Expt.no.3

Determination of percentage of copper in brass sample

Procedure:

Patr A:

Preparation of brass solution

Weigh accurately about 0.1 gram of given sample and transfer it in to a conical flask add 3ml of con

HNO3, and heat till alloy dissolves completely add 2tt of water boil and then add about 1g or urea, and

boil till the brown fumes of oxides of nitrogen are completely expelled , cool in tap water

Part B:

Estimation of copper

Burette : Sodium thio sulphate

Conical flask : Brass + Aqueous Ammonia (till pale blue ppt) + Dilute Acetic acid (till blue ppt disappears)

+ 5 more cc of acetic acid + 100 cc of 10g of KI

Indicator : Starch solution nearly at the end point .

End point : Disappearance of blue color leaving a white ppt

From the volume of sodium thio sulphate solution run down from the burette, calculate the percentage

of copper present in brass sample

Determination of percentage of copper in brass using standard sodium thiosulphate solution

Procedure: - weigh accurately brass sample 0.1 g in to a clean dry conical flask .Add 3ml of con HNO3.

When brass completely dissolves , add 1g of urea , 2tt of distilled water and boil the solution till the

oxides of nitrogen are completely expelled. Cool the solution add drop wise ammonium hydroxide till a

permanent bluish white ppt of cupric hydroxide is obtained. Add acetic acid till the precipitate just

dissolves. Add 1tt of 10% of KI titrate the liberated iodine against the standard sodium thiosulphate

solution taken in the burette tile the color of the solution is pale yellow . Add 1ml of starch indicator and

titrate further till a milky white ppt persists .

Repeat the procedure with another sample of brass. Let the volume of the sodium thiosulphate

consumed be ‘X’ ml

1 2 3Weight of brass(g)FBRIBR 0.0 0.0 0.0Vol of thiosulphate

Weigh of brass =wg, vol of sodium thiosulphate = Xml

Normality of sodium thiosulphate = 0.03X

1000ml of 1N Na2s2O3 = 63.54g of cu(eq.wt of cu)

Xml of 0.05N Na2c2O3 = 63.54 X 0.031000

= Ygm of cu

Wg of brass contains Y gm of cu.

100g of brass contains Y∗100W

Result :- % of cu in given brass sample is __________%

VIVA

1. What are the constituents of brass?

Brass is an alloy of cu (60 – 80%) and Zn(20 – 40%) with small amount of sn, pb and fe

2. What is the purpose of adding urea in the experiment?

To destroy excess HNO3 and completely remove the oxides of nitrogen

3. Why acetic acid is added ?

To neutralize excess of NH4OH and to make the solution slightly acidic

4. What is the role of NH 4OH?

NH4OH is added to neutralize excess of HNO3 is not removed, it would oxidize iodine to iodic

acid and decompose sodium thio sulphate

Expt.no.4

Determination of chemical oxygen demand (COD) from a waste water sample

Procedure :

Prep ration of standard ferrous ammonium sulphate (FAS)

Weigh out given FAS crystals and transfer in to a clean standard flask. To this add 1tt of dilute sulfuric

acid and make it up to mark by adding distill water, mix well.

NFAS = W X 4/m = ____________N = (M is mol.wt = eq.wt of FAS = 392)

Part - A

Burette = FAS solution

Conical flask = 25ml of waste water sample + 10ml of potassium dichromate + 1tt of sulphuric acid

Indicator : Ferroin indicator

End point : Bluish green to reddish brown

Part – B

Blank titration :- (The value will be given by the examiner)=28cc

From the difference in value run down (FAS) from the blank and with waste sample the COD of waste

water sample is determined.

Determination of COD from the waste water sample

Preparation of standard FAS solution

Weigh around 4.8gm of FAS into a 250cm3 volumetric flask. Add 1ml of con H2SO4 and swirl the flask to

dissolve the crystals. Make up the solution with distilled water and calculate the normality of FAS .

Transfer 25ml of waste water sample into 250cm 3 conical flask. Add 10ml of standard

potassium dichromate solution followed by 1tt full or 5ml of 1:1 sulphuric acid containing silver

sulphate with constant shaking of contents of the flask. Add three drops of ferroin indicator and

titrate against FAS solution until the solution turns from blue green to reddish brown.

For balance titration add 10ml to the obtained burette value.

