Reactions in Aqueous Solutions

1
Chapter 5

• Reactions in Aqueous Solutions

2
Chapter goals

• Understand the nature of ionic substances
  dissolved in water.
• Recognize common acids and bases and understand
  their behavior in aqueous solution.
• Recognize and write equations for the common
  types of reactions in aqueous solution.
• Recognize common oxidizing and reducing agents
  and identify common oxidation-reduction
  reactions.
• Define and use the molarity in solution
  stoichiometry.

3
Solution

• homogeneous mixture
• can be gas, liquid, or solid
• solvent component present in highest proportion
• exception - water
• solute component(s) in solution other than
  solvent
• We will mostly study aqueous solutions human
  body is 2/3 water.

4
Examples

• mixture of 35 naphthalene
• and 65 benzene
• solvent - benzene
• solute naphthalene
• mixture of 10 ethanol, 40 methanol, and 50
  propanol
• solvent - propanol
• solute - ethanol and methanol

5
Examples

• mixture of 40 ethanol, 40 methanol, and 20
  butanol
• solvent - ethanol/methanol
• mixed solvent
• solute butanol
• mixture of 40 ethanol, 50 propanol, and 10
  water
• solvent - water
• solute - ethanol and propanol

6
Next

• We will focus on compounds that
• produce ions in aqueous solutions.
• They are salts, acids, and bases.

7
Salts

• Salts ionic compounds made of cations other than
  H and anions other than OH- or O2-
• NaCl Na Cl-
• K2SO4 K SO42-
• FeBr3 Fe3 Br-
• Zn3(PO4)2 Zn2 PO43-
• Ca(HCO3) Ca2 HCO3-

8
Electrolyte

• substance that dissolves to produce an
  electrically conducting medium
• form ions in solution (dissociates/ionizes)
• examples
• soluble ionic compounds
• H2O
• KBr(s) ?? K(aq) Br(aq)
• H2O
• Acids, HCl(g) ?? H(aq) Cl(aq)
• bases, NH3 H2O NH4 OH

9
Nonelectrolytes

• do not form ions in solution
• do not form electrically conducting media upon
  dissolution
• Examples molecular compounds (alcohols, sugars
  acetone)
• H2O
• CH3OH(l) ?? CH3OH(aq) N.R.
• Glucose C6H12O6(s) ? C6H12O6(aq) N.R.
• Sucrose C12H22O11(s) ? C12H22O11(aq) N.R.
• N.R. no reaction
• no dissociation/ionization

10
Types of Electrolytes

• Strong dissociate 100
• most ionic compounds (soluble salts), strong
  acids, and strong bases
• H2O
• KBr(s) ?? K(aq) Br(aq)
• HCl(g) ?? H(aq) Cl(aq)
• Weak insoluble salts, weak acids and bases,
  water, and certain gases (e.g. CO2)
• dissociate only slightly in water
• H2O
• HF(g) H(aq) F(aq)

11
Solubility of Ionic compounds in Water
Solubility Rules

• Soluble Compounds
• 1. alkali metal salts (Li, Na, K, Rb, )
  except potassium perchlorate
• 2. ammonium (NH4) salts
• 3. all nitrates(NO3-), chlorates (ClO3-),
  perchlorates (ClO4-), and acetates (C2H3O2-),
  except silver acetate and potassium perchlorate
• 4. all Cl-, Br-, and l- are soluble except for
  Ag, Pb2, and Hg22 salts
• 5. all SO42- are soluble except for Pb2, Sr2,
  and Ba2 salts

12
Solubility of Ionic compounds in Water Rules

• Insoluble or slightly soluble Compounds
• 6. metal oxides (O2-) except those of the alkali
  metals, Ca2, Sr2, and Ba2
• 7. hydroxides (OH-) except those of the alkali
  metals, Ba2, and Sr2. calcium hydroxide is
  slightly soluble
• 8. carbonates, phosphates, sulfides, and sulfites
  except those of the alkali metals and the
  ammonium ion (NH4)
• 9. for salts of Cr2O72-, P3-, CrO42-, C2O42-,
• assume they are insoluble except for IA metals
  and NH4 salts

13
Precipitation ReactionsA Driving Force in
Chemical Reactions

• formation of insoluble solid (precipitate, ppt)
  is a common reaction in aqueous solutions
• reactants are generally water-soluble ionic
  compounds
• once substances dissolve in water they dissociate
  to give the appropriate cations and anions
• if the cation of one compound forms an insoluble
  compound with the anion of another, precipitation
  will occur

14
Precipitation ReactionA Double Replacement
(Metathesis) Reaction

• Both ionic compounds trade partner ions
• __________
• AB(aq) CD(aq) ?? AD(s) CB(aq)
• _______
• AD is an insoluble or slightly soluble salt
• A, B-, C, and D- are ions
• AgNO3(aq) NaCl(aq) ? AgCl(s) NaNO3(aq)
• weak
  electrolyte
• (unionized
  precipitate)

15
Precipitation ReactionA Double Replacement
(Metathesis) ReactionA (solid) precipitate is
formed.
16
Example complete and balance the equation

• (NH4)3PO4(aq) MgSO4(aq) ? MgPO4(s) NH4SO4(aq)
• we will write the right subscripts later
• Using the solubility rules, predict if at least
  one
• product is going to be insoluble in water.
• According to rule 8, MgPO4 (subscripts not
• right) is not soluble in water.
• Ions are Mg2, PO43-, NH4, and SO42-
• subscripts
• (NH4)3PO4(aq) MgSO4(aq) ? Mg3(PO4)2(s)
  (NH4)2SO4(aq)
• balancing
• 2(NH4)3PO4(aq) 3MgSO4(aq) ? Mg3(PO4)2(s)
  3(NH4)2SO4(aq)

