Question

Oxalic acid is a diprotic acid. calculate the percent of oxalic acid (h2c2o4) in a solid given that a 0.7984-g sample of that solid required 37.98 ml of 0.2283 m naoh for neutralization.

Answer 1

The percent of oxalic acid in a solid is 48.87%.

Step-by-step explanation:

Mass of the solid sample = 0.7984 g

`H_2C_4O_4+2NaOH\rightarrow Na_2C_2O_4+2H_2O`

Volume of NaOH solution = 37.98 mL = 0.03798 L

Concentration or molarity of the NaOH solution = 0.2283 M

Moles of NaOH :

`Molarity* \text{Volume of the solution} = 0.0086708 moles`

According to reaction, 2 moles of sodium hydroxide reacts with 1 mole of oxalic acid .

Then 0.0086708 moles of sodium hydroxide will react with:

`(1)/(2)* 0.0086708 moles=0.0043354 moles` of oxalic acid.

Mass of oxalic acid neutralized = 0.0043354 moles × 90 g/mol =0.390186 g

Percentage of oxalic acid in solid sample :

`\%=\frac{\text{Mass of oxalic acid}}{\text{Mass of sample}}* 100`

`\%=(0.390186 g)/(0.7984 g)* 100=48.87 \%`

The percent of oxalic acid in a solid is 48.87%.