Acids and Bases

1
Chapter 14

• Acids and Bases

2
Naming Acids

• 2 types of acids
• Binary
• ternary (sometimes called oxy)
• binary -H and one other type of atom
• name them hydro _________ ic acid

3
Naming Acids

• Ex1 HCl
• Hydrochloric Acid
• Ex2 HBr
• Hydrobromic Acid
• Ex3 H3P
• Hydrophosphoric Acid

4
Writing formulas from names for Acids

• Criss Cross charges
• Ex4 Hydronitric Acid
• H3N
• Ex5 Hydrosulfuric Acid
• H2S

5
Naming Acids

• ternary (oxy) acids
• H with a polyatomic ion
• Do not start with Hydro-
• Change the ate ending to ic
• Change the ite ending to ous

6
Naming Acids

• Ex6 H2SO4
• Sulfuric Acid
• Ex7 H2SO3
• Sulfurous Acid
• Ex8 HClO4
• Perchloric Acid
• Ex9 HClO
• Hypochlorous Acid

7
Writing Formulas From Names

• Ex10 Nitric Acid
• HNO3
• Ex11 Phosphorous Acid
• H3PO3

8
Some common acids

• Sulfuric used for fertilizer, petroleum,
  production of metal, paper, paint
• HCl stomach acid, food processing, iron, steel
• Acetic acid vinegar, fungicide, produced by
  fermentation
• Nitric acid explosives, rubber, plastics, dyes,
  drugs
• Phosphoric acid beverage flavoring, animal
  feed, detergents

9
Properties of Acids

• Acid comes from Latin meaning acidus, or sour
  tasting.
• Affect the colors of indicators. An indicator
  is a chemical that shows one color in an acid and
  another in a base. Acids turn blue litmus red.

10
Properties of Acids

• Acids react with bases to produce salt and water.
  This is called neutralization.
• HCl(aq) NaOH(aq)? NaCl(aq) HOH (l)
• 3H2SO4(aq) 2Al(OH)3(aq) ? Al2(SO4)3(aq)
  6 HOH (aq)

11
Properties of Acids

• Acids ionize in water. So, they conduct
  electricity (electrolytes).
• Acids react with active metals to produce salts
  and hydrogen.
• Mg 2 HCl (aq) ? MgCl 2(aq) H 2(g)
• Cu(s) HCl(aq) ? NR

12
Arrhenius Acids

• Substances that produces H ions when mixed with
  water.
• HCl(g) H2O(l) ? H1(aq) Cl-1 (aq)
• It is now found that
• H1(aq) H2O(l) ? H3O1 (aq)
• so it is really
• HCl(g) H2O(l) ? H3O1 (aq) Cl-1 (aq)

13
Definitions of Acids

• Bronsted-Lowery Acids proton donors
• Show HCl water and HCl ammonia
• HCl H2O ? H3O(aq) Cl-1(aq)
• HCl NH3 ? NH4(aq) Cl-1(aq)
  (not Cl2!!)

14
Types of Acids

• Strong - HI, HBr, HCl, HNO3, H2SO4, HClO4 (one
  way arrows always!)
• HBr H2O ? H3O (aq) Br-1 (aq)
• Weak HF, H2PO4, H2CO3, H2PO4 (double arrows
  always!)
• HF (aq) H2O (l) ? H3O1 (aq) F-1 (aq)

15
Molecular, Total Ionic, and Net Ionic Equations
for acids

• Molecular Equation
• Zn(s) 2 HCl(aq) ? ZnCl2(aq) H 2(g)
• Total Ionic Equation
• Zn(s) 2H1(aq) 2Cl-1 (aq) ?
  Zn2(aq) 2Cl-1(aq) H2(g)
• Net Ionic Equation
• Zn(s) 2H1(aq) ? Zn2(aq) H2(g)