17
Example complete and balance the equation

• Na2SO4(aq) BaBr2(aq) ?
• Na2SO4(aq) BaBr2(aq) ??BaSO4(s) NaBr
• Na2SO4 BaBr2 ??BaSO4(s) 2NaBr(aq)
• driving force formation of insoluble
  barium sulfate (precipitate)
• Os(NO3)5(aq) Rb2S(aq) ?
• Os(NO3)5 Rb2S ? Os2S5(s) RbNO3(aq)
• 2 Os(NO3)5 5 Rb2S ? Os2S5(s) 10 RbNO3
• driving force form. of insoluble sulfide
• (precipitate)

18
Net Ionic Equations Spectator Ions

• The equation
• AgNO3(aq) NaCl(aq) ? AgCl(s) NaNO3(aq)
• is not quite correct, because three salts are
• dissociated in ions while AgCl is a precipitate.
• Ag(aq) NO3-(aq) Na(aq) Cl-(aq) ? AgCl(s)
  Na(aq) NO3-(aq)
• before reaction
  after reaction
• Na and NO3- are present on both sides of
  equation, i.e.,
• before and after reaction. They are called
  spectator ions do
• not participate in net reaction can be removed
  from the
• equation, but they remain in the solution.
• Ag(aq) Cl-(aq) ? AgCl(s) is the net ionic
  equation

19
Net Ionic Equations Spectator Ions

• For two previous examples
• 2(NH4)3PO4(aq) 3MgSO4(aq) ? Mg3(PO4)2(s)
  3(NH4)2SO4(aq)
• 6NH4(aq) 2PO43-(aq) 3Mg2(aq) 3SO42-(aq) ?
  Mg3(PO4)2(s)
• before reaction
  6NH4(aq) 3SO42-(aq)
  
• 
  after
  reaction
• 3Mg2(aq) 2PO43-(aq) ? Mg3(PO4)2(s) is the
  net equation
• spectator ions are eliminated from the
  equation
• Na2SO4(aq) BaBr2(aq) ??BaSO4(s) 2NaBr(aq)
• 2Na(aq) SO42-(aq) Ba2(aq) 2Br-(aq) ?
  BaSO4(s) 2Na(aq) 2Br-
• Ba2(aq) SO42-(aq) ? BaSO4(s) net ionic
  equation

20
Net Ionic Equations Spectator Ions

• For the metathesis reaction
• 2 KF Pb(NO3)2 ? PbF2(s) 2 KNO3
• formula unit equation
• spectator ions are eliminated
• 2K 2F Pb2 2 NO3 ? PbF2 2K 2NO3
• ionic equation
• PbF2 is the precipitate
• 2F(aq) Pb2(aq) ? PbF2(s)
• net ionic equation

21
Net Ionic Equations Spectator Ions

• NH4Cl(aq) KNO3(aq) ? NH4NO3(aq) KCl(aq)
• NH4 Cl K NO3 ? NH4 NO3
• 
  K Cl
• all ions are spectators all can be cancelled
• no net ionic equation
• no driving force for reaction
• N.R. (no reaction)

22
Acids and Bases

• Acid
• Arrhenius definition
• substance that ionizes in water to produce H,
  hydrogen ion, and hence increases the
  concentration of this ion
• HCl(aq) ? H(aq) Cl(aq)
• Brønsted-Lowry definition
• substance capable of donating H
• HCl H2O ? H3O Cl(aq)

23
Acids and Bases

• Base
• Arrhenius definition
• substance that increases the concentration of OH
  in aqueous solution
• KOH(aq) ? K(aq) OH(aq)
• NH3 H2O NH4 OH
• Brønsted/Lowry definition
• substance capable of accepting H
• KOH(aq) ? K(aq) OH(aq)
• OH H ? H2O (OH from NaOH accepts H)
• NH3 H2O NH4 OH (NH3 accepts
  H)

24
Water can act as both an acid and a base it is
an amphoteric substance

• HClO4 H2O ? H3O ClO4
• acid base
• (accepts H from HClO4)
• NH3 H2O NH4 OH
• base acid
• (donates H to NH3)

25
Strong Acids

• dissociate 100
• HX, X Cl, Br, I hydroic acid
• HNO3 nitric acid
• HClO3 chloric acid
• HClO4 perchloric acid
• H2SO4 (first proton) sulfuric acid
• H2SO4(aq) ? H(aq) HSO4-(aq)
• week HSO4-(aq) H(aq) SO42-(aq)

26
Weak Acids

• dissociate lt100
• most other acids
• HF hydrofluoric acid
• HCN hydrocyanic acid
• HNO2 nitrous acid
• CH3CO2H acetic acid
• H2CO3 carbonic acid (both protons)
• H3PO4 phosphoric acid (all protons)
• H2SO3 sulfurous acid
• oxalic acid H2C2O4(aq) H(aq) HC2O4-(aq)

27
Strong Bases

• dissociate 100
• alkali metal hydroxides
• LiOH, NaOH, KOH, RbOH
• name lithium hydroxide
• hydroxide of
• Ca Ca(OH)2 calcium hydroxide
• Ba Ba(OH)2
• Sr Sr(OH)2

