An acid-base titration question I'm having trouble with, please help?

Question

We're analyzing a sample of impure sodium carbonate, Na2CO3. I need to calculate the amount of it I need to react with about 25mL of approximately 0.1M HCl.
I know it's a simple question, but I've forgotten and can't figure it out.
I know i can obtain the Molar Mass of Na2CO3, but what would I do with that? And do I need the molarity1*volume1 = molarity 2 * volume 2 formula? or something like that. I don't know how to combine it all to figure it out.
Any help would be greatly appreciated!

What do you mean by halve it? I believe it's the second option I'm looking for, but I don't know what to do with the limited information I'm given. Details would be appreciated! thnx

Answer 1

Assuming you've put all the numbers you've got here, this question is actually simpler that you've made it out to be.
For most questions, rather than trying to remember the formula you stated there, I prefere to do it in two stages, first work out the number of moles of the substance you added, and then, using that number, work out the number of moles and therefore concentration of your unknown. (I won't go into any more detail or examples now as that isn't the question, though please feel free to contact me if you want me to go over my method for those calculations in more detail.)

I think, for this question your just looking for the number of grams of Na2CO3 that you need. So, to start of with we need the number of moles of HCl.
This is just moles = concentration x volume. The concentration is 0.1 the volume is 25/1000 or 0.025 (always remember that you want volumes in litres not mililitres)
This gives you 0.0025 moles of HCl.
To work out the next bit you need to write down a balanced formula of the reaction, you should get:
2HCl + Na2CO3 >> 2NaCl +CO2 +H2O
As is always the case in standard titrations, we're not intrested in what products we form, only in the ratio's of the reactants.
Here we can see that it takes 2 HCl atoms to react with 1 Na2CO3 atom, a 2:1. This means that 0.0025 moles of HCl will react with only 0.000125 moles of Na2CO3. You then times this number by the molar mass of Na2CO3 to get it in grams.
(As I mentioned above, if you have missed out information on the Na2CO3, then this answer will be wrong, either way feel free to contact me for more help regarding titration calculations, they are one part of a chemistry course that can really get students stuck if they haven't got a good grasp on the main concepts, but are easy exam marks if you have)

Answer 2

Depends on your product. If you want to convert it to NaHCO3, the bicarbonate, you can use the formula you present. However, if you want to convert it to the "apparent" carbonic acid or CO2, you would halve the amount of moles of Na2CO3.

Answer 3

Na2CO3 + HCl -> NaHCO3 + Na+ + Cl-
NaHCO3 + HCl -> H2CO3 + Cl- + Na+
1mol Na2CO3 = 2mol HCl
0.025L x 0.1 mol/1L HCl x 1mol Na2CO3/2mol HCl x 105.989g/mol Na2CO3 = 0.132 g Na2CO3