16
Some acids donate more than 1 proton.

• Monoprotic (HF) - an acid that donates one proton
  (one hydrogen)
• Ex1 Write the reaction(s) showing the complete
  ionization of HF.
• HF (aq) H2O (l) ? H3O1 (aq) F-1 (aq)

17
Some acids donate more than 1 proton.

• Diprotic (H2SO4) - an acid that donates two
  protons (two hydrogens)
• Ex2 Write the reaction(s) showing the complete
  ionization of H2SO4.
• H2SO4 (aq) H2O (l) ? H3O1 (aq) HSO4-1
  (aq)
• HSO4-1 (aq) H2O(l) ? H3O1(aq) SO4-2 (aq)
• __________________________________
• H2SO4 (aq) 2 H2O(l) ? 2 H3O 1(aq)SO4-2(aq)
• Note When you lose a H1, you gain a negative.

18
Some acids donate more than 1 proton.

• Triprotic Acid an acid that donates three
  protons (three hydrogens).
• Ex3 Write the reaction(s) showing the complete
  ionization of H3PO4.
• H3PO 4(aq) H2O (l) ? H3O 1 (aq) H2PO4
  -1 (aq)
• H2PO4 -1 (aq) H2O (l) ? H3O 1 (aq)
  HPO4 -2 (aq)
• HPO4 -2 (aq) H2O (l) ? H3O 1 (aq)
  PO4 -3 (aq)
• ________________________________________
• H3PO 4(aq) 3 H2O (l) ? 3 H3O 1 (aq) PO4
  -3 (aq)

19
Some acids donate more than 1 proton.

• Diprotic and Triprotic can also be referred to as
  polyprotic.
• 2nd and 3rd ionizations are always weak (so, ?).

20
Bases

• Bases are used in cleaners (floors, drains,
  ovens), react with fats and oils so they become
  water soluble, used to neutralize stomach acid
  (antacids), used as laxatives

21
Properties of Bases

• Bases are electrolytes. They dissociate in
  water. NaOH and KOH are strong electrolytes
  because they are both highly soluble.
•  
• Affect the colors of indicators. An indicator is
  a chemical that shows one color in an acid and
  another in a base. Bases turn red litmus blue.
• Bases react with acids to produce salt and water.
  This is called neutralization.
•  
• Bases taste bitter and feel slippery. Soap is
  an example of a base.

22
Definition of Bases

• A substance that has OH- ions. Bases dissociate
  in water to give OH- positive metal ions.

23
Types of Bases

• Traditional Bases (Arrhenius) a substance that
  contains hydroxide ions and dissociates to give
  hydroxide ions in water.
• NaOH(s) H2O ? Na(aq) OH-(aq)
• Mg(OH)2(s) H2O ?
  Mg2(aq) 2OH-(aq)

24
Types of Bases

• 2. Bronsted- Lowry bases proton acceptors
•  
• NH3(g) H2O(l) ? NH41(aq)
  OH-1(aq)
• Water is amphoteric. It can act as an acid
• or base.

25
Types of Bases
Hydroxides of Column I and II are strong bases

• List of strong bases
• NaOH, KOH, CsOH, Ca(OH)2
• List of weak bases
• many organic compounds with NNH3, C6H5NH2,
  C2H3O2-

26
Neutralization reactions hydronium
hydroxide yields water

• It is a type of double replacement reaction.
• Note H2O HOH
• Acid Base ? Salt and Water
• General Formula
• HX MOH ? MX H2O

27
Neutralization Reaction

• Example hydrochloric acid barium hydroxide (
  molecular, total ionic, net ionic)
• 2HCl(aq) Ba(OH)2(aq) ? BaCl2(aq) 2HOH(l)
• 2H1(aq) 2Cl-1(aq) Ba2(aq) 2OH-1(aq) ?
  Ba2(aq) 2Cl-1(aq) 2HOH(l)
• 2H1(aq) 2OH-1(aq) ? 2HOH(l)

28
Relative Strengths of Acids and Bases

• Some acids and bases are stronger than others.
• Bronsted (Danish) and Lowry (English)
  independently discovered that acids are proton
  donors and bases are proton acceptors. A proton
  is a hydrogen ion.

