Tags:
1. Dec 22, 2016

### pandamonium786

Hey guys! So I have this chemistry problem which I'm finding very difficult.

QUESTION:
You have an unknown triprotic acid, H3X, and have titrated it with 22.53mL of NaOH. The pH of the unknown solution is 2.04 and the concentration is 48.028g/2L. While the concentration of the NaOH is 0.251mol/L. What is the molar mass, as well as the first and second ionization constant (Ka1) for the unknown triprotic acid H3X?

WORK SO FAR:
H3X + 3NaOH --------> Na3X + 3H2O
I know that if I solve for the concentration of the unknown by using the formula 3CaVa=CbVb I get 0.334 mol/L , but then what was the point of being given a concentration?

I really don't understand how to move forward from this point. So any help would be appreciated.

2. Dec 22, 2016

### Staff: Mentor

Hint: both concentrations are in different units. How can you convert between them?

3. Dec 22, 2016

### pandamonium786

So could I do this then:

H3X + 3NaOH --------> Na3X + 3H2O

NaOH = 3[H3X] = 3(0.251mol/L) = 0.753mol/L

H3X=48.028g/2L=24.014g/L

Therefore, 0.753mol/L = 24.014g/L
Molar Mass = 24.014/0.753
= 31.9g/mol

4. Dec 24, 2016

### epenguin

You have titrated what volume of the acid? Not stated.
The volume of NaOH is not mentioned in your calculation. So the calculation is about no experiment.

It is basically quite simple. You can calculate the number of moles of NaOH used in the titration, from its molarity and volume used. This equals the number of moles of protons removed from the acid in the titration. That is probably a 3× 'the number of moles of the tribasic acid originally present that you titrated. (It might be a twice but you give no details of what you mean by 'a titration'.)

You give no information that enables us to say anything about pKa2. Once you know the molarity of the acid, from the pH 2.04 you can give a reasonable figure for pKa1.

If this was a real experiment anyone would need to know more about what it was than you have stated to say anything more.

5. Dec 24, 2016

### pickycat

For the diprotic acid full titration, the steps are outlined in this pdf: http://www.chem.purdue.edu/courses/chm321/lectures/lecture 18 (10-10).pdf
when it says: "The pH of the unknown solution is 2.04 and the concentration is 48.028g/2L." do you mean that the initial volume of H3X was 2L, and there were 48.028 g dissolved in the initial solution that was then titrated?
It says: "You have an unknown triprotic acid, H3X, and have titrated it with 22.53mL of NaOH." I think it means that this was the end-point of the 3 de-protonations, therefore:
mols of H3X = 1/3 * mol NaOH = 1/3 * [NaOH] * Vol NaOH
Also mols H3X = mass H3X / FW
So you can figure out the FW.
For the rest refer to the henderson-Hasselbalch equation as it pertains to the intermediate titration steps.
in the beginning (see step 1 in the pdf above)
[H+] ~= sqrt(Ka1*[H3X])
so you can calculate Ka1.
Then, at the first equivalence point you would have exhausted all the H3X and you are left with H2X- that is starting to deprotonate. Follow the formulas to figure out Ka2, etc.