1. Lab. 1 prepared by : shorog m . altaweel
NORMALITY
MOLARITY
Introduction
MOLE
Lab. 1 prepared by : shorog m . altaweel

2. Review of fundamental concept
Formula weight
It is assumed that you can calculate the formula or molecular weights of compound from respective atomic weight of the elements forming these compounds.
The formula weight of the substance is the sum of the atomic weight of the elements from which this substance is formed from.

3. Find formula weight of CaSO4.7H2O
Atomic weight
Element
40.08 Ca
32.06 S
11 x 16.00
11 O
14 x 1.00
14 H Σ=
FWt

4. The mole concept
The mole is the major word we will use throughout the course. The mole is defined as gram molecular weight which means that:
Grams
Mole
2.00 g
1 mole H2
32.00 g
1 mole O2
16.00 g
1 mole O
58.5 g
1 mole NaCl g
1 mole Na2CO3

5. Assuming approximate atomic weight of 1. 00 , 16. 00 , 23. 00 , 35
Assuming approximate atomic weight of 1.00 , , , 35.5 and atomic mass units of hydrogen, oxygen, sodium, chlorine atom, and carbon respectively.
The number of moles contained in a specific mass of substance can be calculated as :
Mol = g substance / FW substance
The unit of formula weight is gm/mol
In the same manner, the number of mmol of a substance contained in a specific weight of the substance can be calculated as
Mmol = mol/1000 Or
Mmol = mg substance / FW substance

6. Examples
The number of mmol of Na2WO4 (FW = mg/mmol) present in 500 mg of Na2WO4 can be calculated as
?? Mmol of Na2WO4 = 500 mg / (mg/mmol ) = 1.70 mmol
The number of mg contained in 0.25 mmol of Fe2O3 ( FW= mg/mmol) can be calculated as
?? Mg Fe2O3 = 0.25 mmol Fe2O3 X (mg/mmol) = 39.9 mg
Therefore, either the number of mg of a substance can be obtained from its mmols or vice versa

7. Calculation involving solution:
Molarity
Molarity of a solution can be defined as the number of moles of solute dissolved in 1L of a solution.
This means that 1 mole of solute will be dissolved in some amount of water and the volume will be adjusted to 1 L.
The amount of water may be less than 1 L as the final volume of water and solute is exactly 1 L

8. Calculation involving molarity
Molarity = mol/L = mmol/ml This can be further formulated as Number of moles = molarity X volume in L or Mole = M(mol/L) X VL Number of mmole = molarity X volume in mL Mmole = M(mmol/mL) X VmL

9. Examples:
Molarity
Find the molarity of a solution resulting from dissolving 1.26g of AgNO3 (FW = g/mol) in a total volume of 250 ml solution First find mmol AgNO3 = 1.26x10³ mg AgNO3/ mg/mmol = 7.42 mmol
Molarity = mmol/ml
M = 7.42 mmol/250ml = mmol/ml
Let us find the number of mg of NaCl per ml of a 0.25 M NaCl solution
First we should be able to recognize the molarity as 0.25 mol/L or 0.25 mmol/ml. of course, the second term offers what we need directly.
?? mg NaCl in 1 ml = 0.25 mmol NaCl/ml x (58.5 mg NaCl/mmol NaCl) = 14.6 mg NaCl/ml

10. Examples
Molarity
Find the number of grams of Na2SO4 required to prepare 500 ml of 0.1 M solution.
First, we find mmoles needed from the relation
Mmol = M (mmol/ mL) x VmL
Mmol Na2SO4 = 0.1 mmol/mL x 500 ml = 50 mmol
Mmol = mg substance / FW substance
?? Mg Na2SO4 = 50 mmol x 142 mg/mmol = 7100 mg or 7.1 gm.

11. Calculation involving solution:
Normality
The second expression used to describe concentration of a solution is the normality.
Normality can be defined as the number of equivalents of solute dissolved in 1L of solution. Therefore, it is important for us to define what do we mean by the number of equivalents, as well as the equivalents weight of a substance as a parallel term to formula weight.