29
Relative Strengths of Acids and Bases

• Strong Acid Example
• HCl(g) H2O(l) ? H3O1(aq) Cl-1(aq)
• (Acid) (Base)
• Weak Base Example
• NH3(g) H2O(l) ? NH41(aq) OH-1(aq)
• (Base) (Acid)
• Remember water is amphoteric!

30
Conjugate Acids Bases

• Conjugate Acid the substance that was the base
  and now acts as an acid.
• Conjugate Base the substance that was the acid
  and now acts as a base.
• HCl(g) H2O(l) ? H3O1(aq)
  Cl-1(aq)
• (Acid) (Base) (Conjugate Acid)
  (Conjugate Base)
•  

31
Conjugate Acids Bases

• NH3(g) H2O(l) ? NH41(aq)
  OH-1(aq)
•  
• HF(l) H2O(l) ? H3O1(aq)
  F-1(aq)

32
Conjugate Acids and bases

• H2CO3(aq) H2O(l) ? H3O1(aq) HCO3-1(aq)
• The stronger the acid, the weaker the conjugate
  base.
• Proton transfer reactions favor the production of
  the weaker acid and the
• weaker base.

33
Conj. Acid/Base Practice

• Complete the equation and label acid base pairs
• HSO4-1(aq) HCO3-1(aq) ?
• Write an equation showing how NH2-1 is a stronger
  base than HSO4-1

34
Conj. Acid/Base Practice

• Which one is correct?
• HSO4-1(aq) H3O1(aq) ? H2SO4(aq)
  H2O(l)
•   (Base) (Acid) (Conj. Acid)
  (Conj. Base)
• or
• HSO4-1(aq) OH-1(aq) ? SO4-2(aq)
  H2O(l)
• (Acid) (Base) (Conj. Base)
  (Conj. Acid)

The second reaction is favored because a weaker
conjugate acid/base is produced.
35
Stuff to know for Acids and Bases

• 2nd and 3rd ionizations are always weak. This
  means a double yield sign (?).
• Memorize these strong acids. Strong means a
  single yield sign (?).
• HI, HBr, HCl, HNO3, H2SO4, HClO4
• All other acids get double yield signs.
• Strong bases include metals from column 1 and
  column 2 (below magnesium).
• Proton reactions favor the formation of the
  weaker acid and base.

36
Reactions of Acids and Bases

• Neutralization (double replacement)
• Acid Base ? Salt
  Water
• HX MOH ? MX H2O
• Ex1
• HCl(aq) NaOH(aq) ? NaCl(aq) HOH(l)

37
Reactions of Acids and Baes

• Acid Metal (single replacement)
• Metal Acid ? Salt Hydrogen
• Ex2
• Mg(s) 2 HCl(aq) ? MgCl2(aq) H2(g)

38
Reactions of Acids and Bases

• Acid in water
• Acid Water ? Hydronium Ion Negative Ion
• Ex3
• HCl(g) H2O(l) ? H3O1(aq) Cl-1(aq)

39
Reactions of Acids and Bases

• Traditional Base (ends with OH) in water
  (dissociation)
• Base Water ? Positive Ion Hydroxide
• Ex4
• Fe(OH)3(s) H2O(l) ? Fe3(aq) 3 OH-1(aq)

40
Reactions of Acids and Bases

• Formation of acids and bases from anhydrides -
  synthesis (anhydride without water)
• Nonmetal oxide water ? acid
• Ex5a
• CO2(g) H2O(l) ? H2CO3(aq)
• Ex5b
• SO3(g) H2O(l) ? H2SO4(aq)
• Note just add the nonmetal oxide to the water
  to determine the product.

