1. Dr.Shuchita Agrawal BTIRT Sironja,Sagar(MP)
CHEMICAL ANALYSIS OF WATER
Dr.Shuchita Agrawal BTIRT Sironja,Sagar(MP)

2. Alkalinity
It is due to OH- & CO3- - ions known as caustic alkalinity and by HCO3- is temporary hardness.
It is estimated by titrating it with std acid, using phenolphthalein and methyl orange.
OH-+H+→H2O
CO3--+H+→HCO3-
HCO3-+H+→H2O+CO2

3. Possible combinations are-
OH-/ CO3--
HCO3-
OH- &CO3--
CO3-- &HCO3-
OH- & HCO3- (not possible)
They combines and forms CO3--
OH-+ HCO3- → H2O+ CO3--
OH-+ HCO3- +CO3-- (not possible)

4. Procedure- Take 100 ml of water in a clean conical flask.
Add 2-3 drops of phenolphthainlein indicator.
Solution becomes pink .
Run N/50 H2SO4 from burette,till the pink colour discharged.
Reading of burette is V1ml.

5. Now add 2-3 drops of methyl orange to the same solution.
Solution becomes orange.
Run N/50 H2SO4 again from burette, till the Pink colour appears.
Reading of burette is V2ml.
Calculations-

6. P=V1×10 ppm---------(1) M=V1×10 ppm------------(2)

7. S.No.
Alkalinity
OH- (ppm)
CO3--(ppm)
HCO3--(ppm) 1
P=0 M 2
P=1/2M 2P 3
P<1/2M
(M-2P) 4
P>1/2M
(2P-M)
2(M-P) 5
P=M

8. Hardness It is a complexometric method.
EDTA is used in the form of it’s sodium salt.
It produces anion, which combines with Ca++ & Mg++ ions.
EDTA

9. EDTA
Ethylenediaminetetraacetic acid
(disodium salt)

10. Procedure- Preparation of std hard water-
Dissolve 1 gm of pure CaCO3 in dil HCl.
Evaporate the solution up to dryness.
Now dissolve in 1L distilled water.
Solution contains 1mg CaCO3/1ml.
Preparation of EDTA solution-
Dissolve 4gms of pure EDTA & 0.1 gm of MgCl2 in 1L distilled water.
Preparation of indicator-
Dissolve 0.5 gm of Eriochrome Black T in 100 ml of alcohol.

11. Preparation of buffer solution-
Dissolve buffer tablets of 10 pH in distilled water.
Standardization of EDTA solution-
Take 50 ml of std hard water in conical flask.
Add ml buffer solution.
Add 4-5 drops of indicator Eriochrome Black T.
Now titrate it with EDTA solution till wine red colour changes to blue.
suppose the reading of burette is V1 ml.

12. Titration of Hard water-
Take 50 ml of given hard water sample. Titrate it with same procedure.
Suppose the reading of burette is V2 ml.
Titration for permanent hardness-
Take 250 ml of given hard water sample in a beaker.
Boil it till the volume reduces to 50 ml.
Filter the water and make the volume 250 ml with distilled water.

13. Now titrate 50ml of boiled water with same procedure.
Calculations-
suppose the reading of burette is V3 ml.

14. Dissolved Oxygen (DO)
Natural water always contains some amount of dissolved oxygen. The solubility of oxygen is directly proportional to the pressure and inversely proportional to the temperature. The DO level of fresh water ranges from 7 mg/L to 14.6 mg/L at 35⁰C and 1 atm pressure. The DO level of sea water is the least as compare to the other samples.

15. Determination of DO- Winkler’s method-
Manganese sulphat,Pottasium hydroxide, Sulphuric acid & Potassium iodide are added to water sample.
MnSO4 + 2KOH → Mn(OH)2+K2SO4
2Mn(OH)2 + O2 →2 MnO(OH)2
Basic magnic oxide
Basic magnic oxide is carrier of dissolved oxygen. It reacts with sulphuric acid to produce nascent oxygen.
MnO(OH)2 +H2SO4 → MnSO4 +2H2O+[O]

16. Potassium iodide is oxidized by nascent oxygen in acidic medium to produce I2 .
2KI + H2SO4 +[O]→ K2SO4 +H2O+I2
Liberated iodine is titrated by hypo solution (N/50) .
I2 + 2Na2S2O3→ Na2S4O6+2NaI
Suppose the burette reading is A ml.