Observations and calculations :-

Preparation of FAS solution

1. Weight of weighing bottle + FAS crystals = _________gms.

2. Weight of weighing bottle = _________gms.

…………………………

Weight of FAS crystals (W) = _________gms.

NFAS = W X 4 / Eq. wt. = …………… Z [Eq. wt. of FAS=392 ]

Burette readings: - [Y]

Calculations

Normality of FAS = ‘X’ ml

Volume of FAS = ‘Y’ ml

1000ml of 1N FAS = 8000mg of oxygen

Z = balance titrate value

(Z-Y)ml of ‘X’ N FAS = 8000 X (Z−Y )∗X

1000 ‘M’ mg of oxygen

25ml of waste water contains ‘M’ mg of oxygen

1000ml of waste water contains = m∗100025

= ________mg of oxygen.

VIVA

1. Chemical oxygen demand is the amount of oxygen in mg required to oxidize by strong

chemical oxident to organic and inorganic materials present in 1000cc of waste water.

2. BOD is the amount of oxygen required for the biological oxidation for organic matter during

aerobic condition 200c for a period of 5 days.

3. The oxidizable impurities of waste water are straight chain alcohol, acids.

4. The indicator used to determine the COD of waste water is ferroin. The color change at the

equivalence point bluish green to reddish brown.

Trail - 1 Trail - 2 Trail - 3

FBR

IBR

VRD

Determination of dissolved oxygen in water by winkler’s method

Procedure: -

In a winkler’s bottle take drain water apply stopper. Takeout 10cc of water, add 2cc of manganous

sulphate and 2cc of alkaline KI solution stopper and shake well for 10 minutes then 30 drops of con.

sulphuric acid and shake well for 5 minutes . Titrate librated iodine by pipetting 50cc of solution from

the bottle against 0.01Nsodium thiosulphate solution.

1 2 3

FBR

IBR

VRD (Xcc)

1000cc of 1N Na2 S2O3 = 8g of O2

Xcc of 0.01N = 8∗X∗0.011000

= a grams of O2

50cc of water = a grams of O2

106 cc of O2 or ppm (grams of dissolved oxygen in 1 million cc of water ) (8000 mg of oxygen in 106 cc

of water)

VIVA

1. BOD is the number of grams of free oxygen required to oxidize organic matter in a sample of

drain water under aerobic conditions at 20 c. in a period of 5 days.

2. BOD is mentioned as No . of grams or milligrams of oxygen required for the biological oxidation

of impurities present in 1000cc of water or ppm, i.e., No. of grams of oxygen required for

aerobic oxidation of oxidizable impurities present in 1 million or 106cc of water.

Expt No :10

COCLOROMETRIC DETERMINATION OF COPPER

Procedure: Draw out 5,10, 5,20 ml of given copper sulphate solution in to a 4 separate

50ml volumetric flask. Add 5ml of ammonia solution to each one of them and also in to the test solution

of unknown concentration and dilute up to the mark by adding distill water and mix well. Adjust the

absorbency to zero using blank solution (only ammonia and water). Plot a graph of absorbance (OD)

against volume of copper sulphate and determine the volume of given copper sulphate solution and

weight of copper in it.

OBSERVATIONS AND CALCULATIONS.

Volume of copper Sulphate

Absorbency orOptical density

Blank solution 0.00

5ml

10ml

15ml

20ml

Unknown solution[X ml]

1000 cc of copper sulphate contains solution 4 gms of CuSO4

X cc of copper sulphate contains = 4 x X/1000 = _________ a gms of CuSO4

250 gms of copper sulphate contains = 63 gms of Cu.

a gms of copper sulphate contains = 63 X a/250 = _________ gms of copper.

VIVA:

1. When a monochromatic light intensity Io is passed through a transparent medium a part Ia is

absord, a part is reflected Ir and remaining is transmitted It such that

Io = Ia+ Ir+It

For glass air interference Ir is negligible. Therefore, Io = Ia + It

2. Transmittance is It/Io

3. Absorbency or optical density is log [Io/It]

4. Lambert – beer law is A = Log[Io/It] = € ct. A is absorbency, € is a constant, its value depends on

the nature of transparent substance and wavelength; it is also called as molar extinction co-

efficient. ‘C’ is a concentration and t’ is the pass length

5. Pass length is thickness of the medium or the distance of the light passes in the transparent

medium and is also called cell thickness, it’s unit is cm.

6. The units of € is mol-1 cm-1 L.