28
Neutralization Reactions

• acid OH-ctg. base ? salt water
• HF(aq) KOH(aq) ? KF(aq) H2O
• HF(aq) K(aq) OH(aq) ? K(aq) F(aq) H2O
• HF(aq) OH(aq) ? F(aq) H2O net ionic
• spectator ions are eliminated from equation
• HF is a week acid and HCl is a strong acid
• acid non-OH-ctg base ? salt
• HCl(aq) NH3(aq) ? NH4Cl(aq)
• H(aq) Cl(aq) NH3(aq) ? NH4(aq) Cl(aq)
• H(aq) NH3(aq) ? NH4(aq) net ionic equation

29
Formation of a Weak Acid or Base as a Driving
Force

• HNO3(aq) KCN(aq) ? HCN(aq) KNO3(aq)
• H(aq) NO3(aq) K(aq) CN(aq) ? HCN (aq)
• 
  K(aq) NO3(aq)
• H(aq) CN(aq) ? HCN(aq) (a weak acid)
• NH4Cl NaOH(aq) ? NH4OH NaCl(aq)
• NH4(aq) Cl(aq) Na(aq) OH(aq) ? NH4OH
• 
  Na(aq) Cl(aq)
• NH4(aq) OH(aq) ? NH4OH (a weak base)
• NH4OH is NH3 in water, i.e., NH3 H2O

30
When no Weak Electrolytes are Formed

• HNO3(aq) KCl(aq) ? HCl(aq) KNO3(aq)
• H(aq) NO3(aq) K(aq) Cl(aq) ?
• H(aq) Cl(aq) K(aq) NO3(aq)
• There is no net reaction N.R. No driving
  force
• All ions are spectators.
• 
  
• BaCl2(aq) 2NaOH(aq) ? Ba(OH)2(aq) 2NaCl(aq)
• Ba2(aq) 2Cl(aq) 2Na(aq) 2OH(aq) ?
• Ba2(aq) 2OH(aq) 2Na(aq)
  2Cl(aq)
• There is no net reaction N.R. No driving force

31
Gas Forming Reactions (a Driving Force)

• Some of the weak acids and bases that are
• formed at double replacement reactions
• decompose to form a gas and water
• CO2
• Na2CO3(aq) 2HCl(aq) ? H2CO3(aq) 2NaCl(aq)
• H2CO3(aq) ? H2O CO2(g)
• Na2CO3(aq) 2HCl(aq) ? H2O CO2(g) 2NaCl(aq)
• SO2
• Na2SO3(aq) 2HCl(aq) ? H2SO3(aq) 2NaCl(aq)
• H2SO3(aq) ? H2O SO2(g)
• Na2SO3(aq) 2HCl(aq) ? H2O SO2(g) 2NaCl(aq)

32
Redox (Oxidation-Reduction) Reactions

• involve transfer of electron(s)
• oxidation loss of electron(s)
• reduction gain of electron(s)
• some can be identified when an
• uncombined element is a reactant or a product
• eg. Zn(s) Cu2(aq) ? Zn2(aq) Cu(s)
• Zn ? Zn2
• Zn(s) ? Zn2(aq) 2 e, oxidation
• Cu2 ? Cu
• Cu2(aq) 2e ? Cu(s), reduction

33
Single Displacement Reactions

• Zn(s) CuCl2(aq) ? Cu(s) ZnCl2(aq)
• Zn oxidized to Zn2
• Cu2 reduced to Cu
• occurs because zinc is more active than copper
• Cl2(g) CuBr2(aq) ? Br2(l) CuCl2(aq)
• Br oxidized from Br to Br2
• Cl reduced from Cl2 to Cl
• Cl is more active than Br

34
Oxidation Numbers

• also an accounting tool
• very useful
• oxidation numbers of all atoms in substance add
  up to charge on substance
• e.g.
• zero for Al2(SO4)3 and H3PO4
• 1 for NH4
• 2 for Cr2O72

35
Assigning Oxidation Numbers, ON

• ON 0 for all atoms in any substance in most
  elemental form, Na(s), Zn(s), Hg(l) H2(g),
  Cl2(g), I2(s), O2(g), C(s), P4(s), S8(s)
• ON charge for monatomic ions
• ON 1 for F in all compounds
• ON 2 for O in compounds, usually
• exceptions peroxide, O22, ON 1
• superoxide, O2, ON 1/2
• ON 1 for H in compounds, usually
• exception ON 1 in metallic hydrides

36
Assigning Oxidation Numbers, ON

• ON 1 for alkali metals in compounds
• ON 2 for alkaline earth metals in compounds
• ON 3 for Al in compounds
• ON -1 for Cl, Br, and I in binary compounds
  except for those with oxygen

37
Assign ON to Each Atom in the Following Substances

• WCl6
• x 6(1) 0
• 
  x 6 0
• x 6

1
?
38

• ?
• Na2S2O3
• 2
  2x 6 0
• 2x 6 2
• x 2 ON of
  S

2
1
39
2
1

• ?
• Na2S4O8
• 2
  4x 16 0
• 4x 16 2
• 14 7
• x ON of S
• 4 2

40
?
2

• Cr2O72
• 2x 14
  2
• 2x 14 2
• 12
• x 6 ON
  of Cr
• 2

41
1
2

• H2C2O4
• 2
  2x 8 0
• 2x 8 2
• 6
• x 3
  ON of C
• 2

42
?
1

• MoBr5
• x
  5 1
• x 5 1
• x 6 ON of Mo
  6

43
Oxidizing and Reducing Agents

• In every redox reaction there is one reducing
• agent (the one that is oxidized) and one
• oxidizing agent (the one that is reduced)
• ON increases ON
  decreases
• The species is The
  species is
• oxidized
  reduced

