Question

Q.5 A standard hard water contains 15g of CaCO3 per liters . 20ml a this required 25ml of EDTA
solution. 100ml of sample water required 18ml EDTA solution. The same sample after boiling required
12 ml of EDTA solutions. Calculate the temporary hardness of the given sample of water, in terms of
ppm.​

Answer 1

Answer:

this is the answer please kindly check

Answer 2

The 12 mg is temporary hardness of the given sample of water, in terms of ppm.​

• Hardness is the state of the minerals are how hard they are bonded.
• It is about how hard the resistant capacity is and whether it's not easily breakable.
• Hardness is highly resistant to pressure.
• Hardness indicates the amount of calcium and magnesium.
• Not all metals are to fall under the hardness category.
• The unit of hardness is N/mm2.
• An example of a hard substance is a diamond.

To find:

The temporary hardness of the given sample of water, in terms of

ppm.​

Given:

The same sample after boiling required 12 ml of EDTA solutions.

A standard hard water contains 15g of CaCO3 per liters .

20ml a this required 25ml of EDTA solution.

Solution:

1 ml of standard water = 15,000/1000 = 15 mg of CaCO3 equivalent.

25 ml of EDTA solution = 20 ml of SHW contains of CaCO3 equivalent.

1 ml of EDTA solution = 300/25 = 12 mg of CaCO3 equivalent.

Calculation of hardness of water:

100 ml of water and 18 ml of EDTA solution mg of CaCO3 equivalent.

Therefore,  12 mg is temporary hardness of the given sample of water, in terms of ppm.​

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