Posted by cat on Sunday, March 4, 2012 at 8:09pm.
A long problem. I'll get you started.
pH = 2.15. Solve for (H^+) and I obtained 7.08E-3M. Then
...........H2A ==> H^+ + HA^-
initial.....x.......0......0
change..-0.00708..0.00708.0.00708
equil..x-0.00708..0.00708.0.00708
k1 = 0.017 = (H^+)(HA^-)/(H2A)
Plug 0.00708 in for H and HA and x for H2A and solve for x. If I didn't make an error that is 0.01M
You want to be careful here; the x-0.00708 is real and you may not disregard the 0.00708
You know M and volume (0.250L) so you can solve for moles. Then knowing moles and mass you can solve for molar mass.
I don't know what you've been taught for the second equivalence point for diprotic acids but the shortcut I use is (H^+) = sqrt(k1k2)
You know pH, convert that to H^+ and knowing k1 you can calculate k2. There is an expanded version of that equation that your prof may prefer to use and if you've been given that I would use it. Post your work if you need additional help.
I got 82g for the molar mass and I used the same method you demonstrated.
We never learned the (H^+)=sqrt(k1k2) equation so I want to make sure I have the right answer.
Did you get k2=9.3*10^-18
i got the same mm
I tried sqrt k1k2 and I arrived at the same answer you did; however, I don't believe that for a ns. Common sense and a little experience leads me to believe that k2 is between approximately 1E-7 and 1E-8. So let's go another route.
You need to determine the molarity (or grams) of the Ca(OH)2 which can be done with the solubility data. Knowing it is a saturated solution you should be able to calculate the volume added to get to the second eq point. Then the hydrolysis of the salt will be
..........A^2- HOH ==> HA^- + OH^-
initial...
etc.
Kb A^2- = (Kw/k2) = (HA^-)(OH^-)/(A^2-)
You will know Kw, and A^2- from the titration data. HA^- and OH^- you will know from the pH of 9.40. Solve for k2.
This should work; you can see why I opted for the shortcut but I can tell you for sure that the shortcut won't work now that I see the number it produces. The expanded version of the shortcut; however, still might work if you had access to that.
Thank you so much!