A sample of hard water contains96 ppm of SO_(4)^(2-) and 183ppm of HCO_(3)^(-) from1000 kgof 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? (AssumeCaCO_(3) to be completelyexchanged with hydrogen ions , in one litreof the treated water are completely exchanged with hydrogen ions, what will be its pH? (One ppm means one part of the substance in one million parts of water , weight/weight )

A sample of hard water contains96 ppm of SO_(4)^(2-) and 183ppm of HCO

1.	A sample of hard water contains96 ppm of SO_(4)^(2-) and 183ppm of HCO_(3)^(-) from1000 kgof 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? (AssumeCaCO_(3) to be completelyexchanged with hydrogen ions , in one litreof the treated water are completely exchanged with hydrogen ions, what will be its pH? (One ppm means one part of the substance in one million parts of water , weight/weight )
Answer» Solution :`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 assumedto 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 densityof 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 `