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

In 1000 kg of water the mass of `HCO_(3)^(-) = 183 g`
Moles of `HCO_(3)^(-) = (183)/61 = 3`
Moles of `SO_(4)^(2-) = 96/96 = 1`
` :. ` Total moles of `Ca^(2+)` in the solution ` = 1 + 1.5 = 2.5`
`Ca(HCO_(3))_(2) + CaO to 2CaCO_(3) + H_(2)O`
Moles of CaO required to be added to remove all `HCO_(3)^(-) = 1.5`
Now, the `Ca^(2+)` in the solution will be only associated with `SO_(4)^(2-)`. Therefore , moles of `Ca^(+2)` left in the solution = 1 .
ppm of `Ca^(2+) = 1 xx 40 = 40` ppm
Moles of `Ca^(2+) ` ions in 1 L of `H_(2)O = 10^(-3)`
Moles of `H^(+)` ions that `Ca^(2+)` will exchange with `= 2 xx 10^(-3)`
` :. pH = - log (2 xx 10^(k^(3))) = 2.7`