Tuesday

October 6, 2015
Posted by **cat** on Sunday, March 4, 2012 at 8:09pm.

Calculate the molar mass of the acid

Calculate the second dissociation constant for the acid.

- chemistry -
**DrBob222**, Sunday, March 4, 2012 at 8:56pmA 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.

- chemistry -
**cat**, Sunday, March 4, 2012 at 9:09pmI 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

- chemistry -
**katie**, Sunday, March 4, 2012 at 9:12pmi got the same mm

- chemistry -
**DrBob222**, Sunday, March 4, 2012 at 10:20pmI 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.

- chemistry -
**cat**, Monday, March 5, 2012 at 7:02amThank you so much!