The document outlines a lecture on quantitative analysis focusing on equivalent weight, normality, and conversions between various concentration units. Key learning objectives include calculating equivalent weight for acids, bases, and salts, as well as deriving normality from equivalent weights and solution volumes. It also includes sample calculations for normality using sodium carbonate and oxalic acid, along with relations between ppm, ppb, and molarity.

1. Lecture (4)
(242 Chem)
College of Sciences
Department of Chemistry
Level: 4th
Semester BSc
Quantitative analysis-1
Dr Brima Monday 01/02//2016---22/4/1437 1
King Khalid University
Dr Eid Ibrahim Brima

2. 1-To know how to calculate Equivalent Weight (eq. wt)
2- To know how to express concentrations in Normality (N)
3- To know how to convert between ppm, ppb and M.
learning objectives
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3. 3
Expressing concentration in Normality (N)-1
i. Equivalent Weight (eq wt):
Where: R = number of Reactive Units
a. For acids and bases:
R = no of replaceable H+
or OH-
ions
e.g. H2SO4 ; CaOH2
Mw
R
eq wt =
Dr Brima Monday 01/02//2016---22/4/1437

4. 4
i. Equivalent Weight (eq wt):
a. For acids and bases:
● Example:
For HCl R = 1
For H2SO4 R = 2 (Unless otherwise prescribed)
For H3PO4 R = 3 (Unless otherwise prescribed)
For CH3COOHR = 1
Expressing concentration in Normality (N)-2
Dr Brima Monday 01/02//2016---22/4/1437

5. 5
i. Equivalent Weight (eq wt):
a. For acids and bases:
● Example:
For NaOH R = 1
For Cu(OH)2 R = 2
For Al(OH)3 R = 3
Expressing concentration in Normality (N)-3
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6. 6
i. Equivalent Weight (eq wt):
b. In Reduction – Oxidation (Redox)
reactions:
R = no of electrons transferred
(gained or lost)
● Ex.: MnO4
-
+ 8 H+
+ 5 e-
Mn2+
+ 4 H2O
∴ R = 5
Expressing concentration in Normality (N)-4
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7. 7
i. Equivalent Weight (eq wt):
c. For Salts:
R = Total no of +ve or -ve charges
on the cation or the anion of the salt
● Example:
For Na+
Cl-
R = 1
For Na2
2+
CO3
2-
R = 2
Expressing concentration in Normality (N)-5
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8. 8
ii. Gram equivalent weight (g eq wt):
It is the equivalent weight expressed in
grams
● Example:
eq wt of Na2CO3 = Mw / 2 = 106 / 2 = 53
∴1 g eq wt for Na2CO3 = 53 g
Expressing concentration in Normality (N)-6
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9. 9
iii. Number of g eq wt’s:
● Example: How many g eq wt’s are in 106 g of
Na2CO3 (MW = 106)?
- eq wt of Na2CO3 = MW / 2 = 106 / 2 = 53
∴ No of g eq wt’s = 106 / 53 = 2
No of g eq wt =
wt of substance
its eq wt
Expressing concentration in Normality (N)-7
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10. 10
iv. Normality (N):
It is the number of g eq wt’s per one liter
of solution
∴ N x VL = wt of solute / its eq wt
No of g eq wt’s of solute
sol.n Volume in Liters
=
wt / eq wt
VL
∴ N =
Expressing concentration in Normality (N)-8
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11. Dr Brima Monday 01/02//2016---22/4/1437 11
Normality calculation-1
- *Calculate the normality for 0.212g of Na2CO3
dissolved in 100ml water?
- Answer:
[212mg/(106/2)]/100ml = 0.04N
[0.212g/(106/2)]/0.1L = 0.04N
No of g eq wt’s of solute
sol.n Volume in Liters
=
wt / eq wt
VL
∴ N =
83

12. Dr Brima Monday 01/02//2016---22/4/1437 12
Normality calculation-2
- *Calculate the normality for 0.25g/L of Oxalic
acid (H C O₂ ₂ 4) dissolved in 100ml water?
- Answer:
- N = wt(mg)/[Mwt/R]/Vml
= 250mg/[290.4/2]/1000ml = 0.006N
—84-85

13. 13
Relations between ppm, ppb and M
- To convert M to ppm:
∴ ppm = M x MW x 1000
- To convert ppm to M:
∴ M = ppm / (MW x 1000)
- To convert M to ppb:
∴ ppb = M x MW x 106
- To convert ppb to M:
∴ M = ppb / (MW x 106
)
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14. Lecture Summary
• Eq.Wt. :
• Number of g eq wt’s:
• Normality (N):
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Mw
R
eq wt =
No of g eq wt’s of solute
Soln. Volume in Liters
=
wt / eq wt
VL
N =
N x VL = wt of solute / its eq wt
No of g eq wt =
wt of substance
its eq wt

15. Good Luck
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