CHM110H5 Study Guide - Quiz Guide: Calcium Hydroxide, Solubility Equilibrium, Hard Water

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Assignment 3
The overall reaction that occurs in the process of treating hard water with slaked lime is as follows:
Ca(OH)2 + Ca+2 + 2 HCO3-
2 CaCO3 + H2O
The reduction in calcium ions would require more of the Ca(OH) for the reaction to proceed. This would cause the
equilibrium to shift to the right, forming more of the precipitate CaCO, where the solubility of it would be
decreased. This could be seen by doing the following calculations.
First the concentration of the slaked lime, Ca(OH)2, must be determined.
The limiting reagent is Ca(OH)2 and the excess reagent is Ca+2. When the hard water has a Ca2+ concentration of
1.5 x 10-3 M and is to be reduced to 0.7 x 10-3 M, the amount of Ca+2 consumed is:
1.5x 103M0.7 x103M=8.0 x104M
Therefore, the amount of calcium ions consumed in the water is 8.0 x 10-4 M. The molar ratio of Ca2+ and Ca(OH)2
using the above equation is 1:1. The amount of Ca(OH)2 consumed is:
8.0 x104M Ca+2x1mole Ca
(
OH
)
2
1mole Ca+2
Therefore the initial concentration of Ca(OH)2 is
8.0 x104
. This is also the amount that has been consumed in
the reaction.
The concentration of slaked lime, Ca(OH)2, in its saturated solution is now to be calculated. The solubility product
constant for calcium hydroxide is 1.3 x 10-6 . Using the equilibrium constant and expression, the maximum
concentration of slaked lime could be determined.
Ca(OH)2 (s)
Ca+2 + 2OH-
Ksp = [Ca2+] [OH-]2
Using theice table method :
Let x be the concentration of Ca+2 and OH-
Ca(OH)2(s)
Ca+2(aq) 2OH-
- 0 0
- x 2x
- +x +2x
1.3 x 10-6 =[x] [2x]2
1.3 x 10-6 = 4x3
x= 6.9 x 10-3 M= [Ca+2]
Therefore, the molar ratio of Ca2+ and Ca(OH)2 using the equation
Ca(OH)2 + Ca+2 + 2 HCO3-
2 CaCO3 + H2O
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Document Summary

The overall reaction that occurs in the process of treating hard water with slaked lime is as follows: The reduction in calcium ions would require more of the ca(oh) for the reaction to proceed. This would cause the equilibrium to shift to the right, forming more of the precipitate caco , where the solubility of it would be decreased. This could be seen by doing the following calculations. First the concentration of the slaked lime, ca(oh)2, must be determined. The limiting reagent is ca(oh)2 and the excess reagent is ca+2. When the hard water has a ca2+ concentration of. 1. 5 x 10-3 m and is to be reduced to 0. 7 x 10-3 m, the amount of ca+2 consumed is: Therefore, the amount of calcium ions consumed in the water is 8. 0 x 10-4 m. the molar ratio of ca2+ and ca(oh)2 using the above equation is 1:1. 8. 0 x10 4 m ca+2 x 1 moleca(oh )2.