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