100 mL sample of hard water is titrated with 500 " mL of " 0.001 M EDTA solution at `pH=10`, using eriochrome black-T indicator to reach equivalence point. An equal another amount of hard water sample is boiled for 30 min. After filtration and cooling, the same sample is titrated with 200 " mL of " 0.011 M EDTA solution at `pH=10` using Mg-EDTA complex solution and erichrome black-T indicator to reach equivalence point.
(i). Calculate the total hardness of water sample (temporary`+`permanent) in ppm of `CaCO_3`.
(ii). Calculate the permanent hardness of water sample in ppm of `CaCO_3`.
(iii). Calculate the temporary hardness of water sample in ppm of `CaCO_3`

100 mL sample of hard water is titrated with 500 mL of 0.001 MEDTA solution at pH=10 , using eriochrome black -T indicator to reach equivalence point. An equal another amount of hard water sample is boiles for 30 mi n . After filtration and cooling, the same sample is titrated with 200 mL of 0.001 M EDTA solution at pH=10 using Mg-EDTA complex solution and erichrome black -T indicator to reach equivalence point. a. Calculate the total hardness of water sample (temprary +permanent) in p p m of CaCO_(3) . b. Calculate the permanent hardness of water sample is p p m of CaCO_(3) . c. Calculate the temporary hardness of water sample is p p m of CaCO_(3) .

A 50 mL sample of hard water containing Ca^(2+) and Mg^(2+) ions is titrated with 50 mL. 0.005 M EDTA solution at pH=10 , using eriochrome balck -T indicator to reach equivalence point. In a equal amount of hard water sample, Mg^(2+) ions are precipitated as Mg(OH)_(2) by adding suitable amount of NaOH . the solution, after precipitation of Mg(OH)_(2) , is stirred and then titrated with EDTA solution using calcon as indicator, and it requires 10 mL of above EDTA solution to reach equivalence point. a. Calculate the strength of Ca^(2+) and Mg^(2+) ions present in hard water. b. Calculate the hardness due to Ca^(2+) ions in p p m of CaCO_(3) . c. Calculate the hardness due to Mg^(2+) ions in p p m of CaCO_(3) . d. Calculate the total hardness of water in p p m of CaCO_(3) .

A 50 mL sample of hard water containing Ca^(2+) and Mg^(2+) ions is titrated with 50 mL. 0.005 M EDTA solution at pH=10 , using eriochrome balck -T indicator to reach equivalence point. In a equal amount of hard water sample, Mg^(2+) ions are precipated as Mg(OH)_(2) by adding suitable amount of NaOH . the solution, after precipation of Mg(OH)_(2) , is strired and then titrated with EDTA solution using calcon as indicator, and it requires 10 mL of above EDTA solution to reach equivalence point. a. Calculate the strength of Ca^(2+) and Mg^(2+) ions present in hard water. b. Calculate the hardness due to Ca^(2+) ions in p p m of CaCO_(3) . c. Calculate the hardnesss due to Mg^(2+) ions in p p m of CaCO_(3) . d. Calculate the total hardness of water in p p m of CaCO_(3) .

0.0093 g of Na_(2)H_(2)EDTA.2H_(2)O is dissolved in 250 mL of aqueous solution. A sample of hard water containing Ca^(2+) and Mg^(2+) ions is titrated with the above EDTA solution using a buffer of NH_(4)OH+NH_(4)Cl using eriochrome balck- T as indicator. 10 mL of the above EDTA solution requires 10 mL of hard water at equivalent point. another sample of hard water is titrated with 10 mL of above EDTA solution using KOH solution (pH=12) . using murexide indicator, it requires 40 mL of hard water at equivalence point. a. Calculate the ammount of Ca^(2+) and Mg^(2+) present in 1L of hard water. b. Calculate the hardness due to Ca^(2+) , mG^(2+) ions and the total hardness of water in p p m of CaCO_(3) . (Given MW(EDTA sal t)=372 g mol^(-1),MW(CaCO_(3))=100 gmol^(-1) )

100 mL of a sample of tap water requires 50 mL of N // 50- HCl solution for titration. Calculate the temporary hardness of tap water.

Degree of hardness of a water sample is 100ppm. 50 ml of standard soap solution is added to one kilogram of hard water sample. Calculate the degree of hardness of the resulting water sample. (one ml of soap solution is 0.001 g CaCO_3 ).

pH when 100 mL of 0.1 M H_(3)PO_(4) is titrated with 150 mL 0.1 m NaOH solution will be :