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A sample of hard water contains 96 ppm of SO_(4)^(2-) and 183 ppm 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 becompletely insoluble in water). If the Ca^(2+) ions in one litre of 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» <html><body><p></p>Solution :`SO_(4)^(2-)` present in 1000 kgof water `=(96)/(10^(6))xx1000 kg = 96 g = (96)/(96) =1 ` mole <br/> `HCO_(3)^(-)` present in 1000 kg of water `= (183)/(10^(6))xx1000 kg = 183 g = (183)/(61)=3 ` moles <br/> `Ca^(2+)` present along with `SO_(4)^(2-)` ions = 1 mole <br/> `Ca^(2+)` present along with `HCO_(3)^(-) ` as `Ca(HCO_(3))_(2) = (3)/(2) ` mole <br/> `:.` Total `Ca^(2+)` present in 1000 kg of water ` = 1 +(3)/(2) = 2.5` moles <br/> CaO added will react with `Ca(HCO_(3))_(2)` as follows : <br/> `Ca(HCO_(3))_(2) + CaO rarr 2 CaCO_(3) harr + H_(2)O` <br/> But `Ca(HCO_(3))_(2)` present `= (3)/(2)` mole (calculated above) <br/> `:.` CaO required `= (3)/(2) ` mole = 1.5 moles <br/> After treatment with CaO i.e., removal of `Ca(HCO_(3))_(2)`, amount of `Ca^(2+) ` <a href="https://interviewquestions.tuteehub.com/tag/left-1070879" style="font-weight:bold;" target="_blank" title="Click to know more about LEFT">LEFT</a> (<a href="https://interviewquestions.tuteehub.com/tag/due-433472" style="font-weight:bold;" target="_blank" title="Click to know more about DUE">DUE</a> to `CaSO_(4)` only ) in 1000 kgof water = 1 mole = 40 g <br/> `:.` Concentration of residual `Ca^(2+)` (in ppm) = 40 ppm <br/> Now 1000 kg of water contain `Ca^(2+) = 10^(-3)` mole<br/> No. of mole of `H^(+)` <a href="https://interviewquestions.tuteehub.com/tag/exchanged-2075520" style="font-weight:bold;" target="_blank" title="Click to know more about EXCHANGED">EXCHANGED</a> `= 2 xx 10^(-3)` mole<br/> `:. pH= - log (2xx10^(-3))= 2.7`</body></html>


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