1. Introductory Chemistry, 3rd Edition Nivaldo Tro
Chapter 14
Acids and Bases
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
2009, Prentice Hall

2. Tro's Introductory Chemistry, Chapter 14
Types of Electrolytes
Salts are water-soluble ionic compounds.
All strong electrolytes.
Acids form H+1 ions in water solution.
Bases combine with H+1 ions in water solution.
Increases the OH-1 concentration.
May either directly release OH-1 or pull H+1 off H2O.
Tro's Introductory Chemistry, Chapter 14

3. Tro's Introductory Chemistry, Chapter 14
Properties of Acids
Sour taste.
React with “active” metals.
I.e., Al, Zn, Fe, but not Cu, Ag or Au.
2 Al + 6 HCl ® 2 AlCl3 + 3 H2
Corrosive.
React with carbonates, producing CO2.
Marble, baking soda, chalk, limestone.
CaCO3 + 2 HCl ® CaCl2 + CO2 + H2O
Change color of vegetable dyes.
Blue litmus turns red.
React with bases to form ionic salts.
Tro's Introductory Chemistry, Chapter 14

4. Tro's Introductory Chemistry, Chapter 14
Common Acids
Tro's Introductory Chemistry, Chapter 14

5. Tro's Introductory Chemistry, Chapter 14
Structures of Acids
Binary acids have acid hydrogens attached to a nonmetal atom.
HCl, HF
Hydrofluoric acid
Tro's Introductory Chemistry, Chapter 14

6. Tro's Introductory Chemistry, Chapter 14
Structure of Acids
Oxyacids have acid hydrogens attached to an oxygen atom.
H2SO4, HNO3
Tro's Introductory Chemistry, Chapter 14

7. Tro's Introductory Chemistry, Chapter 14
Structure of Acids
Carboxylic acids have COOH group.
HC2H3O2, H3C6H5O3
Only the first H in the formula is acidic.
The H is on the COOH.
Tro's Introductory Chemistry, Chapter 14

8. Tro's Introductory Chemistry, Chapter 14
Properties of Bases
Also known as alkalis.
Taste bitter.
Alkaloids = Plant product that is alkaline.
Often poisonous.
Solutions feel slippery.
Change color of vegetable dyes.
Different color than acid.
Red litmus turns blue.
React with acids to form ionic salts.
Neutralization.
Tro's Introductory Chemistry, Chapter 14

9. Tro's Introductory Chemistry, Chapter 14
Common Bases
Tro's Introductory Chemistry, Chapter 14

10. Tro's Introductory Chemistry, Chapter 14
Structure of Bases
Most ionic bases contain OH ions.
NaOH, Ca(OH)2
Some contain CO32- ions.
CaCO3 NaHCO3
Molecular bases contain structures that react with H+.
Mostly amine groups.
Tro's Introductory Chemistry, Chapter 14

11. Arrhenius Theory
Bases dissociate in water to produce OH- ions and cations.
Ionic substances dissociate in water.
NaOH(aq) → Na+(aq) + OH–(aq)
Acids ionize in water to produce H+ ions and anions.
Because molecular acids are not made of ions, they cannot dissociate.
They must be pulled apart, or ionized, by the water.
HCl(aq) → H+(aq) + Cl–(aq)
In formula, ionizable H is written in front.
HC2H3O2(aq) → H+(aq) + C2H3O2–(aq)
Tro's Introductory Chemistry, Chapter 14

12. Tro's Introductory Chemistry, Chapter 14
Arrow Conventions
Chemists commonly use two kinds of arrows in reactions to indicate the degree of completion of the reactions.
A single arrow indicates that all the reactant molecules are converted to product molecules at the end.
A double arrow indicates that the reaction stops when only some of the reactant molecules have been converted into products.
 in these notes.
Tro's Introductory Chemistry, Chapter 14

13. Arrhenius Theory, Continued
HCl ionizes in water,
producing H+ and Cl– ions.
NaOH dissociates in water,
producing Na+ and OH– ions.
Tro's Introductory Chemistry, Chapter 14

14. Brønsted–Lowry Theory
A Brønsted-Lowry acid–base reaction is any reaction in which an H+ is transferred.
Does not have to take place in aqueous solution.
Broader definition than Arrhenius.
Acid is H+ donor; base is H+ acceptor.
Since H+ is a proton, acid is a proton donor and base is a proton acceptor.
Base structure must contain an atom with an unshared pair of electrons to bond to H+.
In the reaction, the acid molecule gives an H+ to the base molecule.
H–A + :B  :A– + H–B+
Tro's Introductory Chemistry, Chapter 14

15. Comparing Arrhenius Theory and Brønsted–Lowry Theory
HCl(aq) + H2O(l) 
Cl−(aq) + H3O+(aq)
HF(aq) + H2O(l) 
F−(aq) + H3O+(aq)
NaOH(aq) + H2O(l) 
Na+(aq) + OH−(aq) + H2O(l)
NH3(aq) + H2O(l) 
NH4+(aq) + OH−(aq)
Arrhenius theory
HCl(aq) 
H+(aq) + Cl−(aq)
HF(aq)  H+(aq) + F−(aq)
NaOH(aq) 
Na+(aq) + OH−(aq)
NH4OH(aq) 
NH4+(aq) + OH−(aq)
Tro's Introductory Chemistry, Chapter 14

16. Amphoteric Substances
Amphoteric substances can act as either an acid or a base.
They have both transferable H and an atom with a lone pair.
HCl(aq) is acidic because HCl transfers an H+ to H2O, forming H3O+ ions.
Water acts as base, accepting H+.
HCl(aq) + H2O(l) → Cl–(aq) + H3O+(aq)
NH3(aq) is basic because NH3 accepts an H+ from H2O, forming OH–(aq).
Water acts as acid, donating H+.
NH3(aq) + H2O(l)  NH4+(aq) + OH–(aq)
Tro's Introductory Chemistry, Chapter 14

