Question

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?

Answer 1

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 $H^{+}$ ions produced by $H_{2}C_{2}O_{4}$ with total number of moles of $OH^{-}$ ions produced by NaOH.

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

Hence, the total number of moles of $H^{+}$ ions produced are as follows.

      $\frac{(2 \times x)}{90}$ ........... (1)     (We multiply by 2 because Oxalic Acid is a diprotic acid)

So, number of moles of $OH^{-}$ ions produced are as follows.

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

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

                   $\frac{(2 \times x)}{90}$ = 0.05759

                                  x = 2.591

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

Percentage by mass of oxalic acid = $\frac{\text{Mass of oxalic acid in sample}}{\text{Total mass of sample}} \times 100$

                       = $\frac{2.591 g}{0.2816 g} \times 100$

                       = 9.20%

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