1. Chapter 5 Molecular View of Reactions in Aqueous Solutions: Part 2
Chemistry: The Molecular Nature of Matter, 6E
Jespersen/Brady/Hyslop

2. Predicting Precipitation Reactions
Metathesis Reaction
Reactions where anions and cations exchange partners.
Also called Double replacement reaction
Pb(NO3)2(aq) + 2KI(aq)  PbI2(s) + 2KNO3(aq)
Precipitation reactions
Metathesis reactions where precipitate forms
How can we predict if compounds are insoluble?
Must know Solubility Rules

3. Table 5.1 Solubility Rules
Soluble Compounds
All salts of the alkali metals (Group IA) are soluble.
All salts containing NH4+, NO3, ClO4, ClO3, and C2H3O2 are soluble.
All chlorides, bromides, and iodides (salts containing Cl, Br, or I) are soluble except when combined with Ag+, Pb2+, and Hg22+ (note the subscript 2).
All salts containing SO42 are soluble except those of Pb2+, Ca2+, Sr2+, Ba2+, and Hg22+.

4. Table 5.1 Solubility Rules
Insoluble Compounds
5. All metal hydroxides (ionic compounds containing OH) and all metal oxides (ionic compounds containing O2 are insoluble except those of Group IA and those of Ca2+, Sr2+, and Ba2+.
When metal oxides do dissolve, they react with water to form hydroxides. The oxide ion, O2, does not exist in water. For example:
Na2O(s) + H2O  2NaOH(aq)
6. All salts containing PO43, CO32, SO32 and S2 are insoluble except those of Group IA and NH4+

5. Learning Check: Solubility Rules
Which of the following compounds are expected to be soluble in water?
Ca(C2H3O2)2
FeCO3
AgCl
Yes No
soluble : the substance will be aqueous (aq) if water is available.
insoluble : substance will remain a solid (s). even in water.

6. Your Turn
Which of the following will be the solid product of the reaction of
Ca(NO3)2 (aq) + Na2CO3 (aq)  ?
CaCO3
NaNO3
Na(NO3)2
Na2(NO3)2
H2O

7. Metathesis (Double Replacement) Reaction
AB + CD  AD + CB
Cations and anions change partners
Charges on each ion don’t change
Formulas of products are determined by charges of reactants
Occur only if solid, gas, weak electrolyte or non-electrolyte product forms
Otherwise, all ions are spectator ions
What is a metathesis reaction?

8. Predicting Products of Double Replacement Reactions
Identify the ions involved:
Distinguish between subscripts that count and those that are characteristic of a polyatomic ion.
Swap partners and make neutral with appropriate subscripts
Assign states using solubility rules
Balance equation
counting subscript
Characteristic subscripts are the only subscripts that transfer to the product side. 2
HCl(aq)+ Ca(OH)2(aq) 
CaCl H2O
(aq)
ions: H+, Cl– Ca2+, 2OH –

9. Predict if Ionic Reaction Occurs
Write molecular equation for metathesis reaction
Determine which ion combinations form insoluble salt, water, weak electrolyte, or gas.
Translate molecular equation into ionic equation
Cancel spectator ions, to give net ionic equation
Check for driving force: formation of weak electrolyte, solid, gas or non-electrolyte

10. Learning Check: Predict Products
Pb(NO3)2(aq) + Ca(OH)2(aq) 
BaCl2(aq) + Na2CO3(aq) 
2Na3PO4(aq) + 3Hg2(NO3)2(aq) 
2 NaCl(aq) + Ca(NO3)2(aq) 
Pb(OH)2(s) + Ca(NO3)2(aq)
BaCO3(s) + 2 NaCl (aq)
6 NaNO3(aq) + (Hg2)3(PO4)2(s)
Note that Hg22+ is a polyatomic ion.
NR (No reaction)
CaCl2(aq) + 2 NaNO3(aq)

