a solution of HCL is approximately 0.1M and has to be standardized . 1.234g Na2CO3 are dissolved to make 100ml of solution. 20mL of this solution are titrated with the HCL solution, requiring 31.3mL for the second equivalence point. what is the exact concentration of the HCL solution?

I have posted this before but I just want to assure if this is the correct working out or not,

Step 1: moles Na2CO3 = 1.234g/105.89 g/mol = 0.0116 mol.

Step 2: 20mL of Na2CO3 = 0.0116*(20/100) = 0.00232mol/L

Step 3: Balanced equation =
Na2CO3 + 2HCL --> 2NaCl + H2O + CO2

Thus it is a 2:1 ratio, therefore moles HCl = 2* moles Na2CO3 = 2*0.00232 = 0.00464 moles.

Step 4: Conc. of HCl = 0.00464mol/ 0.03130L = 0.1469mol/L

1 answer

Very good. I would offer these thoughts (not suggestions). First I looked up the molar mass of Na2CO3 and obtained 105.989 but we may not have used the same sets of numbers. Second, the number of significant figures you have used is not consistent. If that 1.234 is ok then you're allowed 4 places BUT I see only 3 on the 31.3. Is that 31.30 in the problem and you just omitted the last zero or not. It makes little difference to me but it's a matter of consistency. Finally, you have only 3 places in step 2 and step 1 and partway through in step 4. Other than those minor problems I think you did a great job.
Note: I don't think all of the difference between your answer and mine is due to the difference in molar mass Na2CO3. I think some is rounding error; i.e., caused by rounding each step, writing it down, and re-entering it in the calculator. I do all of that in one step and let the answer from the previous step stay in the calculator.
That's 1.234 g x (1 mol/105.989) x (20.00 mL/100.00 mL) x (2 mol HCl/1 mol Na2CO3) x (1/0.03130) = 0.1487 M.