17. Tro's Introductory Chemistry, Chapter 14
Conjugate Pairs
In a Brønsted-Lowry acid-base reaction, the original base becomes an acid in the reverse reaction, and the original acid becomes a base in the reverse process.
Each reactant and the product it becomes is called a conjugate pair.
The original base becomes the conjugate acid; the original acid becomes the conjugate base.
Tro's Introductory Chemistry, Chapter 14

18. Tro's Introductory Chemistry, Chapter 14
Example—Identify the Brønsted–Lowry Acids and Bases and Their Conjugates in the Reaction.
H2SO H2O  HSO4– + H3O+
When the H2SO4 becomes HSO4, it loses an H+, so H2SO4 must be the acid and HSO4 its conjugate base.
When the H2O becomes H3O+, it accepts an H+, so H2O must be the base and H3O+ its conjugate acid.
H2SO H2O  HSO4– + H3O+
Acid Base Conjugate Conjugate
base acid
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19. Tro's Introductory Chemistry, Chapter 14
Example—Identify the Brønsted-Lowry Acids and Bases and Their Conjugates in the Reaction, Continued.
HCO3– H2O  H2CO3 + HO–
When the HCO3 becomes H2CO3, it accepts an H+, so HCO3 must be the base and H2CO3 its conjugate acid.
When the H2O becomes OH, it donates an H+, so H2O must be the acid and OH its conjugate base.
HCO3– H2O  H2CO3 + HO–
Base Acid Conjugate Conjugate
acid base
Tro's Introductory Chemistry, Chapter 14

20. Practice—Write the Formula for the Conjugate Acid of the Following:
H2O
NH3
CO32−
H2PO41−
H3O+
NH4+
HCO3−
H3PO4
Tro's Introductory Chemistry, Chapter 14

21. Practice—Write the Formula for the Conjugate Base of the Following:
H2O
NH3
CO32−
H2PO41−
HO−
NH2−
Since CO32− does not have an H, it cannot be an acid.
HPO42−
Tro's Introductory Chemistry, Chapter 14

22. Tro's Introductory Chemistry, Chapter 14
Practice—Write the Equations for the Following Reacting with Water and Acting as a Monoprotic Acid. Label the Conjugate Acid and Base.
HSO H2O ® SO H3O+1
Acid Base Conjugate Conjugate
base acid
HSO4-1
CO32− + H2O ® HCO3− + OH−
Base Acid Conjugate Conjugate
acid base
CO32−
Tro's Introductory Chemistry, Chapter 14

23. Tro's Introductory Chemistry, Chapter 14
Practice—Write the Equations for the Following Reacting with Water and Acting as a Monoprotic Acid. Label the Conjugate Acid and Base, Continued.
HBr + H2O ® Br H3O+1
Acid Base Conjugate Conjugate
base acid
HBr
I− + H2O ® HI + OH−
Base Acid Conjugate Conjugate
acid base I−
Tro's Introductory Chemistry, Chapter 14

24. Neutralization Reactions
H+ + OH- H2O
Acid + base salt + water
Double-displacement reactions.
Salt = cation from base + anion from acid.
Cation and anion charges stay constant.
H2SO4 + Ca(OH)2 → CaSO4 + 2 H2O
Some neutralization reactions are gas evolving, where H2CO3 decomposes into CO2 and H2O.
H2SO4 + 2 NaHCO3 → Na2SO4 + 2 H2O + 2 CO2
Tro's Introductory Chemistry, Chapter 14

25. (H+ + NO3−) + (Ca2+ + OH−)  (Ca2+ + NO3−) + H2O(l)
Example 14.2—Write the Equation for the Reaction of Aqueous Nitric Acid with Aqueous Calcium Hydroxide.
1. Write the formulas of the reactants.
HNO3(aq) + Ca(OH)2(aq) 
2. Determine the ions present when each reactant dissociates.
(H+ + NO3−) + (Ca2+ + OH−) 
3. Exchange the ions. H+1 combines with OH-1 to make H2O(l).
(H+ + NO3−) + (Ca2+ + OH−)  (Ca2+ + NO3−) + H2O(l)
Tro's Introductory Chemistry, Chapter 14

26. 4. Write the formulas of the products.
Example 14.2—Write the Equation for the Reaction of Aqueous Nitric Acid with Aqueous Calcium Hydroxide, Continued.
4. Write the formulas of the products.
Cross charges and reduce.
HNO3(aq) + Ca(OH)2(aq)  Ca(NO3)2 + H2O(l)
5. Balance the equation.
May be quickly balanced by matching the numbers of H and OH to make H2O.
Coefficient of the salt is always 1.
2 HNO3(aq) + Ca(OH)2(aq)  Ca(NO3) H2O(l)
Tro's Introductory Chemistry, Chapter 14

27. Ca(NO3)2 is soluble (all NO3- are soluble).
Example 14.2—Write the Equation for the Reaction of Aqueous Nitric Acid with Aqueous Calcium Hydroxide, Continued.
6. Determine the solubility of the salt.
Ca(NO3)2 is soluble (all NO3- are soluble).
7. Write an (s) after the insoluble products and an (aq) after the soluble products.
2 HNO3(aq) + Ca(OH)2(aq)  Ca(NO3)2(aq) + 2 H2O(l)
Tro's Introductory Chemistry, Chapter 14

