Chemistry: Finding Ksp for Ca(OH)2

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Discussion Overview

The discussion revolves around calculating the concentration of hydroxide ions at equilibrium in a solution containing sodium hydroxide and calcium chloride, specifically in the context of the solubility product constant (Ksp) for calcium hydroxide (Ca(OH)2). The problem is framed as a homework question involving chemical equilibria and precipitation reactions.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents the initial problem and references the Ksp value for Ca(OH)2, indicating confusion about how to approach the calculations.
  • Another participant suggests assuming a volume (e.g., 1L) to simplify calculations and notes that there is an excess of calcium in the solution, which should be considered when calculating hydroxide concentration.
  • A subsequent post questions how to set up the equilibrium expressions, suggesting a misunderstanding of how to incorporate changes in concentration due to precipitation.
  • Another participant clarifies that the change in concentration should be represented as a decrease (-x) due to the precipitation of Ca(OH)2, and emphasizes the importance of calculating how much calcium is removed from the solution to determine the remaining hydroxide concentration.

Areas of Agreement / Disagreement

Participants express differing views on how to set up the equilibrium calculations, particularly regarding the treatment of concentrations and the assumptions made about the precipitation process. The discussion remains unresolved as participants explore various approaches without reaching a consensus.

Contextual Notes

There are limitations in the assumptions made about the completeness of the reaction and the stoichiometric relationships involved, which may affect the accuracy of the calculations. The dependence on the chosen volume for calculations is also noted.

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Homework Statement


What is the concentration of hydroxide at equilibrium in a solution that contains a mixture of .10 M sodium hydroxide and .20 M calcium chloride?

I know that the reference value for Ca(OH)2 is 6.10 x 10^-6

Equation: 2NaOH(aq) + CaCl2(aq) ---> 2NaCl(aq) + Ca(OH)2(s)

Ca(OH)2(s)----> Ca + 2OH
I -
C -
E -
Ksp = [Ca][OH]2

precipitate:
Ca: 0.20M x (1/2) = .10 M
OH: .10M x (1/2) = .05 M

My question is, is that what is the volume of the concentrations? Because i don't know how to solve this problem, please help! I have tried different ways but i keep getting a cubed root...I am really confused!
 
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You may assume any volume you like - like 1L - if it helps you. But you don't have to.

There is an excess of calcium, so when the reaction ends you will have still a lot of calcium in the solution. Calculate its concentration, and use it to calculate concentration of OH- that was not precipitated. This is only an approximated method, but good enough.
 
How do I do that? Do I say Ca: .10 M + x
OH: .05 M + 2x

do I disassociate the x?
 
Not +x, but -x - Ca(OH)2 precipitates, so both Ca2+ and OH- are removed.

But the idea is to not use 0.05-2x - but to calculate how much Ca2+ was removed from the solution by precipitation assuming reaction went to completion. This is not exactly true, but knowing Ksp and [Ca2+] you can calculate [OH-] and check if the assumption makes sense (that is, if amount of OH- left in the solution is negligible in stoichiometry calculations).
 

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