IV Normality

A. Equivalent Weight1. Def - is defined differently for different

types of substances - is defined in a way dependent upon

the rxn it is undergoing2. For Acids

a.EW '

FW

number of H %sreacted per acid i n formula

b. NaOH + HCl ÿ NaCl + H2OEWHCl =

c. 2 NaOH + H2SO4 ÿ Na2SO4 + 2 H2OEWH2SO4 =

d. 3 NaOH + H3PO4 ÿ Na3PO4 + 3 H2OEWH3PO4 =

e. H3PO4 + NaOH ÿ H2O + NaH2PO4

EWH3PO4 =

f. H3PO4 + 2 KOH ÿ 2 H2O + K2HPO4

EWH3PO4 =

3. For basesa. -EW '

FW

number OH reacted per formula

b.Ca (OH)2

base%

2 HClacid

Y CaCl2 % 2 H2O

EWCa(OH)2'

FWCa(OH)2

2

c. Ca (OH)2 + HCl ÿ Ca(OH)Cl + H2O

EWCa(OH)2

'

FWCa(OH)2

1

4. In Reduction- Oxidation R x ns

a. EW 'FW

number e& transferred

b. Ag+ + e- ÿ Ag EW 'FWAg

1

c. Ca+2 + 2 e- ÿ Ca EW 'FWCa

2

d. Al+3 + 3e- ÿ Al EW 'FWAl

3

e. EW MnO4-

1. MnO4- + e- ÿ MnO4

-2 EW 'FW1

2. MnO4- +4 H+ + 3e- ÿ MnO2 + 2 H2O

EW 'FW3

3. MnO4- + 8 H+ + 4e- ÿ Mn+3 + 4 H2O

EW 'FW4

5. Precipitate Formation & Complex Formation

a. Salt (ionic)cationr anions

b. EWcation 'FWcation

rcharges on cation

c. EWanion 'FWanion

number of equiv of Metal cation that will react with it

d. EWsalt 'FWsalt

Total number of r charges in 1 formula unit of salt

e. Ba+2 + SO4-2 ÿ BaSO4

EWBa %

2 'FW Ba %2

%2' g/equivalent

Ba (NO3)2

EWBa (NO3)2'

FWBa(NO3)2

%2

+6 -6+3 -2

f. Fe2O3 ÿ complex ion

EWFe2O3'

FWFe2O3

number % charges'

FWFe2O3

6

B. Number of Equivalents1. Def

number e quiv 'number g of Subst

GEW of Subst'

number gnumber g /equiv

# equiv x GEW substance = # g of Subst

C. Normality

Def N 'number e quiv of Solute

number l of Solution'

number me quiv Solutenumber ml Solution

1 equiv of solute in 1.0 liter of Solution

1 equiv/l = 1 Normal = 1 N

# equiv solute = N x #l of Solution (NCV(l))# mequiv solute = N x #ml of Solution (NCV(ml))

D. Value of Equiv & Normality1. Regardless of Stoichiometry

1 equiv of one reactant reacts with 1 equiv ofother reactant to yield 1 equiv of each product

2. 2 NaOH + H3PO4 ÿ Na2HPO4 + 2 H2O2 moles 1 mole 1 mole 2 mole1 equiv 1 equiv 1 equiv 1 equiv

FW- 40 amu 98 amu

GFW- 40 g/mole 98 g/mole

Eq Wt. 40/1= 40 98/2 = 49

# equiv Base = # equiv Acid

ÿ Nb x Vb (l) = Na x Va (l)

E. Conversion Between Molarity & Normality1. N = Mx # H+ that react /formula of acid

# OH- that react/formula of base# e- transferred/formula of redox# + charge’s/formula if cation

2. H3PO4 3M? N

a) H3PO4 + NaOH —> Na H2PO4 + H2ON = 3 M x 1 = 3N

b) H3PO4 + 2 NaOH —> Na2HPO4 + 2 H2ON = 3 M x 2 = 6N

c) H3PO4 + 3 NaOH —> Na3PO4 + 3 H2ON = 3 M x 3 = 9N

d) Hence: If not told the rxn, assume that allH+ react

1. H3PO4 3M? N

N = 3 M x 3 = 9 N

V Other Expressions of Concentration

A. Molality

m 'Number of moles of Solute

Number of Kg of Solvent

B. Mole Fraction

÷ 'Number of moles of Solute

Number of moles Solute % Number of moles Solvent

C. Wt %w/w

Number of g SoluteNumber of g Solute % Number of g Solvent

x 100

Vol %v/v

Number l of SoluteNumber l of Solute % Number l Solvent

x 100

w/vNumber g Solute

Number ml of Solutionx 100

10% (w/w) NaCl

10% (w/v) NaCl

D. Parts per million, ppm

1. or ppm 'mg Solute

kg Solutionmg Solutel Solution

5 mg NaCl in - 1.0 liter solution

5 ppm NaCl

2. Note:

2 NaOH + H3PO4 —> Na2HPO4 + 2H2O5ppm 10ppm

ppm does not work in stoichiometry

3. Can convert ppm to moles/l

ppm 'mgl

' ( gl

x 10&3)