7 6 5 4 3 2 1 0 -1 -2 - 3 - 4 - 5 - 6 - 7
44

• Activity (Electromotive) Series for metals

45

• Li
• K
• Ba
• Ca
• Na
• Mg
• Al
• Mn
• Zn
• Cr
• Fe
• Co
• Ni
• Sn
• Pb
• (H2)
• Cu
• Ag

46
Examples

• Complete and balance each of the following
  chemical equations

47
Li K Ba Ca Na Mg Al Mn Zn Cr Fe Co Ni Sn Pb H2 Cu
Hg Ag Pt Au

• A free and chemically active metal displacing
• a less active metal from a compound
• Mg FeCl3 ?
• Mg FeCl3 ? Fe MgCl2
• 3Mg(s) 2FeCl3(aq) ? 2Fe(s) 3MgCl2(aq)
• Mg ? Mg2, oxidized Mg reducing agent
• Fe3 ? Fe, reduced Fe3 oxidizing agent
• Sn CrF3 ? Sn is less reactive than Cr
• Sn CrF3 ? No Reaction
• Pb(s) Au(ClO3)3(aq) ?
• Pb(s) Au(ClO3)3(aq) ? Au(s) Pb(ClO3)2(aq)
• 3Pb(s) 2Au(ClO3)3(aq) ? 2Au(s)
  3Pb(ClO3)2(aq)
• Pb is oxidized Au3 is reduced

48
Li K Ba Ca Na Mg Al Mn Zn Cr Fe Co Ni Sn Pb H2 Cu
Hg Ag Pt Au

• A free and chemically active metal displacing
• a less active metal from a compound
• Zn CrBr3 ?
• Zn CrBr3 ? Cr ZnBr2
• 3Zn(s) 2CrBr3(aq) ? 2Cr(s) 3 ZnBr2(aq)
• Zn oxidized to Zn2 Zn reducing agent
• Cr3 reduced to Cr Cr3 oxidizing agent
• Ag(s) Hg(NO3)2 ? No Reaction
• Ag is less reactive than Hg

49

• A free and chemically active metal displacing
• Hydrogen from acids or water
• Fe HBr ?
• Fe HBr ? H2 FeBr3
• 2Fe 6HBr ? 3H2 2 FeBr3
• Fe oxidized to Fe3 Fe reducing agent
• H reduced to H2 H oxidizing agent
• Cu HBr ? Cu less active than H2
• Cu HBr ? No Reaction

Li K Ba Ca Na Mg Al Mn Zn Cr Fe Co Ni Sn Pb H2 Cu
Hg Ag Pt Au
50

• A free and chemically active metal displacing
• Hydrogen from acids or water
• K(s) H2O(l) ?
• 2K(s) 2H2O(l) ? 2 KOH(aq) H2(g)
• K oxidized to K K reducing agent
• H reduced to H2 H oxidizing agent
• Ag(s) H2O(l) ? Ag less active than H2
• Ag(s) H2O(l) ? No Reaction
• Ni(s) H2SO4(aq) ?
• Ni(s) H2SO4(aq) ? NiSO4(aq) H2(g)
• Ni oxidized to Ni2 Ni reducing agent
• H reduced to H2 H oxidizing agent

Li K Ba Ca Na Mg Al Mn Zn Cr Fe Co Ni Sn Pb H2 Cu
Hg Ag Pt Au
51

• Which one of the following metals could
• be used safely for lining a tank intended
• for storage of sulfuric acid?
• H2SO4
• aluminum
• iron
• chromium
• mercury
• copper
• tin

Li K Ba Ca Na Mg Al Mn Zn Cr Fe Co Ni Sn Pb H2 Cu
Hg Ag Pt Au
52
Nonmetal Activity Series

• F
• Cl
• Br
• I
• same order as in periodic table

53
Complete and Balance

• Cl2 FeBr3 ?
• Cl2(g) FeBr3(aq) ? Br2(l) FeCl3(aq)
• 3Cl2(g) 2FeBr3(aq) ? 3 Br2(l) 2 FeCl3(aq)
• Cl2 oxidizing agent Br reducing agent
• I2(s) NaF(aq) ? I2 is less active than F2
• I2(s) NaF(aq) ? No Reaction
• F2(g) NaCl(aq) ?
• F2(g) 2 NaCl(aq) ? Cl2(g) 2 NaF(aq)
• F2 oxidizing agent Cl reducing agent
• Br2 FeCl3 ? Br less active than Cl
• Br2 FeCl3 ? No Reaction

54
Identifying Oxidizing and Reducing Agents

• 0 3
  0 2
• 3Zn(s) 2CrBr3(aq) ? 2Cr(s) 3 ZnBr2(aq)
• Zn oxidized to Zn2 Zn reducing agent
• Cr3 reduced to Cr Cr3 oxidizing agent
• 0 1
  2 0
• Ni(s) H2SO4(aq) ? NiSO4(aq)) H2(g)
• Ni oxidized to Ni2 Ni reducing agent
• H reduced to H2 H oxidizing agent
• 0 -1
  0 -1
• F2(g) 2 NaCl(aq) ? Cl2(g) 2 NaF(aq)
• F2 reduced Cl oxidized
• F2 oxidizing agent Cl reducing agent

55
Identifying Oxidizing and Reducing Agents

• The device for testing breath for the presence of
  alcohol is based on the following reaction.
  Identify the oxidizing and reducing agents
• ON 1 6
• 3CH3CH2OH(aq) 2Cr2O72(aq) 16H ?
• ethanol orange-red
• 3
• 3CH3CO2H(aq) 4 Cr3(aq)
  11H2O
• acetic acid green
• ethanol oxidized (reducing agent)
• Cr2O72 (dichromate ion) reduced (oxidizing agt.)