12. Normality
N= eq/L or N= meq/ml Number of equivalent= normality x VL = (eq/L) x L Number of meq= normality x VmL = (meq/mL) x mL Also, Number of equivalents= wt(g)/equivalent weight Number of meq= wt(mg)/eqw

13. N= n x M Equivalent weight= FW/n
Normality
Equivalent weight= FW/n
Meq= mg/eqw substitute for eqw= FW/n gives:
Meq= mg/(FW/n), but mmol= mg/FW, therefore:
Meq = n x mmol
Dividing both sides by volume in mL, we get,
N= n x M
Where n is the number of reacting units (protons, hydroxide or electrons), and if you are forming factors always remember that a mole contains n equivalents.
The factor becomes ( 1 mol/n x eq) or
(n x eq/1mol)

14. (in acid base reaction) (in oxidation reduction reaction)
An equivalent is defined as the weight of substance giving an Avogadro's number of the
Reacting units (n)
protons or hydroxides
(in acid base reaction)
For example, HCl has one reacting unit H+ when reacting with base like NaOH,
Therefore, we say that the equivalent weight of HCl is equal to its formula weight
sulfuric acid has two reacting units (two protons) when reacting completely with a base..
equivalent weight of H2SO4 is equal to one half its formula weight,
Electrons
(in oxidation reduction reaction)
In the reaction where Mn (Vll), in KMnO4. is reduced to Mn(ll), so five electrons are involved and the equivalent weight of KMno4 is equal to its formula weight divided by 5.

15. Normality
Example
Find the equivalent weights of NH3 (FW=17.03), H2C2O4 (FW= 90.04) in the reaction with excess NaOH and KMnO4(FW= )when Mn(Vll) is reduced to Mn(ll).
Solution:
NH3 react with one proton only
Equivalent weights of NH3= FW/1 = g/eq
Two protons of oxalic acid react with the base
Equivalent weight of H2C2O4= FW/2 = 90.04/2=45.02 g/eq
H2C2O4+ 2 NaOH Na2C2O4 + 2 H2O
5 electrons are involved in the reduction of Mn(Vll) to Mn(ll)
Equivalent weight of KMnO4 = FW/5= /5= g/eq

16. Normality
Example
Find the normality of the solution containing g/L of Na2CO3 (FW=105.99), carbonate reacts with two protons. Solution: Normality is the number of equivalents per liter, therefore we first find the number of equivalents. Eq wt= FW/2= /2= Eq= Wt/eq wt= 5.300/53.00= N= eq/L= eq/1L= N

17. Normality
Example
Find the normality of the solution containing 5.267g/L K2Cr2O7 (FW=294.19) if Cr +6 is reduced to Cr+3. Solution: N= eq/L, therefore we should find the number of eq Eq= wt/eq wt, therefore we should find the number of eq wt Eq wt= FW/n, here: each contributes 3 electrons and since the dichromate contains 2 Cr atoms we have 6 reacting units. Eq wt= (294.19g/mol)/(6 eq/mol) Eq= 5.267g/(294.19g/mol)/(6eq/mol) N= eq/L=5.267g/(294.19g/mol)/(6eq/mol)/L=0.1074eq/L

18. Example I2 + 2e 2 I⁻ meq= N x Vml
How many mg of I2 (FW=254 mg/mmol) should you weigh to prepare 250 ml of 0.100N solution in the reaction
I2 + 2e I⁻
meq= N x Vml
meq= 0.1 x 250 = 25meq
mg I2= meq x eq wt= meq x FW/2= 25 X 354/2= 3180mg

19. Example
What volume of concentrated HCL (FW= 36.5g/mol, concentration=32%, density=1.1g/ml) are required to prepare 500ml 0f 2.0 M solution.
Always start with the density and find how many grams of solute in each ml of solution
Density= g solution/ml
Remember that only a 32% of the solution is solute.
Mg HCl/ml = 1.1 x 0.32 x 1000 mg HCL/ml
The problem is now simple as it require conversion of mg HCL to mmol since:
molarity= mmol HCL/ml
Molarity= mmol HCL/ml = {1.1 x 0.32 x 1000mg HCL}/ mg/mmol =9.64 M
Now, we can calculate the volume required from the relation
M1V1 = M2V2
9.64 x Vml = 2.0 x 500 ml
Vml= 10.4ml of the concentrated HCL should be added to dis.H2O and the volume adjusted to 500ml

