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Water Analysis
Calculate the permanent and temporary hardness in ppm of a water sample with the
given data……
1. 50mL of distilled water = 0.5mL of soap solution
2. 50mL of CaCl2 standard solution = 40.5mL of soap solution
3. 50mL of given water = 15.5mL of soap solution
4. 50mL of boiled water = 12.5mL of soap solution
Calculate the permanent and temporary hardness in ppm of a water sample with the
given data……
1. 50mL of distilled water = 0.5mL of soap solution
2. 50mL of CaCl2 standard solution = 40.5mL of soap solution
3. 50mL of given water = 15.5mL of soap solution
4. 50mL of boiled water = 12.5mL of soap solution
V1 mL
V2 mL
V3 mL
V4 mL
Total hardness =
V3 - V1
V2 - V1
X 1000 ppm
Permanent hardness =
V4 - V1
V2 - V1
X 1000 ppm
Temporary hardness = Total Hardness - Permanent Hardness
Now, put the values and find answers…..
Total Hardness = 375 ppm
Permanent Hardness = 300 ppm
Temporary Hardness = 75 ppm
Your answers are ready……
A standard hard water contains 15 gm of CaCO3 per liter, 20 mL of this 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 the temporary hardness of the given sample of water, in terms of ppm.
1. Standardization of EDTA solution
Given, 1 Liter of standard hard water contains 15 gm CaCO3
1000 mL of standard hard water contains 15000 mg CaCO3
1 mL of standard hard water contains 15 mg CaCO3
Now, 25 mL of EDTA = 20 mL of standard hard water
= 20 x 15
= 300 mg of CaCO3
1 mL of EDTA = 300/25 mg of CaCO3 equivalent hardness
= 12 mg of CaCO3 equivalent hardness
2. Calculation of Total Hardness of water
100 mL of sample water = 18 mL of EDTA
= 18 x 12
= 216 mg of CaCO3 equivalent hardness
1000 mL of sample water = 2160 mg of CaCO3 equivalent hardness
1 Liter of sample water = 2160 mg of CaCO3 equivalent hardness
Hence, Total Hardness of Water = 2160 ppm
3. Calculation of Permanent Hardness
100 mL of boiled water = 12 mL of EDTA
= 12 x 12
= 144 mg of CaCO3 equivalent hardness
1000 mL of sample water = 1440 mg of CaCO3 equivalent hardness
1 Liter of boiled water = 1440 mg of CaCO3 equivalent hardness
Hence, Permanent Hardness of Water = 1440 ppm
Temporary hardness = Total Hardness - Permanent Hardness
4. Calculation of Temporary Hardness
Total Hardness = 2160 ppm
Permanent Hardness = 1440 ppm
Temporary Hardness = ??? ppm
Lime Requirement
Soda Requirement
Solved Numerical
on
Zeolite Process
Solved Numericals on Water Analysis: EDTA Method, Clark Method, Lime-Soda Process and Zeolite Process
Solved Numericals on Water Analysis: EDTA Method, Clark Method, Lime-Soda Process and Zeolite Process
Solved Numericals on Water Analysis: EDTA Method, Clark Method, Lime-Soda Process and Zeolite Process

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Solved Numericals on Water Analysis: EDTA Method, Clark Method, Lime-Soda Process and Zeolite Process

  • 1. Water Analysis Calculate the permanent and temporary hardness in ppm of a water sample with the given data…… 1. 50mL of distilled water = 0.5mL of soap solution 2. 50mL of CaCl2 standard solution = 40.5mL of soap solution 3. 50mL of given water = 15.5mL of soap solution 4. 50mL of boiled water = 12.5mL of soap solution
  • 2. Calculate the permanent and temporary hardness in ppm of a water sample with the given data…… 1. 50mL of distilled water = 0.5mL of soap solution 2. 50mL of CaCl2 standard solution = 40.5mL of soap solution 3. 50mL of given water = 15.5mL of soap solution 4. 50mL of boiled water = 12.5mL of soap solution V1 mL V2 mL V3 mL V4 mL Total hardness = V3 - V1 V2 - V1 X 1000 ppm
  • 3. Permanent hardness = V4 - V1 V2 - V1 X 1000 ppm Temporary hardness = Total Hardness - Permanent Hardness Now, put the values and find answers…..
  • 4. Total Hardness = 375 ppm Permanent Hardness = 300 ppm Temporary Hardness = 75 ppm Your answers are ready……
  • 5. A standard hard water contains 15 gm of CaCO3 per liter, 20 mL of this 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 the temporary hardness of the given sample of water, in terms of ppm.
  • 6.
  • 7.
  • 8. 1. Standardization of EDTA solution Given, 1 Liter of standard hard water contains 15 gm CaCO3 1000 mL of standard hard water contains 15000 mg CaCO3 1 mL of standard hard water contains 15 mg CaCO3 Now, 25 mL of EDTA = 20 mL of standard hard water = 20 x 15 = 300 mg of CaCO3 1 mL of EDTA = 300/25 mg of CaCO3 equivalent hardness = 12 mg of CaCO3 equivalent hardness
  • 9. 2. Calculation of Total Hardness of water 100 mL of sample water = 18 mL of EDTA = 18 x 12 = 216 mg of CaCO3 equivalent hardness 1000 mL of sample water = 2160 mg of CaCO3 equivalent hardness 1 Liter of sample water = 2160 mg of CaCO3 equivalent hardness Hence, Total Hardness of Water = 2160 ppm
  • 10. 3. Calculation of Permanent Hardness 100 mL of boiled water = 12 mL of EDTA = 12 x 12 = 144 mg of CaCO3 equivalent hardness 1000 mL of sample water = 1440 mg of CaCO3 equivalent hardness 1 Liter of boiled water = 1440 mg of CaCO3 equivalent hardness Hence, Permanent Hardness of Water = 1440 ppm
  • 11. Temporary hardness = Total Hardness - Permanent Hardness 4. Calculation of Temporary Hardness Total Hardness = 2160 ppm Permanent Hardness = 1440 ppm Temporary Hardness = ??? ppm
  • 12.
  • 14.
  • 15.
  • 16.
  • 17.
  • 18.