The municipal water processing plants of communities with hard water often treat the water supply
with slaked lime, Ca(OH)2
, in order to remove Ca2+
from the water. The slaked lime reacts with
bicarbonate from the dissolved metal bicarbonates in the water according to the following reaction:
Ca(OH)2 + 2 HCO3^- –>CaCO3(s) + CO3^-2+ H2O .
The carbonate produced then reacts with Ca2+
originally in the water to precipitate as CaCO3
(s).
Thus calcium ions from slaked lime are added to the hard water in order to remove calcium ions from
the hard water!
If hard water has a calcium ion concentration of 1.8 x 10^3
M and you want to reduce that
concentration to 0.6 x 10!3
M, what must be the concentration of the slaked lime? If you didn’t do
this calculation and you simply used a saturated solution of slaked lime, what do you think the
consequences might be? Support or refute your prediction by quantitatively assessing the likelihood
of your predicted consequences.