What is the Percent Oxalic Acid Dihydrate in an Unknown Sample?

[ScienceGeek24]

Homework Statement

You are given an impure sample of oxalic acid dihydrate (MW=126.068 g/mole) and asked to determine the percent of oxalic acid dihydrate present in the sample. You place 0.110 grams of the unknown sample in an Erlenmeyer flask and dissolve it in distilled water. Titration with 0.140 M standardized NaOH solution requires 10.10 mLs to reach the titration endpoint. What is the percent of oxalic acid dihydrate in the sample?2. 0.121 grams of oxalic acid dihydrate (MW=126.068 g/mole) is dissolved in water in an Erlenmeyer flask. Remember that each mole of oxalic acid dissolved in water results in 2 moles of hydronium ion so it takes 2 moles of NaOH to neutralize one mole of oxalic acid. It requires 14.22 mLs of your prepared sodium hydroxide solution to reach the endpoint of the titration. What is the molarity of your NaOH solution?

I need to know if this is right at my attemps.

Homework Equations

stochiometry

The Attempt at a Solution

attempt for number 1: (10.10ml)*(1l/1000ml)*(0.140 moles of NaOH/1L)= 1.41*10^-3 moles of Naoh
(0.110 g of oxidic acid)*(1 mole oxidic acid/126.068 g)=8.72*10^-4 MOLE OF OXIDIC ACID
the percetage of oxidic acid, which is what we are looking for i think is 8.72*10^-4moles of oxicid/1.40*10^-3 moles of naOH times 100=61.8 % of oxidic acid

Attempt for number 2: 0.121 g of H2C2O4 *1 mole H2C2O4/126.063 g of H2C2O4=9.59*10^-4 then, 9.59*10^-4 moles of H2C2O4*2moles of NaOH/1 mole of H2C2O4= 1.919*10^-3 moles of NaOH

 

[chemisttree]

You should be more careful with your 'stochiometry' (stoichiometry). Is oxalic acid monoprotic?

 

[ScienceGeek24]

What do you mean by monoprotic?? never heard of this before. hahaha! All i know is that oxalic acid reacts with 2 moles of NaOH to neutralize the concentration.

 

[chemisttree]

Yeah, but if it takes two moles of NaOH to neutralize only one mole of oxalic acid...

 

[ScienceGeek24]

then what does that mean, that is monoprotic??

 

[Borek]

HA <-> H⁺ + A^- - monoprotic (like HCl, or HNO₃).

H₂A <-> 2H⁺ + A^2- - diprotic (like oxalic, or H₂SO₄); note that dissociation is in reality stepwise.

Triprotic (tetra...) left as an exercise for the reader.