28. 1. Write the formulas of the reactants.
Example—When an Aqueous Solution of Sodium Carbonate Is Added to an Aqueous Solution of Nitric Acid, a Gas Evolves.
1. Write the formulas of the reactants.
Na2CO3(aq) + HNO3(aq) 
2. Determine the ions present when each reactant dissociates.
(Na+ + CO32−) + (H1 + NO3−) 
3. Exchange the ions.
(Na+ + CO32−) + (H+ + NO3−)  (Na+ + NO3−) + (H+ + CO32−)
Tro's Introductory Chemistry, Chapter 14

29. Na2CO3(aq) + HNO3(aq)  NaNO3 + H2CO3
Example—When an Aqueous Solution of Sodium Carbonate Is Added to an Aqueous Solution of Nitric Acid, a Gas Evolves, Continued.
4. Write the formulas of the products.
Cross charges and reduce.
Na2CO3(aq) + HNO3(aq)  NaNO3 + H2CO3
5. Check to see of product decomposes – Yes
H2CO3 decomposes into CO2(g) + H2O(l)
Na2CO3(aq) + HNO3(aq)  NaNO3 + CO2(g) + H2O(l)
Tro's Introductory Chemistry, Chapter 14

30. 7. Determine the solubility of other product.
Example—When an Aqueous Solution of Sodium Carbonate Is Added to an Aqueous Solution of Nitric Acid, a Gas Evolves, Continued.
6. Balance the equation.
Na2CO3(aq) + 2 HNO3(aq)  2 NaNO3 + CO2(g) + H2O(l)
7. Determine the solubility of other product.
NaNO3 is soluble (all Na+ are soluble).
8. Write an (s) after the insoluble products and an (aq) after the soluble products.
Na2CO3(aq) + 2 HNO3(aq)  2 NaNO3(aq) + CO2(g) + H2O(l)
Tro's Introductory Chemistry, Chapter 14

31. Practice–Complete Each Reaction.
Ca(OH)2(s) + H2SO3(aq) 
HClO3(aq) + Pb(OH)4(s) 
CaCO3(s) + HNO3(aq) 
Mg(HCO3)2(aq) + HC2H3O2(aq) 
Tro's Introductory Chemistry, Chapter 14

32. Practice–Complete Each Reaction, Continued.
Ca(OH)2(s) + H2SO3(aq)  CaSO3(s) H2O(l)
4 HClO3(aq) + Pb(OH)4(s)  Pb(ClO3)4(s) H2O(l)
CaCO3(s) + 2 HNO3(aq)  Ca(NO3)2(aq) + CO2(g) + 2 H2O(l)
Mg(HCO3)2(aq) + 2 HC2H3O2(aq)  Mg(C2H3O2)2(aq) + 2 CO2(g) + 2 H2O(l)
Tro's Introductory Chemistry, Chapter 14

33. Acid Reactions: Acids React with Metals
Acids react with many metals.
But not all!!
When acids react with metals, they produce a salt and hydrogen gas.
3 H2SO4(aq) + 2 Al(s) → Al2(SO4)3(aq) + 3 H2(g)
Tro's Introductory Chemistry, Chapter 14

34. Acid Reactions: Acids React with Metal Oxides
When acids react with metal oxides, they produce a salt and water.
3 H2SO4 + Al2O3 → Al2(SO4)3 + 3 H2O
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35. Tro's Introductory Chemistry, Chapter 14
Example 14.3a—Write an Equation for the Reaction of Hydroiodic Acid with Potassium Metal.
Write formulas of reactants.
HI(aq) + K(s) 
Identify the type of reaction and predict the pattern.
acid + metal  salt + H2(g)
Determine the charge on the cation.
K  K+
Determine the formula of the salt.
K+ + I−  KI
Tro's Introductory Chemistry, Chapter 14

36. Tro's Introductory Chemistry, Chapter 14
Example 14.3a—Write an Equation for the Reaction of Hydroiodic Acid with Potassium Metal, Continued.
5. Write the skeletal equation.
HI(aq) + K(s)  KI + H2(g)
6. Check the solubility of the salt.
KI is soluble.
7. Write (aq) if soluble; (s) if insoluble.
HI(aq) + K(s)  KI(aq) + H2(g)
8. Balance the equation.
2 HI(aq) + 2 K(s)  2 KI(aq) + H2(g)
Tro's Introductory Chemistry, Chapter 14

37. Tro's Introductory Chemistry, Chapter 14
Example 14.3b—Write an Equation for the Reaction of Hydrobromic Acid with Sodium Oxide(s).
1. Write formulas of reactants.
HBr(aq) + Na2O(s) 
2. Identify the type of reaction and predict the pattern.
Acid + metal oxide  salt + H2O(l)
3. Determine the charge on the cation.
Na+
4. Determine the formula of the salt.
Na+ + Br−  NaBr
Tro's Introductory Chemistry, Chapter 14

38. Tro's Introductory Chemistry, Chapter 14
Example 14.3b—Write an Equation for the Reaction of Hydrobromic Acid with Sodium Oxide(s), Continued.
5. Write the skeletal equation.
HBr(aq) + Na2O(s)  NaBr + H2O(l)
6. Check the solubility of the salt.
NaBr is soluble.
7. Write (aq) if soluble; (s) if insoluble.
HBr(aq) + Na2O(s)  NaBr(aq) + H2O(l)
8. Balance the equation.
2 HBr(aq) + Na2O(s)  2 NaBr(aq) + H2O(l)
Tro's Introductory Chemistry, Chapter 14

39. Practice—Complete and Balance the Following Reactions:
HCl(aq) + CaO(s) 
HCl(aq) + Ca(s) 
Tro's Introductory Chemistry, Chapter 14

