with "Triprotic Acid Titration" Question

AI Thread Summary
The discussion centers on a chemistry problem involving the titration of an unknown triprotic acid, H3X, with NaOH. Key points include the need to calculate the molar mass and the first ionization constant (Ka1) using the provided pH, concentrations, and titration volume. Participants emphasize the importance of determining the moles of NaOH used to find the moles of H3X and suggest using the Henderson-Hasselbalch equation for further calculations. Clarifications are sought regarding the initial conditions of the acid solution and the volume of NaOH used in the titration. The conversation highlights the necessity of precise data to solve for the acid's properties accurately.
pandamonium786
Messages
23
Reaction score
1
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.

Thanks in advance :)
 
Physics news on Phys.org
Hint: both concentrations are in different units. How can you convert between them?
 
Borek said:
Hint: both concentrations are in different units. How can you convert between them?

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
 
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.
 
For the diprotic acid full titration, the steps are outlined in this pdf: http://www.chem.purdue.edu/courses/chm321/lectures/lecture%2018%20(10-10).pdf
About your specific problem:
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.
 
Thread 'Confusion regarding a chemical kinetics problem'
TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...
Back
Top