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.
2 answers
show me your thinking on this.
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)
3 answers