SO

1 mol Na SO

  = 0.00377 mol BaSO4Sodium sulfate is the limiting reactant

Converting from moles of BaSO4 to mass:

4

4

233.4 g BaSO0.0377 moL BaSO

1 mol BaSO

  = 0.879918 = 0.88 g BaSO 4 4.50 Plan: The first step is to write and balance the chemical equation for the reaction To determine which reactant is

limiting, calculate the amount of iron(III) sulfide formed from each reactant, assuming an excess of the other reactant The reactant that produces less product is the limiting reagent Multiply the molarity and volume (in L)

of each of the reactants to determine the moles of each Use the mole ratio from the balanced chemical equation and the molar mass of iron(III) sulfide to determine the mass of iron(III) sulfide formed

Solution:

The balanced chemical equation is:

2FeCl3(aq) + 3CaS(aq)  Fe2S3(s) + 3CaCl2(aq)

Finding the mass of Fe2S3 from the molarity and volume of FeCl3 (if CaS is in excess):

Mass of Fe2S3 from FeCl3 =

Mass of Fe2S3 from CaS =

0.870 g of Fe 2 S 3 are formed

4.51 Plan: A precipitate forms if reactant ions can form combinations that are insoluble, as determined by the solubility

rules in Table 4.1 Create cation-anion combinations other than the original reactants and determine if they are insoluble Any ions not involved in a precipitate are spectator ions and are omitted from the net ionic equation Use the molar ratio in the balanced net ionic equation to calculate the mass of product

Solution:

a) The yellow spheres cannot be ClO4 or NO3 as these ions form only soluble compounds So the yellow sphere must be SO42– The only sulfate compounds possible that would be insoluble are Ag2SO4 and PbSO4 The

precipitate has a 1:1 ratio between its ions Ag2SO4 has a 2:1 ratio between its ions Therefore the blue spheres are

Pb2+ and the yellow spheres are SO42– The precipitate is thus PbSO 4

b) The net ionic equation is Pb2+(aq) + SO42– (aq)  PbSO4(s)

4.52 Plan: A precipitate forms if reactant ions can form combinations that are insoluble, as determined by the solubility

rules in Table 4.1 Create cation-anion combinations other than the original reactants and determine if they are insoluble Any ions not involved in a precipitate are spectator ions and are omitted from the net ionic equation Use the molar ratio in the balanced net ionic equation to calculate the mass of product

Solution:

a) There are 9 purple spheres representing cations and 7 green spheres representing anions In the precipitate, there are 8 purple spheres (cations) and 4 green spheres (anions), indicating a 2:1 ratio between cation and anion

in the compound Only Reaction 3 produces a precipitate (Ag2SO4) fitting this description:

Li2SO4(aq) + 2AgNO3(aq)  2LiNO3(aq) + Ag2SO4(s)

Reaction 1 does not produce a precipitate since all common nitrate and chloride compounds are soluble Reaction

2 does not produce a precipitate since all common perchlorate and chloride compounds are soluble Reaction 4 produces a precipitate, PbBr2, but it has a cation:anion ratio of 1:2, instead of 2:1

Total ionic equation for Reaction 3 =

2Li+(aq) + SO42–(aq) + 2Ag+(aq) + 2NO3(aq)  2Li+(aq) + 2NO3(aq) + Ag2SO4(s)

Net ionic equation = 2Ag+(aq) + SO42–(aq)  Ag2SO4(s)

b) There are 4 unreacted spheres of ions

Number of ions = 4 spheres 2.5x10 mol ions3 6.022x10 ions23

2.5x10 mol SO ions 1 mol Ag SO 311.9 g Ag SO

CaCl2 reacted Write the precipitation reaction using M to represent the alkali metal Use the moles of CaCl2reacted and the molar ratio in the balanced equation to find the moles of alkali metal carbonate Divide the mass

of that carbonate by the moles to obtain the molar mass of the carbonate Subtract the mass of CO3 from the molar mass to obtain the molar mass of the alkali metal, which can be used to identify the alkali metal and the formula and name of the compound

Solution:

