Question

A sample of hard water contains 96 ppm of `SO_(4)^(2-)` and 183 ppm of `HCO_(3)^(-) "with" Ca^(2+)` as the only cation. How many moles of CaO will be required to remove `HCO_(3)^(-)` from 1000 kg of this water ? If 1000 kg of this water is treated with the amount of CaO calculated above, what will be the concentration (in ppm) of residual `Ca^(2+)` ions ? (Assume `CaCO_(3)` to be completely insoluble in water). If the `Ca^(2+)` ions in one litre of the treated water are completely exchanged with hydrogen ions, what will be its pH ? (One ppm means one part of the substance in one million parts of water , weight/weight).

Answer 1

`SO_(4)^(2-)` present in 1000 kg of water `=(96)/(10^(6))xx1000 kg = 96 g = (96)/(96) = 1 ` mole
`HCO_(3)^(-)` present in 1000 kg of water `= (183)/(10^(6))xx1000 kg = 183 g = (183)/(61)=3 ` moles
`Ca^(2+)` present along with `SO_(4)^(2-)` ions = 1 mole
`Ca^(2+)` present along with `HCO_(3)^(-) ` as `Ca(HCO_(3))_(2) = (3)/(2) ` mole
`:.` Total `Ca^(2+)` present in 1000 kg of water ` = 1 +(3)/(2) = 2.5` moles
CaO added will react with `Ca(HCO_(3))_(2)` as follows :
`Ca(HCO_(3))_(2) + CaO rarr 2 CaCO_(3) harr + H_(2)O`
But `Ca(HCO_(3))_(2)` present `= (3)/(2)` mole (calculated above)
`:.` CaO required `= (3)/(2) ` mole = 1.5 moles
After treatment with CaO i.e., removal of `Ca(HCO_(3))_(2)`, amount of `Ca^(2+) ` left (due to `CaSO_(4)` only ) in 1000 kg of water = 1 mole = 40 g
`:.` Concentration of residual `Ca^(2+)` (in ppm) = 40 ppm
Now 1000 kg of water contain `Ca^(2+) = 10^(-3)` mole
No. of mole of `H^(+)` exchanged `= 2 xx 10^(-3)` mole
`:. pH = - log (2xx10^(-3))= 2.7`