11. Learning Check
Predict the reaction that will occur when aqueous solutions of Cd(NO3)2 and Na2S are mixed. Write molecular, ionic and net ionic equations.
Molecular Equation:
Cd(NO3)2(aq) + Na2S(aq) 
Ionic Equation:
Cd2+(aq) + 2NO3–(aq) + 2Na+(aq) + S2–(aq) 
Net Ionic Equation:
Cd2+(aq) + S2–(aq)  CdS(s)
CdS(s) + 2NaNO3(aq)
Dialog: Net ionic eqn. Let’s us generalize.
Any soluble compound of Cd2+ and any soluble compound of S2– will react to form CdS.
Cd(C2H3O2)2(aq) + (NH4)2S(aq)  CdS(s) + 2NH4C2H3O2(aq)
CdS(s) + 2NO3–(aq) Na+(aq)

12. Learning Check
Write molecular, ionic and net ionic equations for the reaction that occurs when Pb(NO3)2 and Fe2(SO4)3 are mixed in solution.
Molecular Equation
3Pb(NO3)2(aq) + Fe2(SO4)3(aq)  PbSO4(s) + 2Fe(NO3)3(aq)
Ionic Equation
3Pb2+(aq) + 6NO3–(aq) + 2Fe3+(aq) + 6SO42–(aq)  Fe3+(aq) + 6NO3–(aq) + PbSO4(s)
Net Ionic Equation
3Pb2+(aq) + 6SO42–(aq)  3PbSO4(s)
Pb2+(aq) + 2SO42–(aq)  PbSO4(s)
Must make lowest possible whole number coefficients 1 2 1

13. Your Turn
Will mixing aqueous solutions of Mg(C2H3O2)2 and CsCl yield a precipitate?
Yes No
Molecular Equation:
Mg(C2H3O2)2(aq) + 2CsCl(aq)  MgCl2(aq) CsC2H3O2(aq)
Ionic Equation:
Mg2+(aq) + 2C2H3O2–(aq) + 2Cs+(aq) + 2Cl–(aq) 
Mg2+(aq) + 2Cl–(aq) + 2Cs+(aq) + 2C2H3O2– (aq)
All ions cancel.
No Net Ionic Equation.
No precipitate formed.

14. 2. Predicting Acid−Base Reactions
Neutralization reaction
When mixed in 1:1 molar ratio, acid and base solutions lose their acidic and basic properties
Combination of H+ and OH– to form salt and water
Salt
Ionic compound without H+, OH–, or O2–
Acid Base  Salt Water
HClO4(aq) + NaOH(aq)  NaClO4(aq) + H2O

15. Neutralization Between Strong Acid and Strong Base
Molecular Equation
2HCl(aq) + Ca(OH)2(aq) →
Ionic Equation
2H+(aq) + 2Cl–(aq) + Ca2+(aq) + 2OH–(aq) → H2O + Ca2+(aq) + 2Cl–(aq)
Net Ionic Equation
2H+(aq) + 2OH–(aq) → 2H2O
H+(aq) + OH–(aq) → H2O
True for any strong acid and strong base
2H2O + CaCl2(aq)

16. Weak Acid with Strong Base
Molecular Equation: HC2H3O2(aq) + NaOH(aq) → Ionic Equation: HC2H3O2(aq) + Na+(aq) + OH–(aq) → H2O + Na+(aq) + C2H3O2–(aq) Net Ionic Equation: HC2H3O2(aq) + OH–(aq) → H2O + C2H3O2–(aq)
H2O + NaC2H3O2(aq)
Write weak electrolytes in “molecular form”

17. Neutralization of Strong Acid with Insoluble Base
Insoluble Hydroxides
Molecular Equation
Mg(OH)2(s) + 2HCl(aq) 
Ionic Equation
Mg(OH)2(s) + 2H+(aq) + 2Cl–(aq)  Mg2+(aq)
+ 2Cl–(aq) + 2H2O
Net Ionic Equation
Mg(OH)2(s) + 2H+(aq)  Mg2+(aq) + 2H2O
MgCl2(aq) + 2H2O
Fig. 5.19

