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 Ex2 HBr Ex3 H3P Hydrochloric Acid
Hydrobromic Acid
Ex 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 Ex7 H2SO3 Ex8 HClO4 Ex9 HClO Sulfuric Acid
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) 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 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)  H+1(aq) + Cl-1 (aq)
It is now found that:
H+1(aq) + H2O(l)  H3O+1 (aq)
so it is really…
HCl(g) + H2O(l)  H3O+1 (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)  H3O+1 (aq) + F-1 (aq)

15. Molecular, Total Ionic, and Net Ionic Equations for acids:
Molecular Equation:
Zn(s) HCl(aq)  ZnCl2(aq) + H 2(g)
Total Ionic Equation:
Zn(s) + 2H+1(aq) + 2Cl-1 (aq)  Zn+2(aq) Cl-1(aq) + H2(g)
Net Ionic Equation:
Zn(s) + 2H+1(aq)  Zn+2(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)  H3O+1 (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)  H3O+1 (aq) + HSO4-1 (aq)
HSO4-1 (aq) + H2O(l)  H3O+1(aq) + SO4-2 (aq)
__________________________________
H2SO4 (aq) + 2 H2O(l)  2 H3O +1(aq)+SO4-2(aq)
*Note: When you lose a H+1, 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  Mg+2(aq) + 2OH-(aq)

24. Types of Bases * Water is amphoteric. It can act as an acid
2. Bronsted- Lowry bases – proton acceptors
NH3(g) + H2O(l)  NH4+1(aq) OH-1(aq)
* Water is amphoteric. It can act as an acid
or base.

25. Types of Bases List of strong bases: List of weak 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 N…NH3, 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)
2H+1(aq) + 2Cl-1(aq) + Ba+2(aq) + 2OH-1(aq)  Ba+2(aq) + 2Cl-1(aq) + 2HOH(l)
2H+1(aq) OH-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)  H3O+1(aq) + Cl-1(aq)
(Acid) (Base)
Weak Base Example:
NH3(g)+ H2O(l)  NH4+1(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)  H3O+1(aq) Cl-1(aq)
(Acid) (Base) (Conjugate Acid) (Conjugate Base)

31. Conjugate Acids & Bases
NH3(g) + H2O(l)  NH4+1(aq) OH-1(aq)
HF(l) H2O(l)  H3O+1(aq) F-1(aq)

32. Conjugate Acids and bases
H2CO3(aq) + H2O(l)  H3O+1(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) H3O+1(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) 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)  H3O+1(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)  Fe+3(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) 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  H3O+1 + 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, [H3O+1] = 1 x 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 = [H3O+1][OH-1]

49. Aqueous Solutions and the Concept of pH
At 25 C0, [H3O+1] = 1 x 10-7
so [OH-1] = 1 x 10-7
so Kw = 1 x
Example: If the [H3O+1] 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 # [H3O+1] [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 4
2.40 x 10-9
4.16 x 10-6

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 =
ex. log of .01 =

54. How to calculate strengths of Acids and Bases
To convert [H3O+1] to pH
pH = - log 10 [H3O+]
log, #, enter, then make it positive – change the sign)
[H3O+1] =
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 [H3O+1]
[H3O+] = 10 –pH
2nd, log, (-), #, enter
[H3O+1] pH
Acid or Base 2 11
5.22

56. How to calculate strengths of Acids and Bases
[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: H2SO NaOH  Na2SO 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 mole of protons or hydroxide ions.
HCl
H2SO4
H3PO4
Moles of Acid 1
1/3
Moles of Hydrogen
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 C = (1)12.0 = 12.0 O = (3)16.0 = 48.0 mw = 62.0 g/mole
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 = g/mole

63. Calculations Step 2: Calculate the # of moles in 9.30 g of H2CO3:
9.30 g H2CO3 x 1 mole H2CO3 = mole g H2CO 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 equivalents calculated above)
eq. wt. = mw/eq
eq. wt. = 62.0 g / .300 equivalents
eq. wt. = 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 g H2CO3 = g 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 = moles g H2C2O H2C2O4
Step 2: Calculate the molarity:
M = 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 / L  x = moles HCl
NaOH  .100 M = x / L  x = 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+  ( moles)(1) = eq of H+
OH-  ( moles)(1) = 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):
eq H eq OH- = 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 4:Calculate the normality of the resulting solution:
N = eq / L
Total liters  50.0 mL mL = 99.0 mL  .099 L
N = eq / L  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
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 / L  x = moles H2SO4
NaOH  1.00 M = x / L  x = moles NaOH

75. Find the pH when mixing 50. 0 mL of. 100 M sulfuric acid with 50
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+  ( moles)(2) = eq
OH-  (.0500 moles)(1) = 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 eq H+ = eq OH-

76. Find the pH when mixing 50. 0 mL of. 100 M sulfuric acid with 50
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 mL = mL  .1000L
N = eq / L  N
Calculate the pH: 1st calculate the pOH:
pOH = -log[OH-]  pOH = -log(.400) 
pOH =.399
Calculate the pH from the pOH
pH = 14 – pOH  pH = 
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 which there are equivalent quantities of H3O+ and OH-
End Point – point in a titration where the indicator changes color
Titration of HCl with NaOH
Titration of 20 mL 0.1 M HCL
High amount of H+1
End point
H+1= OH-
High amount of H+1
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
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)  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 /  x = .157 M

82. Ex2: If 15. 7 mL of sulfuric acid is titrated to the end point by 17
Ex2: If 15.7 mL of sulfuric acid is titrated to the end point by 17.4 mL of M NaOH, what is the Molarity of the acid?
Convert molarity to normality:
N = (M)(n)
N = (.015)(1)  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 = N
Convert normality to molarity:
M = N / n
x = N /  x = M

83. Titration Problems
Ex 3: In a titration of 27.4 mL of 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 mL of .750 N NaOH is required to titrate 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. Step 2: Multiply the normality by the equivalent weight of the acid:
If mL of .750 N NaOH is required to titrate mL of acetic acid, calculate the % acetic acid in solution.
Step 2: Multiply the normality by the equivalent weight of the acid:
.693 N  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
If mL of .750 N NaOH is required to titrate mL of acetic acid, calculate the % acetic acid in solution.
Step 3: Convert the liters to grams:
41.6 g/L  g/1000mL  g/1000g
Step 4: Calculate the % by mass:
(41.6 g / 1000g) 100 = 4.16 %

87. Use the titration formula to find the normality of the acid:
If mL of .750 N NaOH is required to titrate 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 %