Question

how do i calculate the [OH-] and [Ca2+] using the equivalence point? Then I have to calculate the Ksp for calcium hydroxide.

Answer 1

It is easier if you have some numbers to work with. Also some explanation of what you are doing would help.

Answer 2

Well the equivalence point in volume is 11.3 mL and the pH is 7.45.

We did a titration lab of calcium hydroxide. Ksp of calcium hydroxide

Answer 3

Ca(OH)2 ==> Ca^2+ + 2OH^-

Ksp = (Ca^2+)(OH^-)^2

Assuming you titrated with HCl, the equation is
Ca(OH)2 + 2HCl ==> CaCl2 + 2H2O
So at the equivalence point the pH = 7.45. I would convert to pOH (pH + pOH = pKw = 14), then pOH = -log(OH^-) to obtain (OH^-). For Ca^2+, that will be 1/2 moles HCl used to get to the equivalence point; so (MHCl x LHCl)/2 = (Ca^2+)

Answer 4

Kw would be 1.0x10^-15 right? Cant quite remember the exponent

Answer 5

I don't understand this part

(pH + pOH = pKw = 14)

PKw is equal to 14?

Answer 6

Actually forget that. When you say 1/2 the moles of HCl, do you mean take half of the moles of HCl used originally? Times the volume of equivalence point?

Answer 7

I tried doing it and got 1.96x10^-13...I dunno if it's correct

Answer 8

1.96x10^-16 *