Question

Hi guys, I'm looking for a little bit of help with finding the Ksp of Calcium Hydroxide from doing a titration.

So, my class recently did a titration with Ca(OH)2 and 0.05M HCl (aq), with hopes of being able to find the Ksp for the equilibrium equation:
Ca(OH)2 (s) ⇌ Ca2+ (aq) + 2OH- (aq)

I completed 4 tirations, and used an average of 15.75 mL of the 0.05M HCl and 18.63 mL of the Ca(OH)2 to reach the endpoint.

Notably, my teacher said to consider the Ca(OH)2 as a WEAK base, meaning that it does not fully dissociate... So I guess this will change how we solve it (it's now dynamic?)?

From this titration data, I just need to find:
1) [OH-]
2) [Ca2+]
3) Ksp, (which is trivial once [OH-] & [Ca2+] are found :))

Answer 1

For the titration, the reaction is as follows:
Ca(OH)2 + 2HCl ==> CaCl2 + 2H2O. I use millimoles so as to keep all those leading zeros absent.
millimols HCl used = M x mL = 0.05M x 15.75 = 0.78 but you need to recalculate to more places. I'm just showing you how to do it.
Then millimoles Ca(OH)2 = 1/2 that = 0.78/2 = approx 0.39
Then M Ca(OH)2 solution formed from the saturated solution = millimiles/mL = 0.39/18.63 = approx 0.021 M
Now=== Ca(OH)2 = 0.021 M so Ca^2+ = 0.021M and (OH)2 = 0.042 M. Again these are approximate numbers. You should recalculate everything to more significant figures. I assume you can finish. Ksp = (Ca^2+)(OH^-)^2 = ?

Answer 2

DrBob222, I luv u