40. Practice—Complete and Balance the Following Reactions, Continued:
2 HCl(aq) + CaO(s)  CaCl2(aq) + H2O(l)
2 HCl(aq) + Ca(s)  CaCl2(aq) + H2(g)
Tro's Introductory Chemistry, Chapter 14

41. Base Reactions
The reaction all bases have in common is neutralization of acids.
Strong bases will react with Al metal to form sodium aluminate and hydrogen gas.
2 NaOH + 2 Al + 6 H2O → 2 NaAl(OH)4 + 3 H2
Tro's Introductory Chemistry, Chapter 14

42. Tro's Introductory Chemistry, Chapter 14
Titration
Titration is a technique that uses reaction stoichiometry to determine the concentration of an unknown solution.
Titrant (unknown solution) is added from a buret.
Indicators are chemicals that are added to help determine when a reaction is complete.
The endpoint of the titration occurs when the reaction is complete.
Tro's Introductory Chemistry, Chapter 14

43. Tro's Introductory Chemistry, Chapter 14
Titration, Continued
Tro's Introductory Chemistry, Chapter 14

44. Tro's Introductory Chemistry, Chapter 14
Acid–Base Titration
The base solution is the
titrant in the buret.
As the base is added to
the acid, the H+ reacts with
the OH– to form water.
But there is still excess
acid present, so the color
does not change.
At the titration’s endpoint,
just enough base has been
added to neutralize all the
acid. At this point, the
indicator changes color:
[H+] = [OH–]
Tro's Introductory Chemistry, Chapter 14

45. Tro's Introductory Chemistry, Chapter 14
Example 14.4—What Is the Molarity of an HCl Solution if mL Is required to Titrate mL of M NaOH? NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l)
Given:
Find:
12.54 mL NaOH, mL HCl
M HCl
Solution Map:
Relationships:
M = mol/L, 1 mol NaOH = 1 mol HCl, 1 mL = 0.001L
mL HCl
L HCl
L NaOH
mol NaOH M
mol HCl mL
Solve:
Check:
The unit is correct, the magnitude is reasonable.
Tro's Introductory Chemistry, Chapter 14

46. Tro's Introductory Chemistry, Chapter 14
Example 14.4:
The titration of mL of HCl solution of unknown concentration requires mL of M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?
Tro's Introductory Chemistry, Chapter 14

47. Tro's Introductory Chemistry, Chapter 14
Example: The titration of mL of HCl solution of unknown concentration requires mL of M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?
Write down the given quantity and its units.
Given: mL HCl
12.54 mL NaOH
Tro's Introductory Chemistry, Chapter 14

48. Tro's Introductory Chemistry, Chapter 14
Example: The titration of mL of HCl solution of unknown concentration requires mL of M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?
Information:
Given: mL HCl
12.54 mL NaOH
Write down the quantity to find and/or its units.
Find: concentration HCl, M
Tro's Introductory Chemistry, Chapter 14

49. Collect needed equations and conversion factors:
Example: The titration of mL of HCl solution of unknown concentration requires mL of M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?
Information:
Given: mL HCl
12.54 mL NaOH
Find: M HCl
Collect needed equations and conversion factors:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
 1 mole HCl = 1 mole NaOH.
0.100 M NaOH 0.100 mol NaOH  1 L solution.
Tro's Introductory Chemistry, Chapter 14

50. Tro's Introductory Chemistry, Chapter 14
Example: The titration of mL of HCl solution of unknown concentration requires mL of M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?
Information:
Given: mL HCl
12.54 mL NaOH
Find: M HCl
Conversion Factors:
1 mol HCl = 1 mol NaOH
0.100 mol NaOH = 1 L
M = mol/L
Write a solution map: mL
NaOH L
NaOH
mol
NaOH
mol
HCl mL
HCl L
HCl
Tro's Introductory Chemistry, Chapter 14

51. Tro's Introductory Chemistry, Chapter 14
Example: The titration of mL of HCl solution of unknown concentration requires mL of M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?
Information:
Given: mL HCl
12.54 mL NaOH
Find: M HCl
Conversion Factors:
1 mol HCl = 1 mol NaOH
0.100 mol NaOH = 1 L
M = mol/L
Solution Map:
mL NaOH → L NaOH →
mol NaOH → mol HCl;
mL HCl → L HCl & mol  M
Apply the solution map:
= 1.25 x 10-3 mol HCl
Tro's Introductory Chemistry, Chapter 14

52. Tro's Introductory Chemistry, Chapter 14
Example: The titration of mL of HCl solution of unknown concentration requires mL of M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?
Information:
Given: mL HCl
12.54 mL NaOH
Find: M HCl
Conversion Factors:
1 mol HCl = 1 mol NaOH
0.100 mol NaOH = 1 L
M = mol/L
Solution Map:
mL NaOH → L NaOH →
mol NaOH → mol HCl;
mL HCl → L HCl & mol  M
Apply the solution map:
Tro's Introductory Chemistry, Chapter 14

53. Check the solution: HCl solution = 0.125 M
Example: The titration of mL of HCl solution of unknown concentration requires mL of M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?
Information:
Given: mL HCl
12.54 mL NaOH
Find: M HCl
Conversion Factors:
1 mol HCl = 1 mol NaOH
0.100 mol NaOH = 1 L
M = mol/L
Solution Map:
mL NaOH → L NaOH →
mol NaOH → mol HCl;
mL HCl → L HCl & mol  M
Check the solution:
HCl solution = M
The units of the answer, M, are correct.
The magnitude of the answer makes sense since
the neutralization takes less HCl solution than
NaOH solution, so the HCl should be more concentrated.

