Numericals
A standard hard water contains 15 g of CaCO3 per liter. 20 mL of this water sample required 25 mL of
EDTA solution, 100 mL of sample water required 18 mL of EDTA solution. The sample after boiling
required 12 mL EDTA solution. Calculate total, permanent and temporary hardness of the given water
sample in ppm.
Method 1
25 mL of EDTA 18 mL of EDTA 100 mL of water
20 mL of 15 sample (unknown
soln (unknown soln (unknown
gm /L CaCO3 conc.) conc.)
hardness)
Find Total Hardness
Find conc of EDTA 12 mL of EDTA 100 mL of boiled water
soln (unknown sample (unknown
conc.) hardness)
Find Permanent
Hardness
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Numericals
A standard hard water contains 15 g of CaCO3 per liter. 20 mL of this water sample required 25 mL of EDTA solution, 100 mL of sample water required 18 mL of EDTA
solution. The sample after boiling required 12 mL EDTA solution. Calculate total, permanent and temporary hardness of the given water sample in ppm.
Method 1 (Contd…)
𝑔𝑚
For unboiled water sample
𝐶𝑜𝑛𝑐. 𝑜𝑓 𝐶𝑎𝐶𝑂3 𝑖𝑛 𝑠𝑡𝑑. 𝑠𝑜𝑙𝑛 = 15
𝐿 𝑁𝐸𝐷𝑇𝐴 × 𝑉𝐸𝐷𝑇𝐴 = 𝑁𝑊𝑆 × 𝑉𝑊𝑆
15 𝑒𝑞
= 0.24 × 18 = 𝑁𝑊𝑆 × 100
50 𝐿
For 1st Titration 0.24 × 18 𝑒𝑞
𝑁𝑊𝑆 =
100 𝐿
𝑁𝑠𝑡𝑑 𝑠𝑜𝑙𝑛 × 𝑉𝑠𝑡𝑑 𝑠𝑜𝑙𝑛 = 𝑁𝐸𝐷𝑇𝐴 × 𝑉𝐸𝐷𝑇𝐴
0.24 × 18
15 = × 50 𝑔𝑚. 𝐿−1
100
× 20 = 𝑁𝐸𝐷𝑇𝐴 × 25
50
0.24 × 18
15 × 20 = × 50 × 1000 𝑚𝑔. 𝐿−1
100
𝑁𝐸𝐷𝑇𝐴 = 𝑁
50 × 25 = 2160 𝑚𝑔. 𝐿−1
= 0.24 𝑁
= 2160 𝑝𝑝𝑚 𝑜𝑓 𝐶𝑎𝐶𝑂3
This is Total Hardness of given water sample
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Numericals
A standard hard water contains 15 g of CaCO3 per liter. 20 mL of this water sample required 25 mL of EDTA solution, 100 mL of sample water required 18 mL of EDTA
solution. The sample after boiling required 12 mL EDTA solution. Calculate total, permanent and temporary hardness of the given water sample in ppm.
Method 1 (Contd…)
For boiled water sample, Therefore, Temporary hardness of given
𝑁𝐸𝐷𝑇𝐴 × 𝑉𝐸𝐷𝑇𝐴 = 𝑁𝑊𝑆 × 𝑉𝑊𝑆 water sample
0.24 × 12 = 𝑁𝑊𝑆 × 100
= 2160 − 1440 𝑝𝑝𝑚 𝑜𝑓 𝐶𝑎𝐶𝑂3
= 720 𝑝𝑝𝑚 𝑜𝑓 𝐶𝑎𝐶𝑂3
0.24 × 12 𝑒𝑞
𝑁𝑊𝑆 =
100 𝐿
0.24 × 12
= × 50 𝑔𝑚. 𝐿−1
100
0.24 × 12
= × 50 × 1000 𝑚𝑔. 𝐿−1
100
= 1440 𝑚𝑔. 𝐿−1
= 1440 𝑝𝑝𝑚 𝑜𝑓 𝐶𝑎𝐶𝑂3
This is permanent Hardness of
given water sample
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Numericals
A standard hard water contains 15 g of CaCO3 per liter. 20 mL of this water sample required 25 mL of EDTA solution, 100 mL of sample water required 18 mL of EDTA
solution. The sample after boiling required 12 mL EDTA solution. Calculate total, permanent and temporary hardness of the given water sample in ppm.