56
Balancing redox equations

• ON 6 in acidic media
  (H)
• Cr2O72(aq) Fe2(aq) ?
  Cr3(aq) Fe3(aq)
• Cr2O72 6e- ? 2Cr3 (this the
  reduction)
• Fe2 ? Fe3 e- (this is the
  oxidation)
• Cr2O72 6e- 14H ? 2Cr3 (to have 6
  6, charge)
• Cr2O72 6e- 14H ? 2Cr3 7H2O (to
  balance H O)
• Cr2O72 6e- 14H ? 2Cr3 7H2O to
  equal of
• 6?(Fe2 ? Fe3 e-)
  electrons
• Cr2O72 14H 6Fe2 ? 2Cr3 7H2O
  6Fe3

57
Balancing redox equations

• ON 7 4 in basic
  media (OH-) 4 6
• MnO4(aq) SO32-(aq) ?
  MnO2(s) SO42-(aq)
• MnO4 3e- ? MnO2 (this the
  reduction)
• SO32- ? SO42- 2e- (this is the
  oxidation)
• MnO4 3e- ? MnO2 4OH- (to equal
  charges, -4)
• SO32- 2OH- ? SO42- 2e- (to equal
  charges, -4)
• 2?(MnO4 3e- 2H2O ? MnO2 4OH-) to equal
  H, O,
• 3?(SO32- 2OH- ? SO42- 2e- H2O) and
  electrons
• 2MnO4 H2O 3SO32- ? 2MnO2 2OH-
  3SO42-

58
Measuring Concentrations of Compounds in Solutions

• Concentration Terms

59
Parts Per

• Hundred (percent, )
• weight/weight, (w/w), (most common)
• mass solute
• x 100
• mass solution
• volume/volume, (v/v)
• V solute
• x 100
• V solution

60
Parts Per

• Hundred (percent, )
• weight/volume, (w/v)
• mass solute
• x 100
• V solution (mL)

61
Learning Check

• A solution is prepared by mixing 15.0 g of
  Na2CO3 and 235 g of H2O. The final V of solution
  is 242 mL. Calculate the w/w and w/v
  concentration of the solution.
• g solution 15.0 g Na2CO3 235 g H2O 250. g
• 15.0 g solute
• w/w x 100 6.00 Na2CO3
• 250. g solution
• 15.0 g solute
• w/v x 100 6.20 Na2CO3
• 242 mL solution

62
Molarity, M

• The Molarity, M, usually known as the molar
• concentration of a solute in a solution, is the
• number of moles of solute per liter (1000 mL)
• of solution/ or mmoles per mL of solution.
• To calculate it we need moles of solute
• and V(in liters) of solution (or mmol and mL)
• mol solute mmol solute
• M
• V(L) solution V(mL) solution

63
Calculation of Molarity, M

• What is the molarity of 500. mL NaOH solution if
• it contains 6.00 g NaOH?
• FW (NaOH) 40.0 g/mol (from periodic table)
• How many moles of NaOH?
• 1 mol NaOH
• 6.00 g x 0.150 mol NaOH
• 40.0 g NaOH
• mol solute 0.150 mol
• M 0.300 M NaOH
• V(L) solution 0.500 L
• 
  (0.300 mol in 1 L)

64
Formality, F

• is the same as molarity, but referred to ionic
• compounds in aqueous solution
• FW (formula weights) of solute per L of solution
• 1 FW
• FW g solute x of FW
• g solute
• FW ( of
  formula weights)
• F
• V(L) solution (volume in
  liters)
• Formality Molarity

65
Molality, m

• is the amount (moles) of solute per kg of
• solvent (usually but not necessarily water).
• What is the m of a solution prepared by
• dissolving 25.3 g Na2CO3 in 458 g water?
• 458 g 0.458 kg (after dividing by 1000)
• 1 mol Na2CO3
• 25.3 g Na2CO3 x 0.239 mol Na2CO3
• 106.0 g Na2CO3
• mol solute 0.239 mol
• m 0.522 m Na2CO3
• kg H2O 0.458 kg
  
• 
  (0.522 mol in 1 kg H2O)

66
Mole Fraction, X

• is the amount (mol) of a given component of a
• solution per mol of solution
• Here we need moles of every component
• (solute(s) and solvent) and the total
• e.g. for a solution with n1, n2, n3, mol
• ni
  ni
• mol fraction Xi (no units)
• n1 n2 n3
  nt
• ni moles of component i (1, 2, 3, )
• nt total number of moles
  

67
How many g of NaCl are contained in 250.0 mL of
0.2193 M NaCl solution?

• 250.0 mL ? 1000 0.2500 L
• Now, M as a conversion factor
• 0.2193 mol
• 0.2500 L x 0.0548 mol NaCl
• 1 L soln
• grams out of moles and formula weight (58.44)
• 58.44 g NaCl
• 0.0548 mol x 3.204 g NaCl
• 1 mol NaCl

68
Making Solutions

• Consider the making of 1.00 L of 1.00 M NaCl
  solution.
• need 1.00 mol NaCl or 58.5 g NaCl
• dissolve 58.5 g NaCl in 1.00 L water?
• NO!!
• dissolve 58.5 g NaCl in water and dilute to a
  total volume of 1.000 L

69

• volumetric flask calibrated to contain, tc
• pipet calibrated to deliver, td

70
Example Describe the preparation of 300.0 mL of
0.4281 M silver nitrate solution.

• 300.0 mL ? 1000 0.3000 L AgNO3 FW 169.97
  g/mol
• 0.4281 mol
• 0.3000 L x 0.1284 mol AgNO3
• 1 L soln
• 169.97 g
• 0.1284 mol x 21.82 g AgNO3
• 1 mol AgNO3
• Dissolve 21.82 g AgNO3 in 300.0 mL water?
• NO!!
• Dissolve 21.82 g AgNO3 in water and dilute to
  300.0 mL.

