17BEM0049DA1
17BEM0049DA1
17BEM0049DA1
Vedant karnatak
17BEM0049
, 2017
VIT UNIVERSITY
vellore
Digital Assignment – 1
mineral contaminants. They are CaCO3 = 50 mg/L ; MgCO3 = 100 mg/L ; CaCl2 =
111 mg/L; MgCl2 = 47.6 mg/L; NaCl = 5.86 mg/L. If this hard water is to be softened
by lime-soda process, calculate the amount of lime and soda needed or 20,000
Ans-
=74/100[50+238+50]
=250.12
=106/100[100+50]
=159
=5002400=5.0024kg.
For 20,000 litres of water for soda=159*20000
=3180000=3.1800kg.
2. 2.50 ml of a standard hard water containing 1 mg of pure CaCO 3 per 1 ml, consumed 20
mil of EDTA. 50 ml of a water sample consumed 25 ml of EDTA solution using EBT indicator.
3. A water sample contains the following .Mg(HCO3)2 = 95 mg/L; Ca(HCO3)2 = 203 mg/L;
CaSO4 = 1.36 mg/L; CaCl2 = 111 mg/L; MgCl2 = 95 mg/L . Calculate the temporary and
Ans- Given:-
Mg(HCO3)2 = 95mg/L
Ca(HCO3)2 = 203mg/L
CaSO4 = 13.6mg/L
CaCl2 = 111mg/L
Mgcl2 =95mg/L
= 65.06ppm
Ca(HCO3)2 = 20300/162
= 125.3ppm
CaSO4 = 1360/136
= 10ppm
CaCl2= 11100/111
= 100ppm
Mgcl2 = 9500/95
= 100ppm
= 246.31ppm
Permanent hardness is caused by the chlorides and sulphates salts of calcium and
magnesium :-
= 210ppm
4.Calculate the amount of lime required for softening 50,000 liters of hard water
containing Ca(HCO3)2 = 25 ppm, Mg(HCO3)2 = 144 ppm, CaCl2 = 111 ppm, MgCl2
Ans-
NAME OF THE QUANTITY OF MOLECULEAR L OR S
GIVEN THE GIVEN WEIGHT
COMPOUND COMPOUND(mg/L)
Ca(HCO3)2 25 162 2500/162 L
=15.43
Mg(HCO3)2 144 145 14400/145 2L
=99.31
CaCl2 111 111 1100/111 S
=100
MgCl2 95 95 9500/95 L+S
=100
NaSO4 15 142 1500/142 -
=10.56
Fe2O3 25 152 2500/152 -
=64.4
= 314.05Kg
= 15.7Kg
5. How the scale is removed from the boiler water? Explain internal and external
B] Internal Treatment:
In this process an ion is prohibited to exhibit its original character by ‘complexing’ or
converting it into other more soluble salt by adding appropriate reagents.
This is done either by a) precipitating the scale forming impurities in the form of sludge or
b)to convert them into compounds ,which will stay in dissolved form in water and thus not
cause harm.
1-Colloidal conditioning:
2-Phosphate conditioning:
3-Carbonate conditioning:
4-Calgon conditioning:
5-Treating with sodium aluminate:
6.Explain the method of determining the hardness of water?
The estimation is done by adding excess of standard Na2CO3 solution to a given volume of
boiled water containing permanent hardness. The chlorides and sulphates form insoluble
carbonates. The residual Na2CO3 is titrated against standard acid and the difference of
Na2CO3 equivalent gives permanent hardness.
Another method known as EDTA method is also used to determine permanent hardness.
EDTA is used as a complexing agent for Ca2+ and Mg2+. Eriochrome black t is used as an
indicator and pH is maintained at pH 10 using ammonia buffer. EBT forms unstable
complex with Ca2+ and Mg2+ giving wine red colour. When EDTA is added and the total
Ca2+ and Mg2+ forms complex with it and the indicator becomes free, the colour of the
solution changes from wine red to silver blue at the end point.
7. How will you treat the municipal water for domestic purpose?
(1) Sterilization
(2) Dechlorination
(3) Desalination of salt water
(1) Sterilization:-
(a) By addition of bleaching powder:-
Here sterilization is effected by the
HOCl generated by using bleaching
powder.
(2) Dechlorination :-
same EDTA was required for 100 ml of standard hard water containing 1 g
= 1.21 mg of CaCO3
=1210 ppm
9.A sample water on analysis gave the following data: MgCl2 = 95 mg/L; CaSO4 = 13.6
mg/L; Mg(HCO3)2 = 73 mg/L; MgSO4 = 120 mg/L. Calculate the amount of lime (86%)
pure and soda (83% pure) needed for treatment of million litres of water.
Ans-
Lime required=0.74[100+100+128.06]
=242.76kg
=282.27kg
Soda required=1.06[100+100+100]
=318Kg
=383.13Kg
10.Describe the activated carbon filtration method and reverse osmosis with
diagram.
Ans-
1. Granular activated carbon (GAC) is commonly used for removing organic constituents
and residual disinfectants in water supplies.
2.Typical surface area for activated carbon is approximately 1,000 square meters per gram
(m2/gm).
3.However, different raw materials produce different types of activated carbon varying in
hardness, density, pore and particle sizes, surface areas, extractible, ash and pH. These
differences in properties make certain carbons preferable over others in different
applications. 4.Activated carbon filtration can effectively reduce certain organic
compounds and chlorine in drinking water. It can also reduce the quantity of lead,
dissolved radon, and harmless taste- and odor-causing compounds.
5.The two principal mechanisms by which activated carbon removes contaminants from
water are adsorption and catalytic reduction. Organics are removed by adsorption and
residual disinfectants are removed by catalytic reduction.
6. The large surface area of the activated carbon, due to its particle size and pore
configuration, allows for the adsorption to take place.
7.Activated carbon can remove and destroy residual disinfectants (chlorine and chloramine)
through a catalytic reduction reaction.
8.This is a chemical reaction that involves a transfer of electrons from the activated carbon
surface to the residual disinfectant. In other words, activated carbon acts as a reducing
agent. 9. Activated carbon's removal of chlorine reduces the chlorine to a non-oxidative
chloride ion. 10.AC filters will not remove microbial contaminants (such as bacteria and
viruses), calcium and magnesium (hard water minerals), fluoride, nitrate, and many other
compounds.
Reverse Osmosis