18. Neutralization of Strong Acid with Insoluble Base
Insoluble Oxides – Basic Anhydrides
Molecular Equation
Al2O3(s) + 6HCl(aq) 
Ionic Equation
Al2O3(s) + 6H+(aq) + 6Cl–(aq)  2Al3+(aq) +
6Cl–(aq) + 3H2O
Net Ionic Equation
Al2O3(s) + 6H+(aq)  2Al3+(aq) + 3H2O
2AlCl3(aq) + 3H2O

19. Strong Acid with Weak Base
Molecular Equation:
NH3(aq) + HCl(aq) 
Ionic Equation :
NH3(aq) + H+(aq) + Cl–(aq)  NH4+(aq) + Cl–(aq)
Net Ionic Equation :
NH3(aq) + H+(aq)  NH4+(aq)
NH4Cl(aq)
Neutralization doesn’t always form water.
Fig 5.20
A hydronium ion transfers a hydrogen ion to an ammonia molecule.
The products are ammonium ion and water
NH3(aq) + H3O+(aq)  NH4+ (aq) + H2O

20. Challenge Problem
What is the Net Ionic Equation for reaction of an insoluble hydroxide and a weak acid?
Molecular Equation
Mg(OH)2(s) + 2HC2H3O2(aq) 
Ionic Equation
Mg(OH)2(s) + 2HC2H3O2(aq)  Mg2+(aq) + 2H2O C2H3O2–(aq)
There are NO spectator ions!
So Net Ionic and Ionic Equations are the same in this case
Mg(C2H3O2)2(aq)
+ 2H2O

21. Learning Check HNO3(aq) + Ca(OH)2(aq)  H+(aq) + OH–(aq)  H2O
What is the net ionic equation for the reaction between the following reactants:
HNO3(aq) + Ca(OH)2(aq) 
H+(aq) + OH–(aq)  H2O
2. N2H4(aq) + HI(aq) 
N2H4(aq) + H+(aq)  N2H4+(aq)
3. CH3NH2(aq) + HC4H7O2(aq) 
CH3NH2(aq) + HC4H7O2(aq)  CH3NH3+(aq) C4H7O2–(aq)
HCHO2 is a weak acid
N2H4 is a weak base

22. Your Turn Which is the net ionic equation for the reaction:
NaOH(aq) + HF(aq) ?
Na+(aq) + OH–(aq) + H+(aq) + F–(aq)  H2O NaF(aq)
OH–(aq) + H+(aq)  H2O
Na+(aq) + OH–(aq) + HF(aq)  H2O + NaF(aq)
OH–(aq) + HF(aq)  H2O + F–(aq)
No reaction

23. Metathesis and Gas Formation
Metathesis reactions involving certain ions lead to formation of a gas.
Low solubility of gas in solvent (water) leads to escape of gas.
Once escaped, can’t go back in, drives reaction to completion
Many anions that give rise to gases are insoluble
Dissolving in acid followed by gas formation drives reaction to completion
Dissolves insoluble salt

24. Metathesis and Gas Formation
Gases formed by metathesis
H2S, HCN
Unstable compounds—decompose and form gas
H2CO3  H2O and CO2(g)
H2SO3  H2O and SO2(g)
NH4OH  H2O and NH3(g)
The reactions are driven to completion because the gas escapes and is unavailable for back reaction

25. Reactions that Release CO2b)
Acid with Bicarbonate (HCO3–)
NaHCO3(aq) + HI(aq)  NaI(aq) + H2O + CO2(g)
Acid with Carbonate (CO32–)
CaCO3(aq) + 2HCl(aq)  CaCl2(aq) + H2O + CO2(g)
Fig 5.22 and 5.23

26. Reactions that Release Gases
Acid with Sulfites (SO32–) or Bisulfites (HSO32–)
K2SO3(aq) + 2HClO4(aq)  SO2(g) + 2KClO4(aq) + H2O
LiHSO3(aq) + HClO3(aq)  SO2(g) + H2O + LiClO3(aq)
Acid with Sulfides
2HCl(aq) + Na2S(aq)  2NaCl(aq) + H2S(g)
Acid with Cyanides
HNO3(aq) + CsCN(aq)  HCN(g) + CsNO3(aq)
Bases with Ammonium
NaOH(aq) + NH4Cl(aq)  NH3(g) + H2O + NaCl(aq)

