Question

A sample of hard water contains 96 ppm of SO₄^2– and 183 ppm of HCO₃^– with 60 ppm of Ca²⁺ as the only cation. How many moles of CaO will be required to remove HCO₃^2– from 100 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 Ca²⁺ ions ? (Assume CaCO₃ to be completely insoluble in water). If the Ca²⁺ 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 part of water, mass/mass). 

Answer 1

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/96g mol^-1= 1 mol 

and amount of HCO₃^– ions = 183/61 g mol^-1 = 3 mol

These ions are present as CaSO₄ and Ca(HCO₃)₂

Hence, amount of Ca²⁺ ions = (1 +3/2)=  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