54. Tro's Introductory Chemistry, Chapter 14
Practice—What Is the Molarity of a Ba(OH)2 Solution if 37.6 mL Is Required to Titrate 43.8 mL of M HCl? Ba(OH)2(aq) + 2 HCl(aq)  BaCl2(aq) + 2 H2O(l)
Tro's Introductory Chemistry, Chapter 14

55. Tro's Introductory Chemistry, Chapter 14
Practice—What Is the Molarity of a Ba(OH)2 Solution if 37.6 mL Is Required to Titrate 43.8 mL of M HCl? Ba(OH)2(aq) + 2 HCl(aq)  BaCl2(aq) + 2 H2O(l), Continued
Given:
Find:
37.6 mL Ba(OH)2, 43.8 mL HCl
M Ba(OH)2
Solution Map:
Relationships:
M = mol/L, 1 mol Ba(OH)2= 2 mol HCl, 1 mL = 0.001L
mL Ba(OH)2
L Ba(OH)2
L HCl
mol HCl M
mol Ba(OH)2 mL
Solve:
Check:
The unit is correct, the magnitude is reasonable.
Tro's Introductory Chemistry, Chapter 14

56. Tro's Introductory Chemistry, Chapter 14
Strong or Weak
A strong acid is a strong electrolyte.
Practically all the acid molecules ionize, →.
A strong base is a strong electrolyte.
Practically all the base molecules form OH– ions, either through dissociation or reaction with water, →.
A weak acid is a weak electrolyte.
Only a small percentage of the molecules ionize, .
A weak base is a weak electrolyte.
Only a small percentage of the base molecules form OH– ions, either through dissociation or reaction with water, .
Tro's Introductory Chemistry, Chapter 14

57. Tro's Introductory Chemistry, Chapter 14
Strong Acids
The stronger the acid, the more willing it is to donate H.
Use water as the standard base.
Strong acids donate practically all their Hs.
100% ionized in water.
Strong electrolyte.
[H3O+] = [strong acid].
[ ] = molarity.
HCl ® H+ + Cl-
HCl + H2O® H3O+ + Cl-
Tro's Introductory Chemistry, Chapter 14

58. Strong Acids, Continued
Hydrochloric acid
HCl
Hydrobromic acid
HBr
Hydroiodic acid HI
Nitric acid
HNO3
Perchloric acid
HClO4
Sulfuric acid
H2SO4
Tro's Introductory Chemistry, Chapter 14

59. Strong Acids, Continued
Pure water
HCl solution
Tro's Introductory Chemistry, Chapter 14

60. Tro's Introductory Chemistry, Chapter 14
Weak Acids
Weak acids donate a small fraction of their Hs.
Most of the weak acid molecules do not donate H to water.
Much less than 1% ionized in water.
[H3O+] << [weak acid].
HF Û H+ + F-
HF + H2O Û H3O+ + F-
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61. Tro's Introductory Chemistry, Chapter 14
Weak Acids, Continued
Hydrofluoric acid HF
Acetic acid
HC2H3O2
Formic acid
HCHO2
Sulfurous acid
H2SO3
Carbonic acid
H2CO3
Phosphoric acid
H3PO4
Tro's Introductory Chemistry, Chapter 14

62. Tro's Introductory Chemistry, Chapter 14
Weak Acids, Continued
Pure water
HF solution
Tro's Introductory Chemistry, Chapter 14

63. Degree of Ionization
The extent to which an acid ionizes in water depends in part on the strength of the bond between the acid H+ and anion compared to the strength of the bond between the acid H+ and the O of water.
HA(aq) + H2O(l)  A−(aq) + H3O+(aq)
Tro's Introductory Chemistry, Chapter 14

64. Relationship Between Strengths of Acids and Their Conjugate Bases
The stronger an acid is, the weaker the attraction of the ionizable H for the rest of the molecule is.
The better the acid is at donating H, the worse its conjugate base will be at accepting an H.
Strong acid HCl + H2O → Cl– + H3O+ Weak conjugate base
Weak acid HF + H2O  F– + H3O+ Strong conjugate base

65. Example 14.5—Determine the [H3O+] in the Following Solutions:
1.5 M HCl
Since HCl is a strong acid, [H3O+] = [HCl] = 1.5 M.
3.0 M HC2H3O2
Since HC2H3O2 is a weak acid, [H3O+] << [HC2H3O2].
Therefore, [H3O+] << 3.0 M.
Tro's Introductory Chemistry, Chapter 14

66. Practice—Determine the [H3O+] in the Following Solutions:
Strong
Weak
Hydrochloric acid
HCl
Hydrobromic acid
HBr
Hydroiodic acid HI
Nitric
acid
HNO3
Perchloric acid
HClO4
Sulfuric acid
H2SO4
Hydrofluoric acid HF
Acetic
acid
HC2H3O2
Formic
HCHO2
Sulfurous acid
H2SO3
Carbonic acid
H2CO3
Phosphoric acid
H3PO4
0.5 M HI
0.1 M HCHO2

67. Practice—Determine the [H3O+] in the Following Solutions, Continued:
0.5 M HI
Since HI is a strong acid, [H3O+] = [HI] = 0.5 M.
0.1 M HCHO2
Since HCHO2 is a weak acid, [H3O+] << [HCHO2].
Therefore, [H3O+] << 0.1 M.
Tro's Introductory Chemistry, Chapter 14

68. Tro's Introductory Chemistry, Chapter 14
Strong Bases
The stronger the base, the more willing it is to accept H.
Use water as the standard acid.
Strong bases, practically all molecules are dissociated into OH– or accept Hs.
Strong electrolyte.
Multi-OH bases completely dissociated.
[HO–] = [strong base] x (# OH).
NaOH ® Na+ + OH-
Tro's Introductory Chemistry, Chapter 14