27. Learning Check
Write the molecular, ionic and net ionic equations for the reaction of Li2SO3 with formic acid, HCHO2 Molecular Equation: Li2SO3(aq) + 2HCHO2(aq)  2LiCHO2(aq) + SO2(g) + H2O Ionic Equation: 2Li+(aq) + SO32–(aq) + 2HCHO2(aq)  2CHO2–(aq) + 2Li+(aq) + SO2(g) + H2O Net Ionic Equation: SO32 –(aq) + 2HCHO2(aq)  2CHO2–(aq) + SO2(g) + H2O

28. Your Turn
What is the net ionic equation for the reaction of HCl with KHCO3?
HCl(aq) + KHCO3(aq)  KCl(aq) + H2CO3(aq)
H+(aq) + HCO3–(aq)  H2CO3(aq)
HCl(aq) + KHCO3(aq)  KCl(aq) + CO2(g) + H2O
H+(aq) + Cl–(aq) + K+(aq) + HCO3–(aq)  K+(aq) + Cl–(aq) + CO2(g) + H2O
H+(aq) + HCO3–(aq)  CO2(g) + H2O

29. Metathesis Overview Precipitation: Neutralization: Gas-forming:
2 solutions form solid product
Neutralization:
Acid + metal hydroxide or oxide forms water and salt
Gas-forming:
Metathesis reaction forms one of these products:
HCN, H2S, H2CO3(aq) , H2SO3(aq) , NH4OH(aq)
Formation of Weak Electrolyte:
Salt of weak acid reacts with base to form molecule

30. Predicting Reactions and Writing Their Equations
What reaction, if any, occurs between potassium nitrate and ammonium chloride?
Need to know whether net ionic equation exists.
Determine formulas of reactants
KNO3 + NH4Cl  ?
Write molecular equation
KNO3 + NH4Cl  KCl + NH4NO3
Check Solubilities
All are soluble

31. Predicting Reactions and Writing Their Equations
Predicted Molecular Equation
KNO3(aq) + NH4Cl(aq)  KCl(aq) + NH4NO3(aq)
Write Ionic Equation
K+(aq) + NO3–(aq) + NH4+(aq) + Cl–(aq)  K+(aq) + Cl–(aq) + NH4+(aq) + NO3–(aq)
Same on both sides
All ions cancel out
No gases, solids, water, or weak electrolytes formed

32. Learning Check
Determine the net ionic equation for the following reactions.
Co(OH)2 + HNO2
Co(OH)2(s) + 2H+(aq)  Co2+(aq) + 2H2O
KCHO2 + HCl
CHO2–(aq) + H+(aq)  HCHO2(aq)
CuCO3 + HC2H3O2
CuCO3(s) + 2HC2H3O2(aq)  Cu2+(aq) + CO2(g) C2H3O2–(aq) + H2O

33. Your Turn
What is the net ionic reaction when aqueous solutions of NaOH and NiCl2 are mixed?
Ni2+(aq) + 2OH–(aq)  Ni(OH)2(s)
NaOH(aq) + NiCl2(aq)  NaCl(aq) + Ni(OH)2(s)
2NaOH(aq) + NiCl2(aq)  2NaCl(aq) + Ni(OH)2(s)
2Na+(aq) + 2OH–(aq) + Ni2+(aq) + 2Cl–(aq)  Na+(aq) + 2Cl–(aq) + Ni(OH)2(s)
No reaction.

34. Your Turn Which of the following combinations will not react?
Na2CO3(aq) + HCl(aq)
Na2SO3(aq) + CaCl2(aq)
NaCl(aq) + HC2H3O2(aq)
NH4Cl(aq) + HClO4(aq)
KCN(aq) + H2SO4(aq)
Yields a gas (CO2)
Yields a gas (SO2)
All ions are soluble, Acetic acid is a weak acid.
Yields a gas (NH3)
Yields a gas (HCN)

