[t731]

I have a question concerning back titrations.
I need to make a work-plan for a practical assignment. The assignment is about Na2CO3 . 10 H2O.
To calculate the mass percentage of the H2O in the Na2CO3, I need to perform a back titration with HCl.

The formula for that reaction is:
Na2CO3 • 10H2O(s)   +  2 HCl (aq)         ----->         2 NaCl (aq) + CO2(g) + 11 H2O (l)

Now I need to use an abundance of HCl(aq) and do a back titration of the leftovers of the HCl(aq) after the above reaction with NaOH.

My question is: Is that even possible? Don't I need to filter out the NaCl and H2O for that? And more importantly: How do I know what quantities to use?

[Borek]

Is that even possible?

Yes.

Don't I need to filter out the NaCl and H2O for that?

No. Na⁺ and Cl^- are just spectators, you are doing titration using water solutions, so water produced during neutralization is not a problem.

And more importantly: How do I know what quantities to use?

What is given to you? Concentrations? Sample mass? Nothing?

In general, do the calculations assuming the worst/best case scenario, that would tell you minimum and maximum amounts of acid/base to use.

[t731]

Is that even possible?

Yes.

Don't I need to filter out the NaCl and H2O for that?

No. Na⁺ and Cl^- are just spectators, you are doing titration using water solutions, so water produced during neutralization is not a problem.

And more importantly: How do I know what quantities to use?

What is given to you? Concentrations? Sample mass? Nothing?

In general, do the calculations assuming the worst/best case scenario, that would tell you minimum and maximum amounts of acid/base to use.

Thank you, this made everything a lot clearer.
I was able to finish the work plan already. The mods can close this thread:)
As for the quantities: I was only given the molar mass of Na2CO3 . 10 H20(s) specifically, but I was also given a list of which substances were available to me where the concentrations were also given. I chose:
- 1 M HCl(aq)
- 1 M NaOH(aq)
indicator: phenolphtalein

Again, thanks for the help.
Mods can close thread.

(edit: damn, they censor the word "help" in combination with an exclamation mark on this site:D)