71
How many milliliters of 2.00 M HNO3 contain 24.0
g HNO3?

• HNO3 FW 63.0 g/mol How many moles?
• 1 mol HNO3
• 24.0 g HNO3 x 0.381 mol HNO3
• 63.0 g HNO3
• Now the M with the volume on top to get mL
• 1 L soln 1000
  mL
• 0.381 mol x x 191 mL
• 2.00 mol HNO3 1 L sln

72
How many grams of AlCl3 are needed to prepare 25
mL of a 0.150 M solution?

• 25 mL ? 1000 0.025 L FW (AlCl3) 133.5
  g/mol
• All at once
• V(L) of soln and mol and FW
• M to calculate to calculate
• mol of AlCl3 g of AlCl3
• 0.025 L x 0.150 mole x 133.5 g 0.500 g
  AlCl3
• 1 L 1 mole

73
Dilution

• the process of decreasing the concentration
• of solutes in a solution by addition of solvent
• or another solution that does not contain the
• same solutes ChemNow 5.16 final excercise
• volume increases and concentration decreases.
• concentrated diluted solution

74
Example Calculate the concentration of a
solution made by diluting 25.0 mL of 0.200 M
methanol, CH3OH, solution to 100.0 mL.

• Key for calculations moles of solute taken
• from the concentrated solution are the same
• in the diluted solution (only solvent is added.)
• moles concentration x V M x V
• Cc x Vc Cd x Vd Mc x Vc Md x Vd
• one of the four is unknown
• c concentrated d diluted
• L or mL can be used for volume

75
Example Calculate the concentration of a
solution made by diluting 25.0 mL of 0.200 M
methanol, CH3OH, to 100.0 mL.

• Mc x Vc Md x Vd
• Md is the unknown
• 25.0 mL x 0.200 M 100.0 mL x Md
• 25.0 mL x 0.200 M
• Md 0.0500 M
• 100.0 mL
• 
  0.0500 mole/L

76
How many mL of a 0.515 M NaBr solution must be
diluted to produce 500.0 mL of a 0.103 M NaBr
solution?

• Mc x Vc Md x Vd Vc is the unknown
• Vc x 0.515 M 500.0 mL x 0.103 M
• 500.0 mL x 0.103 M
• Vc 100.0 mL
• 0.515 M

77
Serial Dilutions

• Example Calculate the M of NaOH in a solution
• made by diluting 25.0 mL of 0.928 M NaOH to
• 200.0 mL and then diluting 50.0 mL of the
• second solution to 100.0 mL. Mc x Vc Md x
  Vd
• First dilution
• 25.0 mL x 0.928 M
• Md 0.116 M
• 200.0 mL
• Second dilution
• 50.0 mL x 0.116 M
• Md 0.0580 M
• 100.0 mL

78
Solution Stoichiometry

• Use of M, V, and coefficients in equations to
• calculate any amount of reagent or product.
• Example What volume of 0.273 M potassium
  chloride solution is required to react with
  exactly 0.836 mmol of silver nitrate?
• KCl(aq) AgNO3(aq) ?
• KCl(aq) AgNO3(aq) ? KNO3(aq) AgCl(s)
• driving force, formation of precipitate AgCl
• Key work with mmol of KCl and AgNO3

79
Solution Stoichiometry

• What volume of 0.273 M KCl solution is required
• to react with exactly 0.836 mmol of AgNO3 ?
• KCl(aq) AgNO3(aq) ? KNO3(aq) AgCl(s)
• First, calculate mmol of KCl
• 1 mmol KCl
• 0.836 mmol AgNO3 x 0.836 mmol
• 1 mmol AgNO3
  KCl
• Second, calculate volume of KCl solution
• 1 mL
• 0.836 mmol KCl x 3.06 mL of solution
• 0.273 mmol
  (KCl)

80
Titration
Buret
Titrant
Erlenmeyer (conical) Flask
Titrate
81
Titration

• buret calibrated td, fine tip to deliver small
  volumes accurately, stopcock for flow control
• Erlenmeyer flask sloped walls allow swirling of
  solutions without spilling or splashing
• Titrant solution containing reagent that will
  react with sample in well known manner
• Titrate solution containing the sample

82
Titration

• equivalence point point in a titration at which
  the exact stoichiometric amount of titrant has
  been added to react with the titrate
• end point the point in a titration at which the
  indicator changes, titration stopped here and
  volume of titrant read (ChemNow 5.19 Exercise
  Titration)
• ideally, end point equivalence point
• reality, not so, error introduced, hopefully
  small error
• Four parameters V, M of titrant and V, M of
  titrate. Usually one is unknown.

