wiz-icon
MyQuestionIcon
MyQuestionIcon
2
You visited us 2 times! Enjoying our articles? Unlock Full Access!
Question

A sample of hard water contains 96 ppm of SO424 ions, 183 ppm of HCO3 ions and the only cation present is Ca2+. X moles of CaO are required to remove the HCO3 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

A
3.0 mol, 80 ppm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
1.5 mol, 80 ppm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
1.5 mol, 40 ppm
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
3.0 mol, 40 ppm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 1.5 mol, 40 ppm
Density of hard water is approximately 1 g/ml
1L=1000g=1kg
183 ppm of HCO3 represent 183 mg/L or 3×103 moles per 1 kg, So for 1000 kg sample of hard water number of moles of HCO3
=3 moles
We know that ratio of Ca2+ to HCO3 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+CaO1CaCO31+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 SO24 ions which counter ion is Ca2+ of molecular formula CaSO4
As given:
SO24=96 ppm=1×103 moles
1×103 moles of Ca2+=40 ppm
All Ca(HCO3)2 so only 40 ppm of Ca2+ ions left in solution.






flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Balancing of Redox Reactions
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon