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

Step-by-step 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.

`((2 * 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.

`((2 * 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}} * 100`

=
`(2.591 g)/(0.2816 g) * 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% .