A sample of hard water contains `96 pp m."of" SO_(4)^(2-)` and `183 pp m "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 exchange with hydrogen ions, what will be its `pH` (One ppm means one part of the substance in one million part of water, weight`//`weight)?

Explanation

`Ca(HCO_(3))_(2)+CaO = 2CaCO_(3) +H_(2)O`
Mole of CaO = mole of `Ca(HCO_(3))_(2) " in " 10^(6) ` g of solution
` = 1/2 xx " mole of " HCO_(3)^(-)`
` = 1/2 xx 183/61 = 1.5`
As `CaCO_(3)` is assumed to be completely insoluble in water `Ca^(2+)` ions left are , therefore , those associated only with `SO_(4)^(2-)` ion ( 96 ppm)
For `CaSO_(4)` , we have
mole of `Ca^(2+) //10^(6) g = " mole of " SO_(4)^(2-) //10^(6) g = 96/96 " mole/"10^(6) g `
` = 1 " mole /"10^(6) g` 40 ppm.
Now ,assuming density of solution to be 1 g/mL
mole of `Ca^(2+)` is replaced by `H^(+)`
`[H^(+)] = 2xx 10^(-3) M `
pH = `-log (2 xx 10^(-3)) = 2.7 `