69. Strong Bases, Continued
Lithium hydroxide
LiOH
Sodium hydroxide
NaOH
Potassium hydroxide
KOH
Calcium hydroxide
Ca(OH)2
Strontium hydroxide
Sr(OH)2
Barium hydroxide
Tro's Introductory Chemistry, Chapter 14

70. Tro's Introductory Chemistry, Chapter 14
Weak Bases
In weak bases, only a small fraction of molecules accept Hs.
Weak electrolyte.
Most of the weak base molecules do not take H from water.
Much less than 1% ionization in water.
[HO–] << [strong base].
NH3 + H2O Û NH4+ + OH-
Tro's Introductory Chemistry, Chapter 14

71. Tro's Introductory Chemistry, Chapter 14
Weak Bases, Continued
Ammonia
NH3(aq) + H2O(l)  NH4+(aq) + OH−(aq)
Pyridine
C5H5N(aq) + H2O(l)  C5H5NH+(aq) + OH−(aq)
Methyl amine
CH3NH2(aq) + H2O(l)  CH3NH3+(aq) + OH−(aq)
Ethyl amine
C2H5NH2(aq) + H2O(l)  C2H5NH3+(aq) + OH−(aq)
Bicarbonate
HCO3−(aq) + H2O(l)  H2CO3 (aq) + OH−(aq)
Tro's Introductory Chemistry, Chapter 14

72. Example 14.6—Determine the [OH−] in the Following Solutions:
2.25 M KOH
Since KOH is a strong base, [OH−] = [KOH] x 1 = 2.25 M.
0.35 M CH3NH2
Since CH3NH2 is a weak base, [OH−] << [CH3NH2].
Therefore, [OH−] << 0.35 M.
0.025 M Sr(OH)2
Since Sr(OH)2 is a strong base, [OH−] = [Sr(OH)2] x 2 = M.
Tro's Introductory Chemistry, Chapter 14

73. Practice—Determine the [OH−] in the Following Solutions:
Strong
Weak
Lithium hydroxide
LiOH
Sodium hydroxide
NaOH
Potassium hydroxide
KOH
Calcium hydroxide
Ca(OH)2
Strontium hydroxide
Sr(OH)2
Barium hydroxide
Ba(OH)2
Ammonia
NH3
Pyridine
C5H5N
Methyl amine
CH3NH2
Ethyl amine
C2H5NH2
Bicarbonate
H2CO3
0.05 M Ba(OH)2
0.01 M C5H5N

74. Practice—Determine the [OH−] in the Following Solutions:
0.05 M Ba(OH)2
Ba(OH)2 is a strong base.
[OH−] = [Ba(OH)2] x 2 = 0.1 M.
0.01 M C5H5N
C5H5N is a weak base, [OH−] << [C5H5N].
Therefore, [OH−] << 0.01 M.
Tro's Introductory Chemistry, Chapter 14

75. Autoionization of Water
Water is actually an extremely weak electrolyte.
Therefore, there must be a few ions present.
About 1 out of every 10 million water molecules form ions through a process called autoionization.
H2O Û H+ + OH–
H2O + H2O Û H3O+ + OH–
All aqueous solutions contain both H3O+ and OH–.
The concentration of H3O+ and OH– are equal in water.
[H3O+] = [OH–] = 1 x 10-7M at 25 °C in pure water.
Tro's Introductory Chemistry, Chapter 14

76. Tro's Introductory Chemistry, Chapter 14
Ion Product of Water
The product of the H3O+ and OH– concentrations is always the same number.
The number is called the ion product of water and has the symbol Kw.
[H3O+] x [OH–] = 1 x = Kw.
As [H3O+] increases, the [OH–] must decrease so the product stays constant.
Inversely proportional.
Tro's Introductory Chemistry, Chapter 14

77. Acidic and Basic Solutions
Neutral solutions have equal [H3O+] and [OH–].
[H3O+] = [OH–] = 1 x 10-7
Acidic solutions have a larger [H3O+] than [OH–].
[H3O+] > 1 x 10-7; [OH–] < 1 x 10-7
Basic solutions have a larger [OH–] than [H3O+].
[H3O+] < 1 x 10-7; [OH–] > 1 x 10-7
Tro's Introductory Chemistry, Chapter 14

78. Ba(OH)2 = Ba2+ + 2 OH– therefore:
Example—Determine the [H3O+] for a M Ba(OH)2 and Determine Whether the Solution Is Acidic, Basic, or Neutral.
Ba(OH)2 = Ba OH– therefore:
[OH–] = 2 x = = 4.0 x 10−4 M
[H3O+] = 2.5 x M.
Since [H3O+] < 1 x 10−7, the solution is basic.
Tro's Introductory Chemistry, Chapter 14

79. Tro's Introductory Chemistry, Chapter 14
Practice—Determine the [H3O+] Concentration and Whether the Solution Is Acidic, Basic, or Neutral for the Following:
[OH–] = M
[OH–] = 3.50 x 10-8 M
Ca(OH)2 = 0.20 M
Tro's Introductory Chemistry, Chapter 14

80. Tro's Introductory Chemistry, Chapter 14
Practice—Determine the [H3O+] Concentration and Whether the Solution Is Acidic, Basic, or Neutral for the Following, Continued:
[OH–] = M
[OH–] = 3.50 x 10-8 M
Ca(OH)2 = 0.20 M
[H3O+] =
1 x 10-14
2.50 x 10-4
= 4.00 x 10-11
[H3O+] < [OH-1], therefore, base.
[H3O+] =
1 x 10-14
3.50 x 10-8
= 2.86 x 10-7
[H3O+] > [OH-1], therefore, acid.
[OH-1] = 2 x 0.20 = 0.40 M
[H3O+] =
1 x 10-14
4.0 x 10-1
= 2.5 x 10-14
[H3O+] < [OH-1], therefore, base.
Tro's Introductory Chemistry, Chapter 14