Method 2
25 𝑚𝐿 𝑜𝑓 𝐸𝐷𝑇𝐴 𝑠𝑜𝑙𝑛 ≡ 20 𝑚𝐿 𝑜𝑓 15𝑔𝑚. 𝐿−1 𝐶𝑎𝐶𝑂3
15𝑔𝑚
≡ 20 𝑚𝐿 × 𝑜𝑓𝐶𝑎𝐶𝑂3
1000 𝑚𝐿
20 × 15
≡ × 1000 𝑚𝑔 𝑜𝑓 𝐶𝑎𝐶𝑂3
1000
For unboiled water sample
25 𝑚𝐿 𝑜𝑓 𝐸𝐷𝑇𝐴 𝑠𝑜𝑙𝑛 ≡ 20 × 15 𝑚𝑔 𝑜𝑓 𝐶𝑎𝐶𝑂3
20 × 15 This is the equivalent amount of
18 𝑚𝐿 𝑜𝑓 𝐸𝐷𝑇𝐴 𝑠𝑜𝑙𝑛 ≡ × 18 𝑚𝑔 𝑜𝑓 𝐶𝑎𝐶𝑂3 CaCO3 in 100 mL of unboiled water
25
sample
Therefore, Equivalent amount of CaCO3 in 1L(1000 mL of unboiled water sample)
20 × 15 × 18
= × 10𝑚𝑔 = 2160 𝑚𝑔
25
Therefore, Concn of CaCO3 in unboiled Water sample
= 2160 𝑚𝑔. 𝐿−1 = 2160 𝑝𝑝𝑚 𝑜𝑓 𝐶𝑎𝐶𝑂3
This is total hardness of water sample
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Numericals
A standard hard water contains 15 g of CaCO3 per liter. 20 mL of this water sample required 25 mL of EDTA solution, 100 mL of sample water required 18 mL of EDTA
solution. The sample after boiling required 12 mL EDTA solution. Calculate total, permanent and temporary hardness of the given water sample in ppm.
Method 2 (Contd…)
For boiled water sample
25 𝑚𝐿 𝑜𝑓 𝐸𝐷𝑇𝐴 𝑠𝑜𝑙𝑛 ≡ 20 × 15 𝑚𝑔 𝑜𝑓 𝐶𝑎𝐶𝑂3
20 × 15 This is the equivalent amount of
12 𝑚𝐿 𝑜𝑓 𝐸𝐷𝑇𝐴 𝑠𝑜𝑙𝑛 ≡ × 18 𝑚𝑔 𝑜𝑓 𝐶𝑎𝐶𝑂3 CaCO3 in 100 mL of boiled water
25
sample
Therefore, Equivalent amount of CaCO3 in 1L(1000 mL of unboiled water sample)
20 × 15 × 12
= × 10𝑚𝑔 = 1440 𝑚𝑔
25
Therefore, Concn of CaCO3 in boiled Water sample
= 1440 𝑚𝑔. 𝐿−1 = 1440 𝑝𝑝𝑚 𝑜𝑓 𝐶𝑎𝐶𝑂3
This is permanent hardness of water sample
Therefore, temporary hardness of water sample
= 2160 − 1440 𝑝𝑝𝑚 = 720 𝑝𝑝𝑚 𝑜𝑓 𝐶𝑎𝐶𝑂3
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Numericals
A sample of water on analysis has been found to contain
Mg(HCO3)2=73mg/L; Ca(HCO3)2=162mg/L; MgCl2= 95mg/L; CaSO4= 136mg/L; MgSO4=60 mg/L
;NaCl=20mg/L
Calculate the temporary and total hardness of a sample of water in degree French(0Fr).
