A sample of hard water contains 96 ppm of 2 SO4
and 183 ppm of HCO3
, with Ca2+ as the only
cation. How many moles of CaO will be required to remove HCO3
from 1000 kg of this water? If
1000 kg of this water is treated with the amount of CaO calculate above, what will be the conentration
(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, weight/ weights)?
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Answer:
Sample of hard water contains 96 ppm SO
4
2−
and 40 ppm Ca
2+
(CaSO
4
).
Also, it contains 183 ppm HCO
3
−
and 60 ppm Ca
2+
[Ca(HCO
3
)
2
].
To remove Ca(HCO
3
)
2
from 10
3
kg or 10
6
g sample of hard water which contains 243 g Ca(HCO
3
)
2
or
2
3
moles of Ca(HCO
3
)
2
, CaO required is
2
3
moles.
Ca(HCO
3
)
2
+CaO→CaCO
3
+H
2
O+CO
2
Thus, moles of CaO required =
2
3
or 1.5
Also, Ca
2+
ions left in solution are of CaSO
4
=40 ppm
Explanation:
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