41
Reactions of Acids and Bases

• Metal oxide water ? base
• Ex5c
• Na2O(s) H2O(l) ? 2 NaOH(aq)
• Ex5d
• MgO(s) H2O(l) ? Mg(OH)2(aq)

42
Reactions of Acids and Bases

• Acid and metal oxide (really just an acid and a
  base)
• Acid Metal Oxide ? Salt
  Water
• Ex6
• H2SO4(aq) CuO(s) ?
• Turn CuO into Cu(OH)2
• H2SO4(aq) Cu(OH)2 (aq) ? CuSO4(aq)
  HOH(l)
• Now, re-write the original reactants, new
  products, and balance.
• H2SO4(aq) CuO(s) ? CuSO4(aq) H2O(l)

43
Reactions of Acids and Bases

• Base and nonmetal oxide (really just an acid and
  a base
• Base Nonmetal Oxide ? Salt Water
• Ex7a
• CO2(g) NaOH(aq) ? NaHCO3(aq)

44
Reactions of Acids and Bases

• This is a little confusing. So these reactions
  will be done like
• CO2(g) NaOH(aq) ?
• Turn CO2 into H2CO3
• H2CO3(aq) NaOH(aq) ? Na2CO3(aq) HOH(l)
• Now, re-write the original reactants, new
  products, and balance.
• CO2(g) 2 NaOH(aq) ? Na2CO3(aq) H2O(l)

45
Reactions of Acids and Bases

• Ex7b
• 2 CO2(g) Ca(OH)2(aq) ? Ca(HCO3)2(aq)
• CO2(g) Ca(OH)2(aq) ?
• Turn CO2 into H2CO3
• H2CO3(aq) Ca(OH)2(aq) ? CaCO3(aq) HOH(l)
• Now, re-write the original reactants, new
  products, and balance.
• CO2(g) Ca(OH)2(aq) ? CaCO3(aq) H2O(l)

46
Reactions of Acids and Bases

• Metal oxide and nonmetal oxide
• It is like an acid base reaction they yield salt.
  However, it does not produce water since no
  hydrogen is involved.
• Ex8a
• MgO(s) CO2(g) ? MgCO3(s)
• Note just add the nonmetal oxide to the metal
  oxide to determine the product.
• Ex8b
• CuO(s) SO3(g) ? CuSO4(s)

47
Aqueous Solutions and the Concept of pH

• Tap water conducts electricity why? many ions
  present
• examples
• Distilled water appears to not conduct
  electricity, but it does just a little, tiny
  bit
• H2O H2O ? H3O1 OH-1

48
Aqueous Solutions and the Concept of pH

• The normal way to express the quantity of
  hydronium and hydroxide ions is in moles/L (M)
• At 25 C0, H3O1 1 x 10-7 so OH-1
  1 x 10-7
• These numbers are constant in neutral solution,
  so we can multiply them to get a constant
• We call this constant Kw - ionization constant
  for water
• Kw H3O1OH-1

49
Aqueous Solutions and the Concept of pH

• At 25 C0, H3O1 1 x 10-7
• so OH-1 1 x 10-7
• so Kw 1 x 10-14
• Example If the H3O1 is 1 x 10-3M, then what
  is the OH-1?
•  
• The solution is acidic because the hydronium ion
  concentration is greater than the hydroxide
  concentration.

50
Kw Practice
Fill in the table below
Beaker H3O1 OH-1 Acid or Base
1 1 x 10-5 1 x 10-9 Acid
2 1 x 10-12 1 x 10-2 Base
3 2 x 10-4 5 x 10-11 Acid
4 2.40 x 10-9 4.16 x 10-6 Base
51
Aqueous Solutions and the Concept of pH
pH stands for parts per million of Hydrogen ion.
52
How to calculate strengths of Acids and Bases
pH - log 10 H3O
pOH - log 10 OH-
H3O 10 pH
OH- 10-pOH
pH pOH 14
53
How to calculate strengths of Acids and Bases

• Log Review
• Logs are functions of exponents
• ex. log of 1000
• ex. log of .01

54
How to calculate strengths of Acids and Bases

• To convert H3O1 to pH
• pH - log 10 H3O
• log, , enter, then make it positive change the
  sign)

H3O1 pH
1.0 x10 1  
1.0 x 10 2  
3.0 x 10 4  
55
How to calculate strengths of Acids and Bases

• To convert pH to H3O1
• H3O 10 pH
• 2nd, log, (-), , enter

H3O1 pH Acid or Base
2
11
5.22
56
How to calculate strengths of Acids and Bases
H OH- pH pOH Acid or Base
2 x 10-5
3.5 x 10-5
3.25
8.12
8.20
0.0016
2.8 x 10-11
57
Concentration units for Acids and Bases

• Chemical Equivalents quantities of solutes that
  have equivalent combining capactieies.
• Ex1 HCl NaOH ? NaCl
  H2O
• To achieve a balance, 1 mole H needs to cancel
  out 1 mole OH-1.
• So, for the above reaction
• 1 mole of HCl is necessary to balance 1 mole of
  NaOH.
• 1 mole HCl 1 mole NaOH
  ?

58
Concentration units for Acids and Bases

• Ex2 H2SO4 2 NaOH ? Na2SO4 2
  H2O
• To achieve a balance, 1 mole H needs to cancel
  out 1 mole OH-1.
• So, for the above reaction
• 1/2 mole of H2SO4 is necessary to balance 1 mole
  of NaOH.
• ½ mole H2SO4 1 mole NaOH
• ?
•  

59
Concentration units for Acids and Bases

• Ex3 To make H3PO4 chemically equivalent to
  NaOH, 1/3 mole of H3PO4 balances 1 mole NaOH

60
Equivalent Weight

• the of grams of acid or base that will provide
  1 mole of protons or hydroxide ions.

HCl H2SO4 H3PO4
Moles of Acid 1 ½ 1/3
Moles of Hydrogen 1 1 1
Equivalent Weights 36.5 49.05 32.7
61
Equivalent Weight

• Formula for calculating equivalent weight
• Eq. wt. MW / equivalents
• MW molecular weight
• Formula for calculating equivalents
• equivalents (moles)(n)
• n of H or OH in the chemical formula

62
Calculations

• Example 1 How many equivalents in 9.30 g of
  H2CO3?
• Step 1 Calculate the molecular weight of H2CO3
• H (2)1.0 2.0
• C (1)12.0 12.0
• O (3)16.0 48.0
• mw 62.0 g/mole
•  

63
Calculations

• Step 2 Calculate the of moles in 9.30 g of
  H2CO3
• 9.30 g H2CO3 x 1 mole H2CO3 .150 mole
  62.0 g H2CO3 H2CO3
•  
• Step 3 Calculate the number of equivalents
• equivalents (moles)(n)
• n of H or OH in the chemical formula
• eq (.150 moles)(2) ? eq .300 eq

64
Calculations

• Ex 2 Calculate the equivalent weight of 9.30 g
  H2CO3
• (Use 00.300 equivalents calculated above)
• eq. wt. mw/eq
• eq. wt. 62.0 g / .300 equivalents
• eq. wt. 206.7 g/eq

65
Calculations

• Ex 3 How many grams of H2CO3 would equal .290
  equivalents?
• Step 1 Convert to moles
  equivalents (moles)(n)
• moles .29 / 2 ? moles .145
  moles
• Step 2 convert moles to grams
• .145 moles H2CO3 x 62.0 g H2CO3 8.99
  g 1 mole H2CO3 H2CO3

66
Normality

• In the past, we have used M for concentration.
• M moles / L
• A more useful form of concentration for acid/base
  reactions is Normality.
• N eq / L
• equivalents (moles)(n)
• n of H or OH in the chemical formula

67
Normality

• Also, in calculating pH, normality is used over
  molarity.
• Normality is related to Molarity
• N (M)(n)
• M moles/liters (total)
• n of H or OH in the chemical formula

