NUMERICALS
WATER
TECHNOLOGY
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1. Calculate the quantity of lime and soda required to soften 25000 L of water having
Ca(HCO3)2 = 4.86 ppm, Mg (HCO3)2 = 7.3 ppm, CaSO4 = 6.8 ppm, MgCl2 = 5.7 ppm, MgSO4 = 9
ppm, SiO2 = 3.5 ppm, NaCl = 5.85 ppm.
Constituent
Amount (mg/l)
Multiplication
Factor
CaCO3
Equivalents
L&S
Ca(HCO3)2
4.86
100/162
3
L
Mg (HCO3)2
7.3
100/146
5
2L
CaSO4
6.8
100/136
5
S
MgCl2
5.7
100/95
6
L+S
MgSO4
9
100/120
7.5
L+S
Lime = 74/100 [Ca(HCO3)2 + 2 Mg (HCO3)2 + MgCl2 + MgSO4 ] X Volume (lit)
= 74/100 [ 3+2 x 5+ 6+ 7.5 ] X 25,000
= 0.49 Kg
Soda = 106/100 [CaSO4 + MgCl2 + MgSO4 ] X Volume (lit)
= 106/100 [5 + 6 + 7.5 ] X 25000
= 0.49 Kg
2. A totally exhausted zeolite softener required 50 ml of NaCl solution, containing 351 g of NaCl/litre.
How many litres of hard water sample containing 70° Clark can be softened by this processor?
50 ml = 0.05 L
0.05 L of NaCl contains = 0.05 x 351 (gm/L) of NaCl
= 17.55 gm = 17.55 x 103 NaCl
= 17.55 x 103 X 50/58.5 mg CaCO3 eq.
= 1.5 x 104 mg CaCO3 eq.
1 mg/L = 0.07° Clark
70° Clark = 1000mg/L
no of litres of sample water softened = 1.5 x 104 mg / 1000 mg/L = 15 L
3. 1 gm of CaCO3 was dissolved in dil. HCl and diluted to 1 litre. 100 ml of this
solution required 90 ml of EDTA solution for titration. 100 ml of sample hard
water required 40 ml of EDTA solution. In another titration, 100 ml of sample
hard water on boiling, cooling and filtering required 20 ml of EDTA using EBT.
Calculate total, temporary and permanent hardness.
(i) MSHW = w/100V = 1/100 x1 = 0.01 M
(ii) 0.01 x 100 = MEDTA x 90
( SHW)
(EDTA)
0.011 M = MEDTA
(iii) MTH x 100 = 0.011 x 40
(Sample
Water)
(EDTA)
= 4.44 x 10 -3 = 4.44 x 10 -3 x 100 x 1000 ppm = 444 ppm
(iv) MPH x 100 = 0.011 x 40
( Boiled WS) (EDTA)
= 2.22 x 10 -3
= 2.22 x 10 -3 x 100 x1000 ppm
= 222 ppm
(v) T H = 444 – 222 = 222 ppm
4. 100 ml of a water sample on titration with N/50 H2SO4 gave a titre value of 5.8 mL to
phenolphthalein end point and another 100 mL sample on titration with same acid gave a titre value
of 11.6 ml to methyl orange end point. Calculate the type and amount of alkalinity of water sample in
terms of CaCO3.
Phenolphthalein Alkalinity:
Np x 100 = 1/50 x 5.8
(water sample)
( H2SO4)
= 1/50 x 5.8 x 1/100 = 0.00116 x 50 x 1000 ppm = 58 ppm
Methyl orange Alkalinity:
Nm x 100 = 1/50 x 11.6
(water sample)
( H2SO4)
= 1/50 x 11.6 x 1/100 = 0.00232 x 50 x 1000 ppm = 116 ppm
So the type of alkalinity would be,
P=½M
Therefore, Alkalinity due to Carbonate ions = 2P or M
= 116 ppm