(atomic wt. of H=1; C=12; N=14; O=16; Na=23; Mg=24; S=32; Cl=35.5; Ca=40)
Molecular Concn in
Salt Concn in ppm of CaCO3
Weight mg/L
73 𝑚𝑔𝐿−1 × 50
Mg(HCO3)2 146 73 = 50 𝑚𝑔𝐿−1
146/2
162𝑚𝑔𝐿−1 × 50
Cause Temporary Hardness
Ca(HCO3)2 162 162 = 100 𝑚𝑔𝐿−1
162/2
95 𝑚𝑔𝐿−1 × 50
MgCl2 95 95 = 100 𝑚𝑔𝐿−1
95/2
136 𝑚𝑔𝐿−1 × 50
CaSO4 136 136 = 100 𝑚𝑔𝐿−1
136/2 Cause Permanent Hardness
60 𝑚𝑔𝐿−1 × 50
MgSO4 120 60 = 50 𝑚𝑔𝐿−1
120/2
NaCl 58.5 20 Does not cause Hardness
Temporary Hardness = 50 + 100 = 150 ppm = 15 oFr
Permanent Hardness = 100+100+50=250 ppm =25 oFr
Total Hardness = 150 + 250 = 400 ppm =40 oFr
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Numericals
A water sample is alkaline to both phenolphthalein as well as methyl orange. 100 mL of this water sample
required 7.5 mL of N/50 HCl for neutralisation to phenolphthalein end point (P).At this stage a few drops
of methyl orange were added. The acid required further was 14.5 mL of N/50 HCl for neutralisation to
methyl orange end point(M). Calculate alkalinity of water as CaCO3 due to the presence of carbonate and
bicarbonate.
Method 1
𝑁
𝑁𝑎𝑐𝑖𝑑 = = 0.02 𝑁 𝑉𝑎𝑐𝑖𝑑 = 100 𝑚𝐿 𝑃𝑟𝑒𝑎𝑑 = 7.5 𝑚𝐿 𝑀𝑟𝑒𝑎𝑑 = 7.5𝑚𝐿 + 14.5𝑚𝐿
50 = 22 𝑚𝐿
First, we will calculate 𝑃𝑎𝑙𝑘 and 𝑀𝑎𝑙𝑘
𝑃𝑎𝑙𝑘 × 𝑉𝑠𝑎𝑚𝑝𝑙𝑒 = 𝑁𝑎𝑐𝑖𝑑 × 𝑉𝑎𝑐𝑖𝑑 𝑀𝑎𝑙𝑘 × 𝑉𝑠𝑎𝑚𝑝𝑙𝑒 = 𝑁𝑎𝑐𝑖𝑑 × 𝑉𝑎𝑐𝑖𝑑
𝑃𝑎𝑙𝑘 × 100 𝑚𝐿 = 0.02 𝑁 × 7.5 𝑚𝐿 𝑀𝑎𝑙𝑘 × 100 𝑚𝐿 = 0.02 𝑁 × 22 𝑚𝐿
0.02 × 7.5 0.02 × 22
𝑃𝑎𝑙𝑘 = 𝑁 𝑀𝑎𝑙𝑘 = 𝑁
100 100
= 0.0015 𝑒𝑞 𝐿−1 = 0.0044 𝑒𝑞 𝐿−1
= 0.0015𝑒𝑞 𝐿−1 × 50 𝑔𝑚 𝑒𝑞 −1 𝑜𝑓 𝐶𝑎𝐶𝑂3 = 0.0044𝑒𝑞 𝐿−1 × 50 𝑔𝑚 𝑒𝑞 −1 𝑜𝑓 𝐶𝑎𝐶𝑂3
