  Question

A sample of hard water contains 96ppm 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) If the residual Ca2+ ions in 1 L of the above -mentioned water sample is completely replaced by hydrogen ions, what will the pH of the resultant sample of hard water?

A
2  B
2.7  C
3  D
3.3  Solution

The correct option is A 2.7Density of hard water is approximately 1 g/ml  1L=1000g=1kg 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 =3 moles 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 Ca(HCO3)21+CaO1→CaCO31+H2O+CO2 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 As given: 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. One Ca2+ will replaced by 2 H+  ions so for 1000 kg of sample- 1 mole of Ca2+ replaced by  2 mole of H+ or concentration of  [H+]=21000L=2×10−3  pH=−log(2×10−3)=2.6989Chemistry

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