35. Synthesize by Metathesis Reaction
Practical use of metathesis reactions
Desired compound should be easily separated from reaction mixture. Two principal approaches
Desired compound is insoluble in water
Start with 2 soluble reactants
Product isolated by filtration
Desired compound is soluble in water
Acid-base neutralization
Reaction of metal carbonate and acid
Either way, product isolated by evaporation of water
Uncontaminated by Any reactant starting materials, or Any other reaction product
Cation supplied by base. Anion supplied by acid
Cation from metal carbonate. Anion from acid

36. Learning Check
What reaction might we use to synthesize nickel sulfate, NiSO4?
Use solubility rules
NiSO4 is soluble in water
So, there are 2 possible methods
Use acid + base
H2SO4(aq) + Ni(OH)2(s)  NiSO4(aq) + 2H2O
Use acid + carbonate
H2SO4(aq) + NiCO3(s)  NiSO4(aq) + CO2(g) + 2H2O

37. Molar Concentration Dissolve solutes. Make separate solutions. 
Mix Solutions.
Allow Reaction to occur.
Need to know Quantitatively HOW MUCH? of each solute we used.
Define
5.6 Molarity

38. Molarity (M) Number of moles of solute per liter of solution.
Allows us to express relationship between moles of solute and volume of solution
Hence, 0.100M solution of NaCl contains mole NaCl in 1.00 liter of solution
Same concentration results if you dissolve mol of NaCl in liter of solution
moles solute (n)=Volume (L)*Molarity (M)

39. Molarity as Conversion Factor
Often have stoichiometry problems involving amount of chemical and volume of solution
Solve the problem using molarity
Molarity provides conversion factors between moles and volume
M = mole per liter
Molarity
Moles of a Substance
Volume of a solution of Substance

40. Molarity as Conversion Factor
Gives equivalence relationship between “mol NaCl” and “L soln”
Forms two conversion factors
Three basic types of calculations:
0.100 mol NaCl  1.00 L soln

41. Learning Check: Calculating Molarity (from grams and volume)
Calculate the molarity (M) of a solution prepared by dissolving 11.5 g NaOH (40.00 g/mol) solid in enough water to make 1.50 L of solution. g NaOH  mol NaOH  M NaOH

42. Learning Check: Calculating Volume (from Molarity and moles)
How many mL of M NaCl solution are needed to obtain mol of NaCl?
Use M definition
Given molarity and moles, need volume
= 400 mL of M NaCl solution

43. Preparing Solution of Known Molarity
a) b) c) d) e)
a b c d e
Weigh solid and transfer to volumetric flask
Add part of the water
Dissolve solute completely
Add water to reach etched line
Stopper flask and invert to mix thoroughly
Fig 5.24

44. Learning Check: Preparing Solution of Known Molarity from Solid
How many grams of strontium nitrate are required to prepare mL of M Sr(NO3)2 solution?
M × V  mol × MM  g
Convert Molarity and Volume to mole
= mol Sr(NO3)2
Convert mol to g
Strontium nitrate, Sr(NO3)2, is used in fireworks to produce brilliant red colors.
Weigh out 5.29 g Sr(NO3)2 and dilute to mark.
= 5.29 g Sr(NO3)2

45. Your Turn
How many grams of KMnO4 must you weigh out if you want to make 250.mL of a M KMnO4 solution?
7900 g
50.0 g
0.316 g
7.90 g
198 g
M  mol  g
Can do all conversions in series of steps.
Weigh out 7.90 g KMnO4 & dilute to mark.

46. Preparing a Solution of Known Molarity by Dilution
When making solutions, don’t always begin with solute as pure solid, liquid or gas.
Can take solution of high concentration and dilute to a lower concentration.
Amount of MOLES does NOT change! Remains the same
Small
Volume Concentrated Solution
Large
Volume Dilute Solution
Add lots of water

47. Diluting Solutions Vdil  Mdil = Vconc  Mconc
Moles of solute do not change upon dilution
Just changing volume
# moles in dilute = # moles in concentrated
Moles of solute in the dilute solution
Moles of solute in the concentrated solution
Vdil  Mdil = Vconc  Mconc

48. Learning Check: Dilutions
What volume (in mL) of 16.0 M H2SO4 must be used to prepare 1.00 L of M H2SO4?
Rearranging gives
= 125 mL

