InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
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` |
|
Answer» (i). Total hardness (First titration) `M_1V_1(EDTA)=M_2V_2` (Total `Ca^(2+)` and `Mg^(2+)` in Temporary and permanent hardness) `500xx0.001=M_2xx100` `M_2=0.005` (Total `Ca^(2+) and Mg^(2+)`) Total hardness in ppm of `CaCO_3=(0.005xx100xx10^(6))/(10^(3))` `=500ppm` (ii). Permanent hardness (second titration): `M_1V_1(EDTA)=M_2V_2` (total `Ca^(2+)` and `Mg^(2+)` due to permanent hardness) `200xx0.001=M_2xx100` `M_2=0.002` (Total `Ca^(2+)` and `Mg^(2+)` in permanent hardness) Permanent hardness in ppm of `CaCO_3`) `=(0.002xx100xx10^(6))/(10^(3))=200ppm` (iii). Temporary hardness in ppm of `CaCO_3=500-200=300ppm` |
|