Chemistry: Finding Ksp for Ca(OH)2

In summary, the question is asking for the concentration of hydroxide at equilibrium in a solution containing .10 M sodium hydroxide and .20 M calcium chloride. The reference value for Ca(OH)2 is 6.10 x 10^-6 and the equation for the reaction is 2NaOH(aq) + CaCl2(aq) ---> 2NaCl(aq) + Ca(OH)2(s). To solve the problem, one can assume a volume, such as 1L, and calculate the concentration of calcium in the solution. Then, using this concentration, one can approximate the concentration of OH- that was not precipitated. It is important to note that the concentration of Ca2+
  • #1
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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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  • #2
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.
 
  • #3
How do I do that? Do I say Ca: .10 M + x
OH: .05 M + 2x

do I disassociate the x?
 
  • #4
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).
 
  • #5


Based on the information provided, it is not clear what the volume of the solution is. It is important to know the volume in order to calculate the concentration of hydroxide at equilibrium. Once the volume is known, the concentrations of sodium hydroxide and calcium chloride can be used to calculate the initial concentrations of calcium and hydroxide ions in the solution. From there, the Ksp equation can be used to calculate the concentration of hydroxide at equilibrium. It is also important to note that the Ksp value for Ca(OH)2 may vary slightly depending on the temperature and other factors.
 

FAQ: Chemistry: Finding Ksp for Ca(OH)2

1. What is the definition of "Ksp" in chemistry?

Ksp stands for "solubility product constant" and it is a measure of the maximum amount of a compound that can dissolve in a solution at a given temperature.

2. How do you calculate the Ksp for Ca(OH)2?

The Ksp for Ca(OH)2 can be calculated by multiplying the concentrations of calcium ions (Ca2+) and hydroxide ions (OH-) in a saturated solution of Ca(OH)2. The concentrations can be determined experimentally or from the solubility of Ca(OH)2 in water.

3. Why is it important to know the Ksp for a compound?

The Ksp value provides information about the solubility of a compound and its potential to form a precipitate in a solution. It is also useful in understanding the equilibrium of a reaction and predicting the formation of solid products.

4. How does temperature affect the Ksp of Ca(OH)2?

The Ksp of Ca(OH)2 increases with increasing temperature, as higher temperatures lead to an increase in the solubility of the compound. This can be explained by Le Chatelier's principle, which states that an increase in temperature favors the endothermic reaction of dissolving the compound.

5. Can the Ksp of Ca(OH)2 be changed?

The Ksp of Ca(OH)2 is a constant value at a specific temperature, but it can be affected by changing the temperature or the concentrations of Ca2+ and OH- ions in the solution. It can also be altered by adding other compounds that may interact with Ca(OH)2 and affect its solubility.

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