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.
(b). Calculate the " mol of "CaO required to remove `HCO_3^(ɵ)` ion from 1000 kg of the water.
(c). Calculate the concentration of `Ca^(2+)` in ppm remaining in water after adding `CaO`.
(d). If the `Ca^(2+)` ion in 1 L of The treated water are completely exhange with `H^(o+)` ions, what will be its pH.

Explanation

(a). `CaSO_(4)hArrCa^(2+)+SO_(4)^(2-)` ...(i)
`Ca(HCO_(3))_(2)hArrCa^(2+)2HCO_(3)^(ɵ)` ..(ii).
(i). 96 ppm of `SO_4^(2-)=96 g SO_4^(2-)` in `10^(6) mL H_2O`
(Mw of `SO_4^(2-)=96g)`
`=(96)/(96)" mol of "(SO_4^(2-))/(10^(6))mLH_2O`
`1 " mol of "(SO_4^(2-))/(10^(6))mLH_2O`
`1" mol of "(Ca^(2+))/(10^(6))mLH_2O`
(ii). `122 ppm of HCO+3^(ɵ)-=122gHCO_3^(ɵ)` in `10^(6) mL H_2O`
(Mw of `HCO_3^(ɵ)=61g)`
`-=(122)/(61)=2 " mol of "HCO_3^(ɵ)`
`-=1.0" mol of "Ca^(2+)`
[Since 1 mol. of `Ca(HCO_3)_2=2mol." of "HCO_3^(ɵ)`]
Total `Ca^(2+)=1+1=2" mol of "Ca^(2+)=80g` in `10mLH_2O`
`ppmCa^(2+)=80`
(b). `underset(1mol)(CaO)+underset(1mol)(Ca(HCO_3)_2)tounderset(2mol)(2CaCO_3)+H_2O`
`=2 mol HCO_3^(ɵ)` in `10^(6)g H_2O`
2 " mol of "`HCO_3^(ɵ)` is present in `10^(6)mL(=10^(6)g)=1000kg`
of `H_2O `
From equation (iii) 1 " mol of "`CaO` is required of remove 2 " mol of "`HCO_3^(ɵ)` present in 1000 kg of `H_2O`.
(c). Total `Ca^(2+)` already present `=2mol`
`Ca^(2+)` removed in equation (iii)`=1mol`
Thus `Ca^(2+)(l eft)=2-1=1mol=40g=400ppm`
From equatiohn (iii), it is clear that `HCO_3^(ɵ)` ions are removed as `CaCO_3` but `SO_^(4)^(2-)` ions are left in the solution.