The reaction is: M2CO3(aq) + CaCl2(aq) → CaCO3(s) + 2MCl(aq)

Since the alkali metal M forms a +1 ion and carbonate is a –2 ion, the formula is M2CO3

Moles of CaCl2 =   10 L3 0.350 mol CaCl2

1 mol CaCl

  = 0.010885 mol M2CO3Molar mass (g/mol) of M2CO3 = 2 3

mass of M CO = 1.50 gmoles of M CO 0.010885 mol = 137.804 g/mol

Molar mass (g/mol) of M2 = molar mass of M2CO3 – molar mass of CO3

= 137.804 g/mol – 60.01 g/mol = 77.79 g/mol

Molar mass (g/mol) of M = 77.79 g/mol/2 = 38.895 g/mol = 38.9 g/mol

This molar mass is closest to that of potassium; the formula of the compound is K 2 CO 3 , potassium carbonate

4.54 Plan: Multiply the molarity of the AgNO3 solution by its volume in liters to determine the number of moles of

AgNO3 reacted Write the precipitation reaction using M to represent the metal in the chloride compound Use the moles of AgNO3 reacted and the molar ratio in the balanced equation to find the moles of metal chloride Divide the mass of the chloride by the moles to obtain the molar mass of the chloride Subtract the mass of Cl2from the molar mass to obtain the molar mass of the metal, which can be used to identify the metal and the formula and name of the compound

Solution:

The reaction is: MCl2(aq) + 2AgNO3(aq) → 2AgCl(s) + M(NO3)2(aq)

Since the metal M is a +2 ion and chloride is a –1 ion, the formula of the metal chloride is MCl2

Moles of AgNO3 =   10 L3 0.515 mol AgNO3

2 mol AgNO

  = 0.005768 mol MCl2Molar mass (g/mol) of MCl2 = 2

2

mass of MCl = 0.750 gmoles of MCl 0.005768 mol = 130.0277 g/mol

Molar mass (g/mol) of M = molar mass of MCl2 – molar mass of Cl2

= 130.0277 g/mol – 70.90 g/mol = 59.1277 = 59.1 g/mol

This molar mass is closest to that of nickel; the compound is NiCl 2 , nickel(II) chloride

4.55 Plan: Write a balanced equation for the reaction Find the moles of AgNO3 by multiplying the molarity and

volume of the AgNO3 solution; use the molar ratio in the balanced equation to find the moles of Cl– present in the 25.00 mL sample Then, convert moles of Cl– into grams, and convert the sample volume into grams using the given density The mass percent of Cl– is found by dividing the mass of Cl– by the mass of the sample volume and multiplying by 100

Solution:

The balanced equation is AgNO3(aq) + Cl–(aq)  AgCl(s) + NO3(aq)

Moles of AgNO3 =   10 L3 0.2970 mol AgNO3

Mass (g) of Cl– =  3

3

1 mol Cl 35.45 g Cl0.01592811 mol AgNO

1 mol AgNO 1 mol Cl

4.56 Plan: Write the reaction between aluminum sulfate and sodium hydroxide and check the solubility rules to

determine the precipitate Spectator ions are omitted from the net ionic equation Find the moles of sodium hydroxide by multiplying its molarity by its volume in liters; find the moles of aluminum sulfate by converting grams per liter to moles per liter and multiplying by the volume of that solution To determine which reactant is limiting, calculate the amount of precipitate formed from each reactant, assuming an excess of the other reactant, using the molar ratio from the balanced equation The smaller amount of precipitate is the answer

Solution:

a) According to the solubility rules, most common sulfate compounds are soluble, but most common hydroxides are insoluble Aluminum hydroxide is the precipitate

Total ionic equation: Al2(SO4)3(aq) + 6NaOH(aq)  3Na2SO4(aq) + 2Al(OH)3(s)

Net ionic equation: Al3+(aq) + 3OH–(aq)  Al(OH)3(s)

NaOH is the limiting reagent

4.57 Plan: Write the chemical reaction between the two reactants Then write the total ionic equation in which all

soluble ionic substances are dissociated into ions Omit spectator ions in the net ionic equation