81. Complete the Table [H+] vs. [OH-]
Tro's Introductory Chemistry, Chapter 14

82. Complete the Table [H+] vs. [OH-]
Acid
Base
[H+]
OH- H+
[OH-]
Even though it may look like it, neither H+ nor OH- will ever be 0.
The sizes of the H+ and OH- are not to scale
because the divisions are powers of 10 rather than units.
Tro's Introductory Chemistry, Chapter 14

83. Tro's Introductory Chemistry, Chapter 14pH
The acidity/basicity of a solution is often expressed as pH.
pH = ─log[H3O+], [H3O+] = 10−pH
Exponent on 10 with a positive sign.
pHwater = −log[10-7] = 7.
Need to know the [H+] concentration to find pH.
pH < 7 is acidic; pH > 7 is basic; pH = 7 is neutral.
Tro's Introductory Chemistry, Chapter 14

84. pH, Continued
The lower the pH, the more acidic the solution; the higher the pH, the more basic the solution.
1 pH unit corresponds to a factor of 10 difference in acidity.
Normal range is 0 to 14.
pH 0 is [H+] = 1 M, pH 14 is [OH–] = 1 M.
pH can be negative (very acidic) or larger than 14 (very alkaline).

85. pH of Common Substances
1.0 M HCl
0.0
0.1 M HCl
1.0
Stomach acid
1.0 to 3.0
Lemons
2.2 to 2.4
Soft drinks
2.0 to 4.0
Plums
2.8 to 3.0
Apples
2.9 to 3.3
Cherries
3.2 to 4.0
Unpolluted rainwater
5.6
Human blood
7.3 to 7.4
Egg whites
7.6 to 8.0
Milk of magnesia (saturated Mg(OH)2)
10.5
Household ammonia
10.5 to 11.5
1.0 M NaOH 14

86. Example—Calculate the pH of a 0
Example—Calculate the pH of a M Ba(OH)2 Solution and Determine if It Is Acidic, Basic, or Neutral.
Ba(OH)2 = Ba OH− therefore,
[OH-] = 2 x = = 2.0 x 10-3 M.
[H3O+] =
1 x 10-14
2.0 x 10-3
= 5.0 x 10-12M
pH = −log [H3O+] = −log (5.0 x 10-12)
pH = 11.3
pH > 7 therefore, basic.
Tro's Introductory Chemistry, Chapter 14

87. Tro's Introductory Chemistry, Chapter 14
Practice—Calculate the pH of the Following Strong Acid or Base Solutions.
M HCl
M Ca(OH)2
0.25 M HNO3
Tro's Introductory Chemistry, Chapter 14

88. Tro's Introductory Chemistry, Chapter 14
Practice—Calculate the pH of the Following Strong Acid or Base Solutions, Continued.
M HCl therefore, [H3O+] = M.
M Ca(OH)2 therefore, [OH–] = M.
0.25 M HNO3 therefore, [H3O+] = 0.25 M.
pH = −log (2.0 x 10-3) = 2.70
[H3O+] =
1 x 10−14
1 x 10−2
= 1.0 x 10−12
pH = −log (1.0 x 10−12) = 12.00
pH = −log (2.5 x 10−1) = 0.60
Tro's Introductory Chemistry, Chapter 14

89. Tro's Introductory Chemistry, Chapter 14
Complete the Table: pH pH
[H+]
OH- H+
[OH-]
Tro's Introductory Chemistry, Chapter 14

90. Complete the Table: pH, Continued
Acid
Base pH
[H+]
OH- H+
[OH-]
Tro's Introductory Chemistry, Chapter 14

91. Tro's Introductory Chemistry, Chapter 14
Example—Calculate the Concentration of [H3O+] for a Solution with pH 3.7.
[H3O+] = 10-pH
[H3O+] =
means < [H+1] <
[H3O+] = 2 x 10-4 M = M.
Tro's Introductory Chemistry, Chapter 14

92. Practice—Determine the [H3O+] for Each of the Following:
pH = 2.7
pH = 12
pH = 0.60
Tro's Introductory Chemistry, Chapter 14

93. Practice—Determine the [H3O+] for Each of the Following, Continued:
pH = 2.7
pH = 12
pH = 0.60
[H3O+] = 10−2.7 = 2 x 10−3 M = M
[H3O+] = 10−12 = 1 x M
[H3O+] = 10−0.60 = 0.25 M
Tro's Introductory Chemistry, Chapter 14

94. Tro's Introductory Chemistry, Chapter 14
pOH
The acidity/basicity of a solution may also be expressed as pOH.
pOH = ─log[OH−], [OH−] = 10−pOH
Exponent on 10 with a positive sign.
pOHwater = −log[10−7] = 7.
Need to know the [OH−] concentration to find pOH.
pOH < 7 is acidic; pOH > 7 is basic, pOH = 7 is neutral.
Tro's Introductory Chemistry, Chapter 14

95. Tro's Introductory Chemistry, Chapter 14
pOH, Continued
The lower the pOH, the more basic the solution; the higher the pOH, the more acidic the solution.
1 pOH unit corresponds to a factor of 10 difference in basicity.
Normal range is 0 to 14.
pOH 0 is [OH−] = 1 M; pOH 14 is [H3O+] = 1 M.
pOH can be negative (very basic) or larger than 14 (very acidic).
pH + pOH =
Tro's Introductory Chemistry, Chapter 14

