A sample of hard water contains 96 ppm of SO42−4 ions, 183 ppm of HCO−3 ions and the only cation present is Ca2+. X moles of CaO are required to remove the HCO−3 ions from 1000 kg of this sample of water. When the same amount of CaO is added to the water sample, the concentration of residual Ca2+ ions (in ppm) becomes Y. (Assume CaCO3to be completely insoluble in water)
Calculate the value of X and Y
The correct option is C 1.5 mol, 40 ppm
Density of hard water is approximately 1 g/ml
183 ppm of HCO−3 represent 183 mg/L or 3×10−3 moles per 1 kg, So for 1000 kg sample of hard water number of moles of HCO−3
We know that ratio of Ca2+ to HCO−3 is 1:2 in Ca(HCO3)2 , number of moles of Ca2+=1.5 moles.
Intially number of moles of Ca(HCO3)2=1.5
Complete balanced reaction to remove Ca(HCO3)2
One mole of Ca(HCO3)2 replaced by one mole of CaO so 1.5 mole of CaOrequired to remove 1.5 mole of Ca(HCO3)2.
And solution contains SO2−4 ions which counter ion is Ca2+ of molecular formula CaSO4
SO2−4=96 ppm=1×10−3 moles
1×10−3 moles of Ca2+=40 ppm
All Ca(HCO3)2 so only 40 ppm of Ca2+ ions left in solution.