An unknown amount of acid can often be determined by adding an excess of base and then back-titrating the excess. a 0.3471−g sample of a mixture of oxalic acid, which has two ionizable protons, and benzoic acid, which has one, is treated with 97.0 ml of 0.1090 m naoh. the excess naoh is titrated with 21.00 ml of 0.2060 m hcl. find the mass % of benzoic acid.

Explanation:

The given data is as follows.

       Mass of mixture = 0.3471 g

As the mixture contains oxalic acid and benzoic acid. So, oxalic acid will have two protons and benzoic acid has one proton.

This means oxalic acid will react with 2 moles of NaOH and benzoic acid will react with 1 mole of NaOH.

Hence,   moles of NaOH in 97 ml =

                                                       =

Moles of HCl in 21.00 ml =

                                         = mol

Therefore, total moles of NaOH that reacted are as follows.

            -      

                = mol

So, total 3 mole of NaOH will react with 1 mole of mixture. Therefore, number of moles of NaOH reacted with benzoic acid is as follows.

                    = mol

Since, molar mass of NaOH is 40 g/mol. Therefore, calculate the mass of NaOH as follows.

                         = g

                         = 0.0832 g

Whereas molar mass of benzoic acid is 122 g/mol.

Therefore,       40 g NaOH = 122 g benzoic acid

So,            0.0832 g NaOH =

                                             = 0.253 g

Hence, calculate the % mass of benzoic acid as follows.

                           = 73.10%

Thus, we can conclude that mass % of benzoic acid is 73.10%.

? you expect us to answer this? ? this is your schoolwork, i don't think we can with this, i think it would considering being