Understanding the Constant Nature of Ksp in Solution Calculations

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SUMMARY

The discussion centers on the constancy of the solubility product constant (Ksp) for calcium hydroxide in various solutions, specifically in the presence of potassium chloride and calcium chloride. Participants clarify that Ksp remains constant regardless of the concentration of these salts due to its dependence on temperature and solvent conditions. The primary focus is on calculating molar solubility by determining calcium ion concentration and dividing by the total volume, rather than altering Ksp values. The common ion effect is also highlighted as a significant factor in these calculations.

PREREQUISITES
  • Understanding of Ksp (solubility product constant) and its implications in solubility calculations.
  • Knowledge of the common ion effect and its impact on solubility.
  • Familiarity with molar solubility calculations and ion concentration measurements.
  • Basic principles of chemical equilibrium and temperature effects on solubility.
NEXT STEPS
  • Research the common ion effect and its applications in solubility calculations.
  • Learn about the temperature dependence of Ksp values for various compounds.
  • Explore methods for measuring ion concentrations in solution accurately.
  • Investigate the use of activities versus molarities in high concentration scenarios.
USEFUL FOR

Chemistry students, laboratory researchers, and educators involved in solubility and equilibrium studies, particularly those focusing on the behavior of ionic compounds in solution.

kevinnn
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My lab group and I are a bit confused. We are doing a lab where we need to calculate the Ksp of calcium hydroxide in pure water and then in different concentrations of calcium chloride and potassium chloride to analyze the diverse and common ion effect. I was absent one day but my lab partner told me that the teacher said Ksp for calcium hydroxide will remain constant for the solutions of potassium chloride and calcium chloride. She wants us to make a graph where we compare the molar solubilities of the different solutions, not the Ksp. My question is how on Earth can something that tells us about solubility, Ksp, remain constant when more is forced to dissolve, KCl, and less is forced to dissolve, CaCl2? Does this mean the molar solubility calculation is really as easy as just determining the calcium ion concentration in each solution and dividing it by the total volume? Gracias!
 
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kevinnn said:
My lab group and I are a bit confused. We are doing a lab where we need to calculate the Ksp of calcium hydroxide in pure water and then in different concentrations of calcium chloride and potassium chloride to analyze the diverse and common ion effect. I was absent one day but my lab partner told me that the teacher said Ksp for calcium hydroxide will remain constant for the solutions of potassium chloride and calcium chloride. She wants us to make a graph where we compare the molar solubilities of the different solutions, not the Ksp. My question is how on Earth can something that tells us about solubility, Ksp, remain constant when more is forced to dissolve, KCl, and less is forced to dissolve, CaCl2? Does this mean the molar solubility calculation is really as easy as just determining the calcium ion concentration in each solution and dividing it by the total volume? Gracias!

Tell us what you exactly measure and how. Write the equation for the solubility constant What are you asked to make a graph of? If not told what seems a good idea?

Then if doing this does not already indicate to you the answers we have a chance of helping.
 
Last edited:
It is called a constant for a reason, because it remains constant. It will change with temperature though, pretty much everything changes with temperature, as well as with solvents and such. The numbers may also look funny if you are working at very high concentrations/ionic strengths which require the use of activities instead of molarities in equilibrium concentrations.

Assuming you performed all trials under similar conditions then, yes, you do the fairly simple calculations using the common ion effect when necessary and calculate molar solubility based on how much of the CaOH went into solution.
 

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