Oxalic acid, a diprotic acid having the formula h2c2o4, is used to clean the rust out of radiators in cars. a sample of an oxalic acid mixture was analyzed by titrating a 0.2816 g sample dissolved in water with 0.0461 m naoh. a volume of 11.49 ml of the base was required to completely neutralize the oxalic acid. what was the percentage by mass of oxalic acid in the sample?

Explanation:

It is given that a sample of a mixture of oxalic acid with water which has a wight of 0.2816 g.

This sample is titrated with 11.49 mL or 0.01149 L (as 1 ml = 0.001 L) of 0.0461 M NaOH.

Therefore, for complete neutralization, equate the total number of moles of ions produced by with total number of moles of ions produced by NaOH.

Assuming that weight of oxalic acid in the given sample is 'x'.

Hence, the total number of moles of ions produced are as follows.

      ........... (1)     (We multiply by 2 because Oxalic Acid is a diprotic acid)

So, number of moles of ions produced are as follows.

        (0.0461 + 0.01149) = 0.05759  ............. (2)

Now, equating both we get the value of x as follows.

                    = 0.05759

                                  x = 2.591

Therefore, weight of oxalic acid in the sample is 2.591 g.

Percentage by mass of oxalic acid =

                       =

                       = 9.20%

Thus, we can conclude that mass by percent of oxalic acid in given sample of mass 0.2816 g is 9.20% .

answers: to make sure that all of the broken pieces of tablet were added to the water, to allow for fair comparison of all three trials, with different particle sizes to avoid experimental error