A water sample is found to contain 96 ppm of `SO_4^(2-)` and 122 ppm of `HCO+3^(ɵ)` with `Ca^(2_)` ion as the only cation.
(a). Calculate the ppm of `Ca^(2+)` in water.

Explanation

(a) `CaSO_(4)hArrCa^(2+)+SO_(4)^(2-)...(i)`
`Ca(HCO_(3))_(2)hArrCa^(2+)+2HCO_(3)^(ɵ)....(ii)`
(i) `96 p p m of SO_(4)^(2-)=96 gSO_(4)^(2-)` in `10^(6) mL H_(2)O`
(`MW` of `SO_(4)^(2-)=96 g`)
`=96/96` mol of `SO_(4)^(2-)//10^(6)mLH_(2)O`
`=1 mol` of `SO_(4)^(2-)//10^(6)mLH_(2)O`
`=1 mol` of `Ca^(2+)//10^(6)mL H_(2)O`
`(ii) 122` p p m of `HCO_(3)^(ɵ)-=122gHCO_(3)^(ɵ)` in `10^(6)mL H_(2)O`
(MW of `HCO_(3)^(ɵ)`=61 g)
-=122/61=2 mol of `HCO_(3)^(ɵ)`
`-=0.1`mol of `Ca^(2+)`.
[Since `1` mole of `Ca(HCO_(3))_(2)=2` mol of `HCO_(3)^(ɵ)`]
Total `Ca^(2+) =1+1=2` mol of `Ca^(2+)=80 g ` in `10^(6)mLH_(2)O`
`pp mCa^(2+)=80`
(b) `underset("1 mol")(CaO)+underset("1 mol")(Ca(HCO_(3))_(2)tounderset("2 mol")(2CaCO_(3)+H_(2)O`
`=2 mol HCO_(3)^(ɵ)` in `10^(6)g H_(2)O`
`2 mol of HCO_(3)^(ɵ)` is present in `10^(6) mL(=10^(6)g)=1000 kg` of `H_(2)O`.
From eq. (iii). `1` mol of `CaO` is required to remove `2` mol of `HCO_(3)^(ɵ)` present in `1000 kg of H_(2)O`.
(c) Total `Ca^(2+)` already present `=2`mol
`Ca^(2+)` already present `=2` mol
Thus `Ca^(2+)` (left) `=2-1=1` mol `=40 g=40p p m`
From eq. (iii), it is clear that `HCO_(3)^(ɵ)` ions are removed as `CaCO_(3)`, but `SO_(4)^(2-)` ions are left in the solution.