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

This method of quantitative determination of percent purity is titrimetric reactions. These reactions most commonly involve neutralization reactions between an acid and a base. Then, we look at the neutralization reaction:

H₂C₂O₄ + 2 NaOH ⇒ Na₂C₂O₄ + 2 H₂O

So, we do the stoichiometric calculations. The important data we should know is the molar mass of oxalic acid which is equal to 90 g/mol.

(0.2283 mol/L NaOH * 0.3798 L * 1 mol H₂C₂O₄/ 2mol NaOH * 90 g/mol H₂C₂O₄) ÷ 0.7984 g *100%
= 488%

This is impossible. The purity can't be more than 100%. Looking at our calculations and the balance reaction, all steps were done correctly. So, I think there is some typographical error in the given. The mass of the sample should be 7.984 g. Then, the answer would be 48.87% purity.

Answer 2

Answer:

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

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 :

$Molarity\times \text{Volume of the solution} = 0.0086708 moles$

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:

$\frac{1}{2}\times 0.0086708 moles=0.0043354 moles$ of oxalic acid.

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

Percentage of oxalic acid in solid sample :

$\%=\frac{\text{Mass of oxalic acid}}{\text{Mass of sample}}\times 100$

$\%=\frac{0.390186 g}{0.7984 g}\times 100=48.87 \%$

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