83
Titration

• Example Calculate the concentration of
• hydrochloric acid in a solution if 35.0 mL of
• it required 28.9 mL of 0.178 M potassium
• hydroxide solution for titration.
• HCl titrated, V known, M unknown
• (in this problem)
• KOH titrant, V and M known

84
Titration
Buret
0.178 M KOH
HCl KOH ? KCl H2O
Erlenmeyer (conical) Flask
35.0 mL sample of HCl Soln, unknown M drops
indicator
85
Titration

• Strategy mmol of KOH are calculated first.
• Second, by using stoichiometry coefficients
• mol of HCl are calculated.
• Finally, M of HCl is calculated with mmol and
  V(mL)
• 0.178 mmol KOH 1 mmol HCl
• 28.9 mLx x 5.14 mmol
• 1 mL 1 mmol
  KOH HCl
• Based on the end point concept (mol ratio)
• 5.14 mmol
• MHCl 0.147 M (mol/L)
• 35.0 mL
  (mmol/mL)

86
Example Calculate the concentration of arsenic
acid in a solution given that a 25.0 mL sample of
that solution required 42.2 mL of 0.274 M
potassium hydroxide for titration.

• H3AsO4 KOH ? K3AsO4 H2O
• H3AsO4 3 KOH ? K3AsO4 3 H2O
• M of H3AsO4 is the unknown.
• 1 mol H3AsO4 reacts with 3 mol KOH. That is
• the mole ratio.
• We can use mL and mmol instead of L and mol

87
Example Calculate the concentration of arsenic
acid in a solution given that a 25.0 mL sample of
that solution required 42.2 mL of 0.274 M
potassium hydroxide for titration.

• H3AsO4 3 KOH ? K3AsO4 3 H2O
• 0.274 mmol KOH 1 mmol H3AsO4
• 42.2 mLx x 3.85 mmol
• 1 mL 3 mmol
  KOH H3AsO4
• 3.85 mmol
• M H3AsO4 0.154 M (mol/L)
• 25.0 mL
  (mmol/mL)

88
Purity Analysis A 1.034 g-sample of impure
oxalic acid is dissolved in water and an
acid-base indicator added. The sample requires
34.47 mL of 0.485 M NaOH solution to reach the
equivalence point. What is the mass of H2C2O4 and
what is its mass percent in the sample?

• H2C2O4 2 NaOH ? Na2C2O4 2 H2O
• Strategy mol of H2C2O4 out of mol of NaOH (by
• using V and M). Then, mol of H2C2O4 will be
• used to calculate g of H2C2O4 and, hence, of it
• in the sample.
• Coefficients are 1 for H2C2O4 and 2 for NaOH.

89
Oxalic acid Oxac M W 90.04 g/mol

• 0.485 mol 1 mol Oxac
• 34.47 mLx x 0.00836 mol
• 1000 mL 2 mol NaOH
  Oxac
• Based on the end point concept (mol ratio)
• 90.04 g
• 0.00836 mol x 0.753 g oxalic acid
• 1 mol
• 0.753 g oxac
• Oxac x 100 72.8
• 1.034 g of sample

90
Molar Mass of an Acid A 1.056 g of a pure acid,
HA, is dissolved in water and an acid-base
indicator added. The solution requires 33.78 mL
of 0.256 M NaOH solution to reach the equivalence
point. What is the molar mass of the acid?
We dont know what A is

• HA NaOH ? NaA H2O
• Strategy mol of HA out of mol of NaOH (by
• using V and M). Then, mol of HA and g of HA
• will be used to calculate the molar mass.
• Coefficients are 1 for HA and 1 for NaOH.

91
Molar Mass of HA

• 33.78 mL NaOH solution 0.03378 L
• 0.256 mol NaOH 1 mol HA
• 0.03378 Lx x 8.65x10-3
• 1 L soln 1 mol
  NaOH mol HA
• 1.056 g
  HA
• Molar mass, M W 122 g/mol
• 8.65x10-3
  mol HA
• 122 grams per mol

92
Vinegar A 25.0-mL sample of vinegar (which
contains the weak acetic acid, CH3CO2H) requires
28.33 mL of a 0.953 M solution of NaOH for
titration to the equivalence point. What mass of
the acid, in grams, is in the vinegar sample and
what is the M of acetic acid in the vinegar?

• CH3CO2H NaOH ? NaCH3CO2 H2O
• Strategy mol of CH3CO2H (HOAc) out of mol of
• NaOH (by using V and M). Then, g of HOAc will
• be calculated with the molar mass. The M will
• be calculated by using the volume of vinegar.
• Coefficients are 1 for HOAc and 1 for NaOH.

93
CH3CO2H HOAc M W 60.05 g/mol

• 0.953 mol 1 mol HOAc
• 28.33 mLx x 0.0270 mol
• 1000 mL 1 mol NaOH
  HOAc
• 60.05 g
• 0.0270 mol x 1.62 g acetic acid in vinegar
• 1 mol
• For the molarity, Vvinegar 25.0 mL 0.0250 L
  (soln)
• 0.0270 mol HOAc
• M HOAc 1.08 M
• 0.0250 L soln

94
Problem 75 of textbook To analyze an
iron-containing compound, you convert all the
iron into Fe2 in aqueous solution and then
titrate the solution with standardized KMnO4. The
balanced-net ionic equation isMnO4- 5Fe2
8H ? Mn2 5Fe3 4H2OA 0.598-g sample
of the iron compound requires 22.25 mL of 0.0123
M KMnO4 solution for titration to the equivalence
point. What is the mass of iron in the sample?
Strategy

