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.

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
Dr Brima Monday 01/02//2016---
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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
Dr Brima Monday 01/02//2016---
22/4/1437

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
Dr Brima Monday 01/02//2016---
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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

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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
)
Dr Brima Monday 01/02//2016---
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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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22/4/1437
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