= 0.075 𝑔𝑚 𝐿−1 𝑜𝑓 𝐶𝑎𝐶𝑂3 = 0.22 𝑔𝑚 𝐿−1 𝑜𝑓 𝐶𝑎𝐶𝑂3
= 0.075 × 1000 𝑚𝑔 𝐿−1 𝑜𝑓 𝐶𝑎𝐶𝑂3 = 0.22 × 1000 𝑚𝑔 𝐿−1 𝑜𝑓 𝐶𝑎𝐶𝑂3
= 75 𝑝𝑝𝑚 𝑜𝑓 𝐶𝑎𝐶𝑂3 = 220 𝑝𝑝𝑚 𝑜𝑓 𝐶𝑎𝐶𝑂3
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Numericals
A water sample is alkaline to both phenolphthalein as well as methyl orange. 100 mL of this water sample required 7.5 mL of N/50 HCl for
neutralisation to phenolphthalein end point (P).At this stage a few drops of methyl orange were added. The acid required further was 14.5
mL of N/50 HCl for neutralisation to methyl orange end point(M). Calculate alkalinity of water as CaCO3 due to the presence of carbonate and
bicarbonate.
Method 1 (Contd…)
𝑃𝑎𝑙𝑘 = 75 𝑝𝑝𝑚 𝑜𝑓 𝐶𝑎𝐶𝑂3 𝑀𝑎𝑙𝑘 (220 𝑝𝑝𝑚)
𝑃𝑎𝑙𝑘 (75 𝑝𝑝𝑚) <
𝑀𝑎𝑙𝑘 = 220 𝑝𝑝𝑚 𝑜𝑓 𝐶𝑎𝐶𝑂3 2
Therefore, the sample contains 𝐻𝐶𝑂3− and 𝐶𝑂32− only
Therefore, using the table
𝑂𝐻 − = 0 𝑝𝑝𝑚
𝐻𝐶𝑂3− = 𝑀𝑎𝑙𝑘 − 2 × 𝑃𝑎𝑙𝑘
= 220 𝑝𝑝𝑚 − 2 × 75 𝑝𝑝𝑚
= 70 𝑝𝑝𝑚 𝑜𝑓 𝐶𝑎𝐶𝑂3
𝐶𝑂32− = 2 × 𝑃𝑎𝑙𝑘
= 2 × 75 𝑝𝑝𝑚
= 150 𝑝𝑝𝑚 𝑜𝑓 𝐶𝑎𝐶𝑂3
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Numericals
A water sample is alkaline to both phenolphthalein as well as methyl orange. 100 mL of this water sample required 7.5 mL of N/50 HCl for
neutralisation to phenolphthalein end point (P).At this stage a few drops of methyl orange were added. The acid required further was 14.5
mL of N/50 HCl for neutralisation to methyl orange end point(M). Calculate alkalinity of water as CaCO3 due to the presence of carbonate and
bicarbonate.