20. An easy short cut: M= Density x percentage of solute x 1000
Formula weight
The percentage is a fraction : ( i.e : is written as 0.35)

21. Stoichiometric Calculations: Volumetric Analysis
In this section we look at calculations involved in titration processes as well as general quantitative reactions. In a volumetric titration, an analyte of unknown concentration is titrated with a standard in presence of a suitable indicator.

22. Volumetric Analysis Titration type Titration method Back titration
Acid base
Redox
Precipitation
Volhard's method
Mohr's method
Complexometric
Back titration
Titration method
In-direct method
Direct method

23. For a reaction to be used in titration the following characteristics should be satisfied:
stoichiometry of the reaction should be exactly known. This means that we should know the number of moles of A reacting with 1 mole of B.
reaction should be rapid and reaction between A and B should occur immediately and instantly after addition of each drop of titrant
(the solution in the burette).
There should be no side reactions. A reacts with B only.
There should exist a suitable indicator which has distinct color change.
The reaction should be quantitative. A reacts completely with B.

24. Standard solution: Primary Standard solution Primary standard solution
A standard solution is a solution of known and exactly defined concentration. Usually standards are classified as either primary standards or secondary standards. There are not too many secondary standards available to analysts and standardization of other substances is necessary to prepare secondary standards
Primary Standard solution
Should have a purity of at least 99.98%
Stable to drying, a necessary step to expel adsorbed water molecules before weighing
Should possess the same properties as that required for a titration
Should be non hygroscopic
have a relatively large molar mass (to minimise errors during weighing)
Primary standard solution

25. Non Primary Standards solution
Standard solution:
Non Primary Standards solution
NaOH, KOH
Readily absorb CO2, from the atmosphere
Readily absorb H2O, from the atmosphere
HCl, and H2SO4,
The % written on the reagent bottle is a claimed value and should not be taken as guaranteed.
HNO3
It always contains a little HNO2, which has a destructive action on many acid-base indicators.

26. Primary Standards solution
anhydrous sodium carbonate, Na2CO3, weak base
Prepare a dilute solution of HCL,H2SO4 with an approximate concentration, and then determine the concentration accurately by titration with a primary standard solution of Na2CO3
potassium hydrogen phthalate, KH(C8H4O4), a weak acid
you can prepare a NaOH solution with an approximate concentration. This solution can then be titrated with an acidic primary standard solution, such as KH(C8H4O4).
Standard solution:
Because the accurate concentration of acid or base has been derived using a primary standard, the acid or base is referred to as a secondary standard

27. Molarity volumetric analysis
What you really need is to use the stoichiometry of the reaction to find how many mmol of A as compared to the number of mmoles of B
Moles of Compound A
Molar ratio
Moles of Compound B
Relates moles of reactants and products in balanced chemical equation

28. Example NaHCO3 + HCl NaCl + H2CO3
A g sample containing NaHCO3 (FW = mg/mmol) was dissolved and titrated with M HCl requiring mL. Find the percentage of bicarbonate in the sample.
We should write the equation in order to identify the stoichiometry
NaHCO3 + HCl NaCl + H2CO3 
Now it is clear that the number of mmol of bicarbonate is equal to the number of mmol HCl
Mmol NaHCO3 = mmol HCl
Mmol = M x VmL
Mmol NaHCO3 = ( mmol/ml ) x mL = mmol
Now get mg bicarbonate by multiplying mmol times FW
Mg NaHCO3 = mmol x (84.01 mg/mmol) = 365.0
% NaHCO3 = (365.0 x 10-3 g/ g) x 100 = 78.14%

