Asked by Andrea
A sample of oxalic acid (H2C2O4, with two acidic protons), of volume 37.09 mL was titrated to the stoichiometric point with 22.41 mL of 0.163 M NaOH(aq). What is the molarity of the oxalic acid?
Answer in units of M.
Answer in units of M.
Answers
Answered by
bobpursley
show me your thinking on this.
Answered by
DrRebel
If one assumes that the 22.41mL of 0.163M NaOH(aq) was used to titrate through both equivalence points, then 2 moles of NaOH(aq) are used for each mole of H₂C₂O₄(aq) neutralized. That is…
H₂C₂O₄(aq) + 2NaOH(aq) => Na₂C₂O₄(aq) + H₂O(l)
2 moles NaOH > 1 mole H₂C₂O₄
=> 1 moles NaOH = 2 mole H₂C₂O₄ for problem solution
Since Molarity x Volume in Liters of Solution = moles, then …
[Molarity x Volume(L)] NaOH used = 2[Molarity x Volume(L)] H₂C₂O₄ neutralized
(0.163 mol/L NaOH)(0.02241L)] used = 2·M(0.03709L)
M(H₂C₂O₄) = [(0.163)(0.02241)/(2 x 0.03709)]M H₂C₂O₄(aq) = 0.0492M H₂C₂O₄(aq)
H₂C₂O₄(aq) + 2NaOH(aq) => Na₂C₂O₄(aq) + H₂O(l)
2 moles NaOH > 1 mole H₂C₂O₄
=> 1 moles NaOH = 2 mole H₂C₂O₄ for problem solution
Since Molarity x Volume in Liters of Solution = moles, then …
[Molarity x Volume(L)] NaOH used = 2[Molarity x Volume(L)] H₂C₂O₄ neutralized
(0.163 mol/L NaOH)(0.02241L)] used = 2·M(0.03709L)
M(H₂C₂O₄) = [(0.163)(0.02241)/(2 x 0.03709)]M H₂C₂O₄(aq) = 0.0492M H₂C₂O₄(aq)