68
Normality

• Ex 1 Find the normality of a solution that
  contains 1 mole H2SO4 in 1 L solution
• Step 1 Calculate the molarity
• 1 mole H2SO4 1 M
• 1 L
• Step 2 Calculate the normality
• N (M)(n)
• N 1 M x 2
• N 2 N

69
Normality

• Ex2 Calculate the Normality if 1.80 g of H2C2O4
  is dissolved in 150 mL of solution.
• Step 1 convert grams to moles
• 1.80 g H2C2O4 x 1 mole H2C2O4 .0200
  moles 90.0 g H2C2O4
  H2C2O4
• Step 2 Calculate the molarity
• M .0200 moles / .150 L ? .133 M
• Step 3 Calculate the normality
• N (M)(n)
• N (.133)(2) ? .267 N

70
Problems involving mixing unequal amounts of acid
and base

• Ex1 Find the pH of a solution made by mixing
  50.0 mL of .100 M HCl with 49.0 mL of .100 M
  NaOH.
• Step 1 Find the moles of HCl and moles of NaOH
• HCl ? .100 M x / .0500 L ? x
  .00500 moles HCl
• NaOH ? .100 M x / .0490 L ? x .00490
  moles NaOH

71
Find the pH of a solution made by mixing 50.0 mL
of .100 M HCl with 49.0 mL of .100 M NaOH.

• Step 2 Find the equivalents of H in HCl and the
  equivalents of OH- in NaOH
• eq (moles)(n)
• H ? (.00500 moles)(1) .00500 eq of H
• OH- ? (.00490 moles)(1) .00490 eq of
  OH-
• Step 3 Find the equivalents of H or OH- left
  over by subtracting the equivalents of H and
  equivalents of OH- from each other (absolute
  value)
• .00500 eq H - .00490 eq OH- .00010 eq
  H

72
Find the pH of a solution made by mixing 50.0 mL
of .100 M HCl with 49.0 mL of .100 M NaOH.

• Step 4Calculate the normality of the resulting
  solution
• N eq / L
• Total liters ? 50.0 mL 49.0 mL 99.0 mL ? .099
  L
• N .00010 eq / .0990 L ? .00101
  N
• Step 5 Calculate the pH
• pH - log H ? pH - log (.00101) ?
  
• 
  pH 3.00

73
Example 2

• Find the pH when mixing 50.0 mL of .100 M
  sulfuric acid with 50.0 mL of 1.00 M NaOH
• Step 1 Find the moles of H2SO4 and moles of NaOH
• Step 2 Find the equivalents of H in H2SO4 and
  the equivalents of OH- in NaOH
• Step 3 Find the equivalents of H or OH- left
  over by subtracting the equivalents of H and
  equivalents of OH- from each other
• Step 4 Calculate the normality of the resulting
  solution
• Step 5 Calculate the pH

74
Find the pH when mixing 50.0 mL of .100 M
sulfuric acid with 50.0 mL of 1.00 M NaOH

• Find the moles of H2SO4 and moles of NaOH
• H2SO4 ? .100 M x / .0500 L ? x .00500
  moles H2SO4
• NaOH ? 1.00 M x / .0500 L ? x .0500
  moles NaOH

75
Find the pH when mixing 50.0 mL of .100 M
sulfuric acid with 50.0 mL of 1.00 M NaOH

• Find the equivalents of H in H2SO4 and the
  equivalents of OH- in NaOH
• eq (moles)(n)
• H ? (.00500 moles)(2) .0100 eq
• OH- ? (.0500 moles)(1) .0500 eq
• Find the equivalents of H or OH- left over by
  subtracting the equivalents of H and equivalents
  of OH- from each other (absolute value)
• .0500 eq OH-1 - .0100 eq H .0400 eq OH-

