4016 g sample of impure sodium carbonate (soda ash) is dissolved in 50 mL of distilled water. Phenolphthalein was added and 11.40 mL of 0.09942 M HCL…

A 0.4016 g sample of impure sodium carbonate (soda ash) is dissolved in 50 mL of distilled water. Phenolphthalein was added and 11.40 mL of 0.09942 M HCL was required to reach the first end point. Excess volume of HCL was added according to the lab procedure (if x is the amount of acid needed to complete the first titration, you will add a total of 2x + 10mL of acid to ensure excess). After boiling, the excess acid was titrated with 0.1020 M NaOH. It was found that 10.78 mL of the NaOH was required to reach the end point of this titration.

1. How many total mL of the 0.09942 M HCL will have been added if the lab procedure is followed?

I calculated this and got 32.8 mL
2. Based on your answer to #1 above, how many total moles of HCL were added to the sample?
3.26 x 10^-3 moles
3. How many moles of NaOH were used in the back titration?

1.099 x 10^-3 moles of NaOH
4. How many moles of the HCL in problem #2 were used up in reacting with sodium carbonate in the sample of unknown?
I just subtracted the two (not sure if this is correct?) and got .0021614 so .002 moles HCI
5. What was the % sodium carbonate in the original unknown sample?

And I’m not sure about this one and are the others correct?