A sample of hard water contains 96 ppm of SO4 2- and 183 pppm of HCO3- and Ca 2+ as the only cation.How many moles of CaO will be required to remove HCO3-from 1000 kg of this water?If this water is treated with the amount of CaO calculated above,what will be the concentration (inppm) of residual Ca 2+ ions in one litre of the treated water are completely exchanged with hydrogen ions,what will be the pH ?

Dear student,

In 106 g(= 1000 kg) of the given hard water, we have amount of SO4 2– ions = 96 g
amount of HCO3 – ions = 183 g
So amount of SO4 2– ions = 96 g mol  = 1 mol
and amount of HCO3 – ions = 61 g mol
 = 3 mol
These ions are present as CaSO4 and Ca(HCO3)2.
Hence, amount of Ca2+ ions =  2.5 mol
The addition of CaO causes the following reactions:
CaO + Ca(HCO3)2 → 2CaCO3 + H2O
1.5 mol of CaO will be required for the removal of 1.5 mol of Ca(HCO3)2 in form of CaCO3. In the treated water, only CaSO4 is present now.
Thus, 1 mol of Ca2+ ions will be present in 106 g of water. Hence, its concentration will be 40 ppm.

Molarity of Ca2+ ions in the treated water will be 10–3 mol l–1. If the Ca2+ ions are exchanged by H+ ions then,
Molartiy of H+ in the treated water = 2 × 10–3 M
Thus, pH = – log(2 × 10–3) = 2.7