titration question, please help (1 Viewer)

[xanny]

Hey guys, i am having some trouble with this titration question, i hope someone will figure it out. thanks!

3.00 g of oxalic acid (C2H2O4) reacts with 42.0 mL of 1.586 M NaOH in a titration. Calculate how many moles of oxalic acid there are in the 3.00 g sample and the molar mass of oxalic acid.

 

[Dreamerish*~]

I'm making an educated guess that the equation of the reaction is:

C₂H₂O₄ + 2NaOH → Na₂C₂O₄ + 2H₂O

Since oxalic acid is diprotic:

(C₁V₁)/(C₂V₂) = 1/2

(With ₁ being oxalic acid and ₂ being sodium hydroxide)

I'm assuming that 3.00 g means 3.00 mL, because if you meant 3.00 g of pure oxalic acid, then you won't need the titration information to calculate how many moles are present.

(C₁ x 3)/(42 x 1.586) = 1/2

C₁ = 11.102

. : The concentration of the oxalic acid is 11.102 mol L^-1

In 3.00 mL, there are 0.003 x 11.102 = 0.0331 moles (3 sig. fig.s)

The molar mass of oxalic acid is 90.016 g.