chemistry
posted by cat .
a 205 mg sample of diprotic acid is dissolved in enough water to make 250 ml of solution. The pH of this solution is 2.15. A saturated solution of calcium hyrdoxide (Ksp=1.3*10^6) is prepared by adding excess calcium hydroxide to water and then removing the undissovled solid by filtration. Enough of the calucium hydroxide solution is added to the solution of the acid to reach the second equivalence point. The pH at the second equivalence is 9.40. The first dissociation constant for the acid is 1.7*10^2. Assume that the volumes of solutions are additive and that the first K for the acid is at least 1000 times greater than the second K.
Calculate the molar mass of the acid
Calculate the second dissociation constant for the acid.

A long problem. I'll get you started.
pH = 2.15. Solve for (H^+) and I obtained 7.08E3M. Then
...........H2A ==> H^+ + HA^
initial.....x.......0......0
change..0.00708..0.00708.0.00708
equil..x0.00708..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 x0.00708 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 1E7 and 1E8. 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!
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