96. Complete the Table: pOH
Tro's Introductory Chemistry, Chapter 14

97. Complete the Table: pOH, Continued
Acid
Base
[H+]
OH- H+
[OH-]
pOH
Tro's Introductory Chemistry, Chapter 14

98. Example—Calculate the pH of a 0
Example—Calculate the pH of a M Ba(OH)2 Solution and Determine if It Is Acidic, Basic, or Neutral.
Ba(OH)2 = Ba OH− therefore,
[OH−] = 2 x = = 2.0 x 10−3 M.
pOH = -log [OH−] = -log (2.0 x 10−3)
pOH = 2.70
pH = pOH =
pH = 11.30
pH > 7 therefore, basic.
Tro's Introductory Chemistry, Chapter 14

99. Tro's Introductory Chemistry, Chapter 14
Practice—Calculate the pOH and pH of the Following Strong Acid or Base Solutions.
M KOH
M Ca(OH)2
0.25 M HNO3
Tro's Introductory Chemistry, Chapter 14

100. Tro's Introductory Chemistry, Chapter 14
Practice—Calculate the pOH and pH of the Following Strong Acid or Base Solutions, Continued.
M KOH therefore, [OH–] = M.
M Ca(OH)2 therefore, [OH–] = M.
0.25 M HNO3 therefore, [H3O+] = 0.25 M.
pOH = −log (2.0 x 10-3) = 2.70
pH = – 2.70 = 11.30
pOH = −log (1.0 x 10-2) = 2.00
pH = – 2.00 = 12.00
pH = −log (2.5 x 10-1) = 0.60
pOH = – 0.60 = 13.40
Tro's Introductory Chemistry, Chapter 14

101. Tro's Introductory Chemistry, Chapter 14
Buffers
Buffers are solutions that resist changing pH when small amounts of acid or base are added.
They resist changing pH by neutralizing added acid or base.
Buffers are made by mixing together a weak acid and its conjugate base.
Or weak base and its conjugate acid.
Tro's Introductory Chemistry, Chapter 14

102. Tro's Introductory Chemistry, Chapter 14
How Buffers Work
The weak acid present in the buffer mixture can neutralize added base.
The conjugate base present in the buffer mixture can neutralize added acid.
The net result is little to no change in the solution pH.
Tro's Introductory Chemistry, Chapter 14

103. How Buffers Work, Continued
New HA HA HA A− A− 
H3O+ +
Added
H3O+
Tro's Introductory Chemistry, Chapter 14

104. How Buffers Work, Continued
New A− A− HA A− HA 
H3O+ +
Added
HO−
Tro's Introductory Chemistry, Chapter 14

105. A Buffer Made from Acetic Acid and Sodium Acetate
A buffer solution with a pH of 4.75 can be made by mixing equal volumes of 1 M HC2H3O2 and 1 M NaC2H3O2.
Adding 10 mL of 0.1 M HCl to 1 L of this solution will give a solution with a pH of 4.75.
Adding 10 mL of 0.1 M HCl to 1 L of distilled water will give a solution with pH of 3.0.
Adding 10 mL of 0.1 M NaOH to 1 L of this solution will give a solution with a pH of 4.75.
Adding 10 mL of 0.1 M NaOH to 1 L of distilled water will give a solution with pH of 11.0.
Tro's Introductory Chemistry, Chapter 14

106. Acetic Acid/Acetate Buffer
Tro's Introductory Chemistry, Chapter 14

107. Nonmetal Oxides Are Acidic
Nonmetal oxides react with water to form acids.
Causes acid rain.
CO2 (g) + H2O(l) → H2CO3(aq)
2 SO2(g) + O2(g) + 2 H2O(l) → 2 H2SO4(aq)
4 NO2(g) + O2(g) + 2 H2O(l) → 4 HNO3(aq)
Tro's Introductory Chemistry, Chapter 14

108. Tro's Introductory Chemistry, Chapter 14
What Is Acid Rain?
Natural rain water has a pH of 5.6.
Naturally slightly acidic due mainly to CO2.
Rain water with a pH lower than 5.6 is called acid rain.
Acid rain is linked to damage in ecosystems and structures.
Tro's Introductory Chemistry, Chapter 14

109. Tro's Introductory Chemistry, Chapter 14
What Causes Acid Rain?
Many natural and pollutant gases dissolved in the air are nonmetal oxides.
CO2, SO2, NO2.
Nonmetal oxides are acidic.
CO2 + H2O  H2CO3
2 SO2 + O2 + 2 H2O  2 H2SO4
Processes that produce nonmetal oxide gases as waste increase the acidity of the rain.
Natural—volcanoes and some bacterial action.
Man-made—combustion of fuel.
Weather patterns may cause rain to be acidic in regions other than where the nonmetal oxide is produced.
Tro's Introductory Chemistry, Chapter 14

110. pH of Rain in Different Regions

111. Sources of SO2 from Utilities

112. Tro's Introductory Chemistry, Chapter 14
Damage from Acid Rain
Acids react with metals and materials that contain carbonates.
Acid rain damages bridges, cars, and other metallic structures.
Acid rain damages buildings and other structures made of limestone or cement.
Acidifies lakes affecting aquatic life.
Dissolves and leaches more minerals from soil.
Making it difficult for trees.
Tro's Introductory Chemistry, Chapter 14

113. Tro's Introductory Chemistry, Chapter 14
Damage from Acid Rain
circa 1995
circa 1935
Tro's Introductory Chemistry, Chapter 14

114. pH of Rain in Different Regions
Tro's Introductory Chemistry, Chapter 14

115. Sources of SO2 from Utilities

116. Tro's Introductory Chemistry, Chapter 14
Acid Rain Legislation
1990 Clean Air Act attacks acid rain.
Forces utilities to reduce SO2.
The result is acid rain in the northeast is stabilized and begins to be reduced.
Tro's Introductory Chemistry, Chapter 14