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Question

A sample of hard water contains 96 ppm of SO24 and 183 ppm of HCO3 with Ca2+ as the only cation. Few moles of CaO will be required to remove HCO3 from 1000 kg of this water. If 1000 kg of this water is treated with the amount of CaO calculated above, the concentration (in ppm) of residual Ca2+ ions will be :
(Assume CaCO3 to be completely insoluble in water. The Ca2+ ions in one litre of the treated water are completely exchanged with hydrogen ions. One ppm means one part of the substance in one million part of water, mass/mass)

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Solution

Sample of hard water contains 96 ppm SO24 and 40 ppm Ca2+(CaSO4).
Also, it contains 183 ppm HCO3 and 60 ppm Ca2+[Ca(HCO3)2].
To remove Ca(HCO3)2 from 103 kg or 106 g sample of hard water which contains 243 g Ca(HCO3)2 or 32 moles of Ca(HCO3)2, CaO required is 32 moles.
Ca(HCO3)2+CaOCaCO3+H2O+CO2
Thus, moles of CaO required =32 or 1.5
Also, Ca2+ ions left in solution are of CaSO4 =40 ppm

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