Solution:

The molecular equation is:

H2SO4(aq) + Sr(OH)2(aq)  SrSO4(s) + 2H2O(l)

The total ionic equation is:

2H+(aq) + SO42–(aq) + Sr2+(aq) + 2OH–(aq)  SrSO4(s) + 2H2O(l)

According to the solubility rules, SrSO4 is insoluble and therefore does not dissociate into ions

Since there are no spectator ions, the total and net ionic equations are the same

4.58 Plan: H+ is used to represent an acid while OH– is used to represent the base In a neutralization

reaction, stoichiometric amounts of an acid and a base react to form water

Solution:

The general equation for a neutralization reaction is H+(aq) + OH–(aq)  H2O(l)

4.59 Plan: Review the section on acid-base reactions

Solution:

a) Any three of HCl, HBr, HI, HNO3, H2SO4, or HClO4

b) Any three of NaOH, KOH, Ca(OH)2, Sr(OH)2, Ba(OH)2

c) Strong acids and bases dissociate 100% into ions in aqueous solution

4.60 Plan: Review the section on acid-base reactions

Solution:

a) There are many possibilities including: acetic acid (HC2H3O2), chlorous acid (HClO2), and nitrous acid

(HNO2) All acids are weak except for the six strong acids listed in the text

c) Strong acids and bases dissociate 100% into ions and are therefore strong electrolytes; weak acids and bases dissociate much less than this (typically less than 10%) in aqueous solution and are therefore weak electrolytes The electrical conductivity of a solution of a strong acid or base would be much higher than that of a weak acid or base

of equal concentration

4.61 Plan: Formation of a gaseous product or a precipitate drives a reaction to completion

Solution:

a) The formation of a gas, SO2(g), and formation of water drive this reaction to completion, because both products

remove reactants from solution

b) The formation of a precipitate, Ba3(PO4)2(s), will cause this reaction to go to completion This reaction is one

between an acid and a base, so the formation of water molecules through the combination of H+ and OH– ions also drives the reaction

4.62 Plan: Since strong acids and bases dissociate completely in water, these substances can be written as ions in a

total ionic equation; since weak acids and bases dissociate into ions only to a small extent, these substances appear undissociated in total ionic equations

Solution:

a) Acetic acid is a weak acid and sodium hydroxide is a strong base:

Molecular equation: CH3COOH(aq) + NaOH(aq)  CH3COONa(aq) + H2O(l)

Total ionic equation: CH3COOH (aq) + Na+(aq) + OH–(aq)  Na+(aq) + CH3COO–(aq) + H2O(l)

Net ionic equation (remove the spectator ion Na+): CH3COOH(aq) + OH–(aq)  CH3COO–(aq) + H2O(l)

Hydrochloric acid is a strong acid:

Molecular equation: HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)

Total ionic equation: H+(aq) + Cl–(aq) + Na+(aq) + OH–(aq)  Na+(aq) + Cl–(aq) + H2O(l)

Net ionic equation (remove the spectator ions Na+ and Cl–): H+(aq) + OH–(aq)  H2O(l)

The difference in the net ionic equation is due to the fact that CH3COOH is a weak acid and dissociates very little while HCl is a strong acid and dissociates completely

b) When acetic acid dissociates in water, most of the species in the solution is un-ionized acid, CH3COOH(aq);

the amounts of its ions, H+ and CH3COO– , are equal but very small: [CH 3 COOH] >> [H + ] = [CH 3 COO–]

4.63 Plan: 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 HClO4, HNO3, and HCl each produce one mole of

H+ upon dissociation, so moles H+ = moles acid Calculate the moles of acid by multiplying the molarity (moles/L) by the volume in liters

Solution:

a) HClO4(aq) → H+(aq) + ClO4(aq)

Moles H+ = mol HClO4 =   

mol0.25L40

b) HNO3(aq) → H+(aq) + NO3(aq)

Moles H+ = mol HNO3 = 6.8 mL 10 L3 0.92 mol

c) HCl(aq) → H+(aq) + Cl–(aq)

Moles H+ = mol HCl = 2.6 L 0.085 mol