49. Preparing Solution of Known M
Using volumetric glassware ensures that the volumes are known precisely
Use a volumetric pipette to transfer the stock solution
Use a volumetric flask to receive the final solution
Fig 5.27

50. Your Turn
What volume of 12.1 M HCl is needed to create mL of 3.2 M HCl?
66 mL
800 mL
3025 mL
945 mL
9680 mL
Vconc = 66 mL

51. Your Turn
25 mL of 6.0 M HCl are diluted to 500 mL with water. What is the molarity of the resulting solution?
150 M
3.0 M
0.120 M
120 M
0.30 M
Mdil = 0.30 M

52. Solution Stoichiometry
Often work with solutions when running reactions
How do we determine amounts needed to completely react one compound?
Like any other stoichiometry problem
Now use Volume and Molarity to obtain moles of each substance.
5.7 Solution Stoichiometry

53. Learning Check: Solution Stoichiometry
How many milliliters of M H3PO4 could be completely neutralized by 45.0 mL of M KOH? The balanced equation for the reaction is H3PO4(aq) + 3KOH(aq)  K3PO4(aq) + 3H2O Strategy:
Coefficients of
Balanced equation
mol KOH
mol H3PO4
mol and M of
H3PO4 soln
Vol and M of
KOH soln
KOH solution
H3PO4 soln

54. Learning Check: Solution Stoichiometry
1. Calculate moles of KOH 2. Use coefficients to calculate the moles H3PO4 required 3. Calculate volume of H3PO4 needed
= 4.50 × 10–3 mol KOH
= 1.50 × 10–3 mol H3PO4
= 31.6 mL H3PO4

55. Stoichiometry of Ionic Equations
Calculating Concentrations of Ions in Solutions of Electrolytes.
When using solutions of electrolytes, need to know [ions] in solution.
Ions don’t stay together, [ions] may not be same
Can easily calculate from M electrolyte in molecular form.
Concentration of particular ion equals concentration of salt multiplied by number of ions of that kind in one formula unit of salt.
Note: [XX] = concentration of whatever species is in square bracket

56. Learning Check: Ion Concentrations
If you have M Na2CO3 (aq), what is the concentration of each type of ion in solution? Means Na2CO3(aq)  2 Na+(aq) + CO32–(aq) Concentration of Na+ ions is: Concentration of CO32– ions is:

57. Your Turn
If the solution concentration of sulfate ion is M, what is the concentration of Al2(SO4)3, assuming that all of the sulfate ion comes aluminum sulfate?
0.750 M
2.25 M
0.250 M
1.50 M
0.500 M
= M Al2(SO4)3

58. Learning Check: Net Ionic Eqns in Solution Stoichiometry Calculations
What volume, in mL, of M KOH is needed to react completely with 60.0 mL of M FeCl2 to form Fe(OH)2 solid?
1. Write Balanced Net Ionic Equation
Fe2+(aq) OH–(aq)  Fe(OH)2(s)
2. Determine the game plan
mol Fe2+
mol OH–
M Fe2+
V OH–
M FeCl2
V KOH

59. Learning Check: Net Ionic Eqns in Solution Stoichiometry Calculations
3. Convert M FeCl2  M Fe2+  mol Fe2+ 4. Convert mol Fe2+  mol OH– 5. Convert mol OH–  V OH–  V KOH
= mol Fe2+
= 60.0 mL KOH

60. Learning Check: Solution Limiting Reagent Problem
How many grams of PbI2 (461.0) will form if mL of M FeI3 (436.5) is mixed with 50.0 mL of M Pb(NO3)2 (269.2)?
3Pb(NO3)2(aq) + 2FeI3(aq)  3PbI2(s) + 2Fe(NO3)3(aq)
Net Ionic Eqn: Pb2+(aq) + 2I(aq)  PbI2(s)
Strategy
Vol Pb(NO3)2  mol Pb(NO3)2  mol Pb2+  mol PbI2  g PbI2
Vol FeI3  mol FeI3  mol I  mol PbI2  g PbI2
Whichever is less determines how much is formed and which reagent is limiting

61. Limiting Reagent Problem
Starting with Pb(NO3)2 Starting with FeI3
= 6.92 g PbI2
= g PbI2
Pb(NO3)2 is limiting and only 6.92 g of PbI2 can be made.

