how do i calculate the [OH-] and [Ca2+] using the equivalence point? Then I have to calculate the Ksp for calcium hydroxide.
8 answers
It is easier if you have some numbers to work with. Also some explanation of what you are doing would help.
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
We did a titration lab of calcium hydroxide. Ksp of calcium hydroxide
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+)
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+)
Kw would be 1.0x10^-15 right? Cant quite remember the exponent
I don't understand this part
(pH + pOH = pKw = 14)
PKw is equal to 14?
(pH + pOH = pKw = 14)
PKw is equal to 14?
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?
I tried doing it and got 1.96x10^-13...I dunno if it's correct
1.96x10^-16 *
2 answers