• mL and M of MnO4- to ? mol of MnO4- and Fe2
• Coefficients are 1 and 5 for MnO4- and Fe2
  respectively

95
Fe2 and MnO4- (mole ratio is 5 to 1)

• 0.0123 mol MnO4- 5 mol Fe2
• 22.25 mLx x 1.37x10-3
• 1000 mL 1 mol
  MnO4- mol Fe2
• 
  55.85 g Fe
• g of iron 1.37 x 10-3 mol Fe2 x
  0.0765
• 
  1 mol Fe2 g Fe
• Now, for the mass of iron
• 0.0765 g Fe
• x 100 12.8 Fe
• 0.598 g sample

96
Normality

• equivalents (eq) solute per liter solution
• milliequivalents (meq) solute per mL solution
• eq solute meq solute
• N (calculated by dividing)
• V(L) soln V(mL) soln
• equivalent 1 equivalent weight
• Equivalent Weight (EW) given in g/eq
• acid/base the mass of a substance required to
  furnish or react with exactly 1 mol H
• redox reactions the mass of substance able to
  gain or lose 1 mol e-

97
Equivalent weight

• HCl
• HCl ? H Cl
• 1 mol HCl ? 1 mol H, ? EW MW
• H2SO4
• H2SO4 ? 2H SO42
• 1 mol H2SO4 ? 2 mol H, ??EW MW/2
• HnA (in general)
• EW MW/n, n H in molecule of acid
• KOH
• KOH ? K OH
• OH H ? H2O
• ?, 1 mol KOH reacts with 1 mol H, ? EW FW
• continued with polyhydroxides

98
Equivalent weight

• Fe(OH)3
• 1 mol Fe(OH)3 ? 3OH
• 1 mol reacts with 3 mol H, ? EW FW/3
• M(OH)n (in general)
• EW FW/n, n OH in formula unit of base
• Redox
• 7
  0
• MnO4 5e ? Mn2 Zn ? Zn2
  2 e
• FW MnO4
  AW Zn
• EW EW
  
• 5
  2

99
Example Calculate the normality of barium
hydroxide in a solution made by dissolving 0.991
g of barium hydroxide in water and diluting to
100.0 mL.

• Ba(OH)2
• FW 171.32 g
  mg
• EW 85.66 85.66
• 2 2
  eq meq
• 1 eq Ba(OH)2
• 0.991 g Ba(OH)2 x 0.0116 eq Ba(OH)2
• 85.66 g Ba(OH)2
• 0.0116 eq Ba(OH)2
• N Ba(OH)2 0.116 N Ba(OH)2
• 0.1000 L soln

100
Example Describe the preparation of 250.0 mL of
0.100 N oxalic acid solution from the solid.
Oxalic acid is H2C2O4.

• 1 mol oxalic acid ? 2 mol H
• EW 1/2 MW 1/2(90.04) 45.03g/eq
• 0.100 N indicates 0.100 eq OA/L soln
• 0.100 eq OA
• 0.2500 L x 0.0250 eq OA
• 1 L soln
• 45.03 g OA
• 0.0250 eq OA x 1.13 g OA
• 1 eq OA
• Dissolve 1.13 g of oxalic acid in water and
  dilute to a total volume of 250.0 mL

101
Example What is the normality of a 0.300 M
arsenic acid solution?

• H3AsO4
• 
  MW
• 1 mol H3AsO4 ? 3 mol H EW
• 
  3
• g H3AsO4 g H3AsO4 g
  H3AsO4
• eq 3 x 3 mol
• EW H3AsO4 MW MW H3AsO4
  H3AsO4
• 3
• eq mol
• N M Then,
• V(L) V(L)
• N H3AsO4 3 x M H3AsO4 3 x 0.300 0.900 N

102
CONCLUSION For an Acid HnA or a Base M(OH)n

• N n x M
• n H in molecule of acid or OH in formula
  unit of base

103
By definition, 1 eq of anything will react with
exactly 1 eq of anything else

• Equivalence Point
• The point in a titration at which
• eq titrant eq titrate
• meq titrant meq titrate

104
Example Calculate the concentration of
phosphoric acid in a solution given that a 25.0
mL sample of that solution required 42.2 mL of
0.274 N potassium hydroxide for titration.

• H3PO4 3 KOH ? K3PO4 3 H2O
• at Eq. Pt., meq H3PO4 meq KOH
• meq H3PO4 meq KOH
• (mL H3PO4)(N H3PO4) ( mL KOH)(N KOH)
• (25.0 mL)(X ) (42.2 mL)(0.274 N )
• X 0.463 N
• Now, N 3x M, then M N/3
• X 0.463/3 0.154 M

105
Example Calculate the oxalic acid (H2C2O4) in
a solid given that a 1.709 g sample of the solid
required 24.9 mL of 0.0998 N potassium hydroxide
for titration.

• H2C2O4 2KOH ? K2C2O4 2H2O
• Due to the two protons of H2C2O4 (OA),
• EW MW/2 45.03g/eq 45.03mg/meq
• g OA
• OA x 100
• g sample
• Strategy calculate eq of NaOH that are the same
• for OA then, calculate g of OA with eq and EW
• of OA

106
At equivalence point,

• eq OA eq KOH data of
  KOH soln
• 
  
• 0.0998 eq KOH
• eq OA 0.0249 L x 0.00249 eq
• 1 L soln
• 45.03 g OA
• 0.00249 eq OA x 0.112 g OA
• 1 eq OA
• 0.112 g OA
• OA x 100 6.55
• 1.709 g sample