molesl

' ( gl

× 10&3) × 1FW

'g

FW× 10&3

l

To get moles/l must ÷ (ppm x 10-3) by FW

ppmFW

× 10&3 'mole ×10&3

l

E. Titre1. Definition of Titer

T 'mg Solute Reacted by Solution Being Standardized

ml of Standardized Solution Required

2. NaOH + HCl —> NaCl + H2O88 8888 Want to standardize

“Weigh out” 4.00 mg NaOH

Add HCl until neutralize all NaOH

Record # ml of HCl required

3. If 1.00 ml of HCl neutralized 4.00 mgNaOH

HCl + NaOH —> NaCl + H2O

Calculate the NaOH Titre

TNaOH '4.00 mg NaOH

1.00 ml HCl' 4.00

mg NaOH

ml HCl

4. Suppose you have an unknown NaOHsample and itRequires 3.0 ml of the Std. HCl solutionHow much NaOH is present?

#mg = TNaOH x #ml of Std. HCl

Number mg NaOH '4.00 mg NaOH

ml HCl× 3.0 ml HCl

= 12.0 mg NaOH

5. May relate Titre & Normality

a) T = N x EW88 88 88

Expressed in termsof the substancewith which thesolution reacts

NormalConcentrationof standardsolution

of thesubstance withwhich thesolution reacts

b) Ca (OH)2 + 2 HCl —> CaCl2 + 2 H2O

1) TCa(OH)2' NHCl × EWCa(OH)2

NHCl 'TCa(OH)2

EWCa(OH)2

EWCa(OH)2'

FWCa(OH)2

2

2) Calculate the Ca(OH)2 titre of a.1000 N HCl solution

6. Suppose you have a 0.20 N solution of HCl. What is its

a) NaOH titre? FW = 40

TNaOH ' NHCl × EWNaOH ' 0.20 × 401

' 8.0 mgml

b) Ca(OH)2 titre? 74 = FW

TCa(OH)2' NHCl × EWCa(OH)2

' .20 ×74

2' 7.4

mg

ml

c) NH4OH titre ? FW=35

TNH4OH ' NHCl x EWNH4OH ' .20 x 351

' 7.0 mgml

d) Al (OH)3 titre FW= 78

TAl(OH)3'NHCl x EWAl(OH)3

' .20 x 783

' 5.2 mgml

7. Suppose that a solution of HCl has a NaOHtitre of 0.20 mg/ml. If it takes 10.00 ml ofthe HCl solution to react with an unknownsolution containing NaOH, how much NaOHis present? ( #mg, #g, #moles NaOH)

#mg NaOH = TNaOH x #ml of HCl

' 0.20 mg NaOH1 ml HCl

x 10.00 ml HCl

= 2.0 mg NaOH

#g = .002 g

# moles = .002 g40 g/mole

Determination of Concentration of Acid FromInformation Given on Bottle

M 'Density x %Acid x 10

FW of Acid'

Density x Fraction of solution that is acid x 1000FW of Acid

Eq For HClSp. Grav = Density = 1.1836.5 - 38.0 % HClFW of HCl = 36.5

M '1.18 g of solution

ml of solutionx .38 g HCl

g of solutionx 1000 ml solution

l of solution

36.5 g HClmole HCl

M ' 12.3 moles HCll of solution

' 12.3 M

Eq For H3PO4 Sp. Grav = Density = 1.6985 % H3PO4

H3PO4 FW = 98

M = 1.69 g solutionml solution

x.85 g H3PO4

g solutionx 1000 ml solution

l of solution98 g H3PO4

mole H3PO4

M = 14.7 M H3PO4

Composition of Commercial Acids & Bases

Acid or Base Specific Gravity Percentage by Mass Molarity Normality

HCl 1.19 38 12.4 12.4

HNO3 1.42 70 15.8 15.8

H2SO4 1.84 95 17.8 35.6

HC2H3O2 1.05 99 17.3 17.3

H3PO4 1.69 85 14.7 44.1

HClO4 1.68 71 11.9 11.9

NH3(aq) 0.90 28 14.8 14.8