Chemistry: Finding Ksp for Ca(OH)2

[madeeeeee]

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!

 

[Borek]

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.

 

[madeeeeee]

How do I do that? Do I say Ca: .10 M + x
OH: .05 M + 2x

do I disassociate the x?

 

[Borek]

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).