17. Calculations-
Strength =

18. Biological oxygen demand (BOD)
The organic matter in waste water is of two kinds-
Can be oxidized by bacteria- biologically active .
Can not be oxidized by bacteria-
biologically inactive.

19. The demand of O2 for various biological processes is known as BOD.
The pure drinking water should have a
(BOD)5 in the range of mg/L .
(BOD)5 is the amount of o2 required by micro-organisms to decompose the organic matter in a water sample over a period of 5-days at 20⁰c

20. Determination of BOD-
Polluted water sample is diluted with special dilution water.
The special dilution water contains micro-organisms.
The diluted water sample is divided in two equal volumes in two bottles.
The amount DO in first bottle is determined immediately by winkler’s method.

21. The amount DO in second bottle is determined after 5 days incubation
The amount DO in second bottle is determined after 5 days incubation. by winkler’s method.
Calculations-
(BOD)5 mg/L = (D1-D2) × f
D1 = DO of diluted sample before incubation.
D2 = DO of diluted sample after 5 days incubation.
f = Dilution factor=

22. Chemical oxygen demand (COD)
It is the measure of O2 demand for biologically active and inert matter.
COD is the amount of oxygen used for chemical oxidation of organic content of a sample with the help of strong oxidizing agent in acidic medium.

23. Determination of COD-
100 ml of water sample is refluxed with K2Cr2O7 and conc. H2SO4.
The organic matter is oxidized in CO2 and H2O.
The un reacted K2Cr2O7 is titrated with FAS using Ferroin as indicator.
Sample titration- 0.5 gm of mercuric sulphat is taken in 250 ml round bottom flask & dissolved in 20 ml water sample.

24. Now add some porcelain pieces.
Add 30 ml silver sulphate (H2SO4) reagent & 15 ml of 0.25 N K2Cr2O7 ,then heat for one hour.
Now cool the sample & make the volume 150 ml with distilled water.
Add Ferroin as indicator and titrate it with 0.25 N FAS solution.
At the end point solution changes bluish green to wine red.
suppose the reading of burette is V1 ml.

25. Blank titration- Same procedure is repeated with distilled water.
suppose the reading of burette is V2 ml.
Calculations-
COD of water sample
=0.25×(V2-V1)×8×1000 mg/L 20

26. Determination of Chlorides-
Water sample is titrated with the std solution of AgNO3 using K2CrO4 as indicator.
The solubility product of AgCl is lower then that of Ag2CrO4.
As long as Cl- ions are available, less soluble AgCl is precipitated.
At this stage Ag+ ions are not sufficient to precipitate the Ag2Cro4.

27. As soon as all Cl- ions are finished as AgCl, a drop of AgNO3 gives red ppt of Ag2Cro4.
Ag+ + Cl- → AgCl (white ppt)
2Ag+ + CrO4-- → Ag2CrO4 (red ppt)
100 ml watrer is taken in conical flask & add 1 ml K2CrO4 solution.
Run N/50 AgNO3 solution from burette.
Initially white ppt comes, continue the addition of AgNO3 solution till reddish brown colour obtained.
Note the reading of burette, suppose the reading of burette is V2 ml.

28. Calculations-
suppose the reading of burette is V2 ml.
Strength of Cl- ions
=V2×N×1×35.5×1000
50×100
=7.1×V2 ppm

29. Determination of free Chlorine-
It is based on the oxidation of KI by free Cl2.
Chlorine liberates the equal amount of I2
The liberated I2 is titrated with hypo (Na2S2O3) using starch as indicator.
Cl2 + KI → 2KCl + I2
I2 + KI → KI3
I2 + starch → Deep Blue
I2 + 2Na2S2O3 → 2NaI + Na2S4O6

30. 50 mL of water is taken in conical flask & add 10 mL of 10% KI solution.
Now shake well and titrate it with N/50 hypo solution.
Initially solution becomes yellow.
Now add starch solution, solution becomes blue.
Titrate it again with hypo till white ppt comes.
The volume of hypo used is V2 mL.

31. Calculations-
Strength of Cl2
=V2 × N × 35.5 × 1000 mg / L
50 × 50
= V2 × 35.5 × 1000 mg / L
2500
= 14.2 × V2 mg/L s
=14.2 × V2 ppm

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