Method 2
𝑁
𝑁𝑎𝑐𝑖𝑑 = = 0.02 𝑁 𝑉𝑎𝑐𝑖𝑑 = 100 𝑚𝐿 𝑃𝑟𝑒𝑎𝑑 = 7.5 𝑚𝐿 𝑀𝑟𝑒𝑎𝑑 = 7.5𝑚𝐿 + 14.5𝑚𝐿
50 = 22 𝑚𝐿
First, we will compare 𝑃𝑟𝑒𝑎𝑑 and 𝑀𝑟𝑒𝑎𝑑
𝑀𝑟𝑒𝑎𝑑 (22 𝑚𝐿)
𝑃𝑟𝑒𝑎𝑑 7.5 𝑚𝐿 <
2
Therefore, the sample contains 𝐻𝐶𝑂3− and 𝐶𝑂32− only
Using the table,
Volume of acid required to neutralize 𝑂𝐻 − = 0 𝑚𝐿
Volume of acid required to neutralize 𝐻𝐶𝑂3− = 𝑀𝑟𝑒𝑎𝑑 − 2 × 𝑃𝑟𝑒𝑎𝑑 = 22 𝑚𝐿 − 2 × 7.5 𝑚𝐿 = 7 𝑚𝐿
Volume of acid required to neutralize 𝐶𝑂32− = 2 × 𝑃𝑟𝑒𝑎𝑑 = 2 × 7.5 𝑚𝐿 = 15 𝑚𝐿
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Numericals
A water sample is alkaline to both phenolphthalein as well as methyl orange. 100 mL of this water sample required 7.5 mL of N/50 HCl for
neutralisation to phenolphthalein end point (P).At this stage a few drops of methyl orange were added. The acid required further was 14.5
mL of N/50 HCl for neutralisation to methyl orange end point(M). Calculate alkalinity of water as CaCO3 due to the presence of carbonate and
bicarbonate.
Method 2 (Contd…)
Therefore,
𝑁𝐻𝐶𝑂3− × 𝑉𝑠𝑎𝑚𝑝𝑙𝑒 = 𝑁𝑎𝑐𝑖𝑑 × 𝑉𝑎𝑐𝑖𝑑(𝐻𝐶𝑂3− ) 𝑁𝐶𝑂32− × 𝑉𝑠𝑎𝑚𝑝𝑙𝑒 = 𝑁𝑎𝑐𝑖𝑑 × 𝑉𝑎𝑐𝑖𝑑(𝐶𝑂32− )
𝑁𝐻𝐶𝑂3− × 100 𝑚𝐿 = 0.02 𝑁 × 7 𝑚𝐿 𝑁𝐶𝑂32− × 100 𝑚𝐿 = 0.02 𝑁 × 15 𝑚𝐿
0.02 × 7 0.02 × 15
𝑁𝐻𝐶𝑂3− = 𝑁 = 0.0014 𝑒𝑞 𝐿−1 𝑁𝐶𝑂32− = 𝑁 = 0.003 𝑒𝑞 𝐿−1
100 100
= 0.0014 𝑒𝑞 𝐿−1 × 50 𝑔𝑚 𝑒𝑞 −1 𝑜𝑓 𝐶𝑎𝐶𝑂3 = 0.003 𝑒𝑞 𝐿−1 × 50 𝑔𝑚 𝑒𝑞 −1 𝑜𝑓 𝐶𝑎𝐶𝑂3
= 0.07 𝑔𝑚 𝐿−1 𝑜𝑓 𝐶𝑎𝐶𝑂3 = 0.15 𝑔𝑚 𝐿−1 𝑜𝑓 𝐶𝑎𝐶𝑂3
= 0.07 × 1000 𝑚𝑔 𝐿−1 𝑜𝑓 𝐶𝑎𝐶𝑂3 = 0.15 × 1000 𝑚𝑔 𝐿−1 𝑜𝑓 𝐶𝑎𝐶𝑂3
= 70 𝑚𝑔 𝐿−1 𝑜𝑓 𝐶𝑎𝐶𝑂3 = 150 𝑚𝑔 𝐿−1 𝑜𝑓 𝐶𝑎𝐶𝑂3
= 70 𝑝𝑝𝑚 𝑜𝑓 𝐶𝑎𝐶𝑂3 = 150 𝑝𝑝𝑚 𝑜𝑓 𝐶𝑎𝐶𝑂3
𝐻𝐶𝑂3− = 70 𝑝𝑝𝑚 𝑜𝑓 𝐶𝑎𝐶𝑂3 𝐶𝑂32− = 150 𝑝𝑝𝑚 𝑜𝑓 𝐶𝑎𝐶𝑂3
𝑂𝐻 − = 0 𝑝𝑝𝑚 𝑜𝑓 𝐶𝑎𝐶𝑂3
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