29. Example
A g sample containing Na2CO3 (FW = 106mg/mmol) was dissolved and titrated with M HCl requiring mL. Find the percentage of carbonate in the sample.   
The equation should be the first thing to formulate
Na2CO3 +2 HCl = 2NaCl + H2CO3
Mmol Na2CO3 = ½ mmol HCl
Now get the number of mmol Na2CO3 = ½ x ( MHCl x VmL (HCl) )
Now get the number of mmol Na2CO3 = ½ x x = mmol
Now get mg Na2CO3 = mmol x FW = x 106 = 230 mg
% Na2CO3 = (230 x 10-3 g/ g ) x 100 = 49.3 %

30. Example Mmol NaOH = 2 mmol H2SO4
How many mL of 0.25 M NaOH will react with 5.0 mL of 0.10 M H2SO4.
H2SO4 + 2 NaOH = Na2SO4 + 2 H2O
Mmol NaOH = 2 mmol H2SO4
Mmol NaOH = 2 {M (H2SO4) x VmL (H2SO4)}
Mmol NaOH = 2 x 0.10 x 5.0 = 1.0 mmol
mmol NaOH = MNaOH x VmL (NaOH)
0.1mmol NaOH = 0.25 x VmL
VmL = 4.0 mL

31. Example 5H2O2 + 2KMnO4 + 6H+ 2Mn+2 +5O2 + 8H2O
Find the volume of 0.1 M KMnO4 that will react with 50 ml of 0.2 M H2O2 according to the following equation:
5H2O2 + 2KMnO4 + 6H Mn+2 +5O2 + 8H2O
The following step is to formulate the relationship between the number of moles of the two reactants.
We always start with the one we want to calculate, that is
Mmol KMnO4 = 2/5 H2O2
Mmol = molarity x Vml
0.1 x Vml = (2/5) x 0.2 x 50
Vml = 40 ml KMnO4

32. Example
Find the volume of 0.1 M KMnO4 that will react with 50 ml of 0.2 M MnSO4 according to the following equation:
3MnSO4 + 2KMnO4 + 4 OH MnO2 + 2H2O + 2K+ + 3SO4 -2
mmol KMnO4 = 2/3 mmol MnSO4
0.1 M x Vml = 2/3 x 0.2 M X 50
Vml = 66.7 ml KMnO4

33. Normality volumetric calculation
We have seen previously that solving volumetric problems required setting up a relation between the number of mmoles or reacting species. In case of normality, calculation is easier as we always have the number of meq of substance A is equal to meq of substance B, regardless of the stoichiometry in the chemical equation. Of course this is accounted for in the calculation of meqs. Therefore, the first step in a calculation using normalities is to write down the reaction meqA = meq B

34. Example
A g sample containing sodium bicarbonate NaHCO3 was titrated with HCl requiring ml. the HCL was standardized by titrating g of sodium carbonate Na2CO3 (FW= 106 mg/mmol) requiring ml of the acid. Find the percentage of NaHCO3 (FW= mg/mmol) in the sample.
Eq wt Na2CO3 = FW/2= 53.0
Eq wt NaHCO3 = FW/1=84.0
meq HCl = meq Na2CO3
Normality x volume (ml) = wt (mg) / eq wt
N x = mg / (53 mg/meq)
N HCL = eq/L
Now we can find meq of NaHCO3
meq NaHCO3 = meq HCl
Mg NaHCO3/ eq wt = N X Vml
Mg NaHCO3/ 84 = x 40.72
Mg NaHCO3 = 84 x x 40.72
% Mg NaHCO3= (0.67) x 100 = 67.2%

35. Example
Use normality to calculate how many ml of a 0.1 M H2SO4 will react with 20 ml of 0.25 M NaOH
We can first convert molarities to normalities:
N= n x M
N (H2SO4)= 2 x 0.1= 0.20
N (NaOH)= 1 x 0.25= 0.25
Meq H2SO4 = meq NaOH
Substitution for meq as usual (either N xVml or mg/eq wt)
0.2 x Vml = 0.25 x 20
Vml= 25 ml