62. Chemical Analysis Qualitative Analysis Quantitative Analysis
What substances are present in a sample
Quantitative Analysis
Measure the amounts of various substances in a sample
Chemical reactions  useful strategy
Convert all of an element present in a sample into a substance of known formula
Use the amount of this known to determine amount of element present in the orginial sample (unknown)
5.8 Chemical Analysis and Titration

63. Learning Check: Chemical Analysis
Insecticide contains H, C, Cl. Carry out reactions on g sample to convert all chlorine to Cl– in H2O. Resulting solution treated with AgNO3 solution in excess. AgCl ppts and is collected, dried, and weighed. Mass AgCl = g. What is the percentage by mass of Cl in original sample? Strategy: g AgCl  mol AgCl  mol Cl  g Cl

64. How Much Cl in 2.022 g of AgCl? g AgCl  mol AgCl  mol Cl  g Cl
% Cl in original Sample?
= 50.02% Cl

65. Titrations Widely used analytical technique
Used to determine concentration of solute
Used daily to monitor:
Water purity
Quality control in food industry
How it Works:
Must know precise reaction that occurs
Reaction must be rapid and complete
Must know exact quantity of one reactant
Use stoichiometry to find exact amount of any other substance in solution

66. Titration
Controlled addition of 1 reactant to known quantity of another until reaction is complete
Acid-Base Titration
Very common type of titration
Ex. Analysis of citric acid in orange juice by neutralization with NaOH
Know MNaOH and measure exact VNaOH needed to completely neutralize citric acid
MNaOH × VNaOH = mol NaOH
mol NAOH  mol citric acid
mol citric acid × MM citric acid = g citric acid

67. Titration in practice:
Buret
Volumetric measuring device with 0.10 mL markings
Stopcock
Permits flow of titrant to stop when reaction is complete
Volume titrant
used = Vf –Vi
Figure 5.26

68. Titration: Definitions
Titrant
Solution in the buret
Known concentration
Can be either acid or base depending on nature of the analyte
Analyte
Solution being analyzed
Solution in flask
Solution of unknown concentration
Equivalence Point
Volume of titrant where moles of titrant and moles of analyte are exactly equal .

69. Titration Indicator Endpoint:
Dye that is 1 color in acid and 2nd color in base
Ex. phenolphthalein
Colorless in acid and bright pink in base
Color change signals end point of titration
Endpoint:
Volume of titrant required to complete reaction monitored by color change of indicator
Choose indicator so endpoint and equivalence point are
the same

70. Learning Check
Suppose that mL of a solution of oxalic acid, H2C2O4, extracted from rhubarb leaves, is titrated with M NaOH(aq) and that the stoichiometric point is reached when 37.5 mL of the solution of base is added. What is the molarity of the oxalic solution?
Step 1: Write the balanced equation.
H2C2O4(aq) + 2NaOH(aq)  Na2C2O4(aq) + 2H2O

71. Learning Check: Oxalic acid + NaOH
Step 2: Calculate moles of Base used
Step 3: Calculate moles of Oxalic Acid
Step 4: Calculate M H2C2O4
= M H2C2O4

72. Summary of Stoichiometry CalculationsNA

73. Your Turn
25.00 mL of HNO3 are titrated with mL of M Ca(OH)2. What is the concentration of HNO3 in the initial sample?
2HNO3(aq) + Ca(OH)2(aq)  2AgBr(s) + Ca(NO3)2(aq)
0.433 M
1.95 M
0.867 M
3.90 M
7.80 M
= 7.80 M HNO3

74. Your Turn
A sample of metal ore is reacted according to the following reaction:
Fe(s) + 2H+(aq) → Fe2+(aq) + H2(g)
If mL of 2.3 M HCl are used, what mass of Fe was in the ore? (Atomic mass of Fe is g/mol)
0.515 g
1.03 g
1.21 g
1.61 g
3.20 g
= 1.61 g