76
Find the pH when mixing 50.0 mL of .100 M
sulfuric acid with 50.0 mL of 1.00 M NaOH

• Calculate the normality of the resulting
  solution
• N eq / L
• total L ? 50.0 mL 50.0 mL 100.0 mL
  ? .1000L
• N .0400 eq / .1000 L ? .400 N
• Calculate the pH 1st calculate the pOH
• pOH -logOH- ? pOH -log(.400) ?
• 
  pOH .399
• Calculate the pH from the pOH
• pH 14 pOH ? pH 14 - .399 ?
• 
  pH 13.6

77
Titrations

• A controlled addition and measurement of the
  amount of a solution of a known concentration
  that is required to react completely with a
  measured amount of a solution of unknown
  concentration.

78
Titrations
79
Titrations

• Equivalence point In a neutralization reaction,
  the point _at_ which there are equivalent quantities
  of H3O and OH-
• End Point point in a titration where the
  indicator changes color

Titration of 20 mL 0.1 M HCL
Titration of HCl with NaOH
High amount of H1
End point
H1 OH-
High amount of H1
mL of 0.1 M NaOH Titrant
80
Normality

• Normality and Titration
• Equation
• NaVa NbVb
• N normality
• V volume in Liters

81
Ex1 A 15.5 mL sample of .215 M KOH requires 21.2
mL of acetic acid to titrate to the end point.
What is the Molarity of the acid?

• Step 1 Convert molarity to normality
• N (M)(n)
• n of H or OH in the chemical formula
• N (.215)(1) ? .215 N KOH
•  Step 2 Use the titration formula to find the
  normality of the acid
• NaVa NbVb
• (x)(21.2 mL) (.215 N)(15.5 mL) ? x
  .157 N
• Step 3 Convert normality to molarity
• M N / n ? x .157 N / 1 ? x
  .157 M

82
Ex2 If 15.7 mL of sulfuric acid is titrated to
the end point by 17.4 mL of .0150 M NaOH, what is
the Molarity of the acid?

• Convert molarity to normality
• N (M)(n)
• N (.015)(1) ? .0150 N KOH
• Use the titration formula to find the normality
  of the acid
• NaVa NbVb
• (x)(15.7 mL) (.0150 N)(17.4 mL) x .0166
  N
• Convert normality to molarity
• M N / n
• x .0166 N / 2 ? x .00831 M

83
Titration Problems

• Ex 3 In a titration of 27.4 mL of 0.0154 M
  Ba(OH)2 solution is added to a 20 mL sample of an
  HCl solution. What is the Molarity of the HCl
  solution?

84
Percent Problems

• Long way.
• If 18.75 mL of .750 N NaOH is required to titrate
  20.30 mL of acetic acid, calculate the acetic
  acid in solution.
• Step 1 Use the titration formula to find the
  normality of the acid
• NaVa NbVb
• (x)(20.30 mL) (.750 N)(18.75 mL) ? x
  .693 N

85
If 18.75 mL of .750 N NaOH is required to titrate
20.30 mL of acetic acid, calculate the acetic
acid in solution.

• Step 2 Multiply the normality by the equivalent
  weight of the acid
• .693 N ? .693 eq/L
• (.693 eq/L)(60.0g/eq) 41.6 g/L

86
If 18.75 mL of .750 N NaOH is required to titrate
20.30 mL of acetic acid, calculate the acetic
acid in solution.

• Step 3 Convert the liters to grams
• 41.6 g/L ? 41.6 g/1000mL ? 41.6 g/1000g
•  
• Step 4 Calculate the by mass
• (41.6 g / 1000g) 100 4.16

87
If 18.75 mL of .750 N NaOH is required to titrate
20.30 mL of acetic acid, calculate the acetic
acid in solution.

• The short way ?
• Use the titration formula to find the normality
  of the acid
• NaVa NbVb
• (x)(20.30 mL) (.750 N)(18.75 mL) x .693 N
• Use (N)(eq wt)/10
• (.693)(60.0) / 10 x
• x 4.16
•