If I understand the post you have calculated the volume to reach the first equivalence point as well as the volume to reach the second equivalence point. Did you find 38.57 mL for the first equivalence point (I suppose you would round that to 38.6 mL) and 77.14 mL (which I suppose would be rounded to 77.1 mL) for the second equivalence point. [Note: the rounding depends upon how many significant figures you have and I don't know that since you typed 30 and you probably meant 30.0 or something like that.]. Here is what I would do for the remainder of the problem.
30 mL x 0.360M = 10.80 mmoles acid
ymL (e.g. 10 mL) x 0.280M = 2.8 mmoles NaOH added.
To simplify, let's call ascorbic acid, H2C.
.............H2C + OH^- ==>HC^- + H2O
initial.....10.80...0........0......0
add...............2.80...........
change.....-2.80..-2.80.....+2.80...
equil.....8.00......0........2.80...
pH = pK1 + log[(base)/(acid)
pH = 4.167 + log[(2.80)/(8.00)]
So you divide your calculations into parts.
a. zero mL. This is pure ascorbic acid. Calculate the pH based on the ionization of the first H (ignore the second H) and proceed as if you had a simple acid, such as acetic acid. I know you've done those problems a hundred times. This is no different.
b. Everything between zero mL and the first equivalence point is done as I've set up the ICE chart above. That equation I've used is the Henderson-Hasselbalch equation in case you don't recognize it.
c. At the first equivalence point you have a solution of the semi-salt; i.e., the half neutralized ascorbic acid (sodium hydrogen ascorbate). The pH for that ONE point, and only this one point, is 1/2(pK1 + pK2).
d. Everything between the first and second equivalence point is done with the HH equation as above. The ONLY difference is that pK2 is not the same as pK1. You start with 10.80 mmoles HC^- and add OH^- ===> C^2- + OH^-
At 48.57 total volume (that's just 10.00 mL past the 1st equivalence point), you have 10.80 mmoles HC^-, you add 2.80 mmoles OH^- and continue.
e. At the second equivalence point, the pH is that of a solution of disodium ascorbate (and I know you've done a hundred of those--just like sodium acetate at the equivalence point of acetic acid and NaOH).
f. Everything past the second equivalence point is just exxcess OH^- added. Don't forget to take into account the dilution factor.
And that gets all of the points for the titration.
To summarize.
a. zero mL. pure ascorbic acid
b. buffer (HH) equation.
c. 1st eq. pt is 1/2(pK1 + pK2)
d. buffer equation.
e. 2nd eq. pt. is hydrolysis of the diprotic salt.
f. excess hydroxide.
Post your work if you get stuck.
I have a titration of 30ml of .360 M H2C6H6O6 Its K1 =6.8E-5 K2 =2.8 E-12 With .280 M NaOH. It is a weak acid being titrated by a strong base Its a two step process with the ice table to find the eqpts. Ineed to list the volumes of NaOH needed for the titration as I go. Been working on it for days now can someone please give me some qulitative answer in an understandble way as how to calculate the volume needed to titrate the two protons from the ascorbic acid Please. I Know i find the concentration and from those values at eqpt find the volume needed to titrate. I also know that at eqpt the number of moles of NaOh needed is the number of moles of acid that was initally present. having found the conc at eqpt1 and eqpt2 what then follows to find volume needed to titrate at eqpt2 thanks
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