Question

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)?

Answer 1

Sample of hard water contains `96 pp m SO_(4)^(2-)` and `40 pp m Ca^(2+)` ions left in solution are `CaSO_(4),i.e., 40pp m`
Now `1 "litre"` water contains `Ca^(2+)` after removal of
`Ca(HCO_(3))_(2)=(40xx10^(3))/(10^(6))=40xx10^(-3)g`
or `[Ca^(2+)]=(40xx10^(-3))/(40)=10^(-3)`
If these `Ca^(2+)` are exchanged with `H^(+)`, then `[H^(++)]` in solution `=2xx10^(-3)`
`:. pH= - log2xx10^(-3)=2.6989`