What volume of a 0.3300-m solution of sodium hydroxide would be required to titrate 15.00 ml of 0.1500 m oxalic acid? c2 o4 h2(aq) + 2naoh(aq) ⟶ na2 c2 o4(aq) + 2h2 o(l)?

13.64 ml

Explanation:

1) Titration will be done when all the moles of the oxalic acid have reacted with the volume of sodium hydroxide.

2) The balanced chemical equation is:

C₂O₄H₂ (aq) + 2 NaOH(aq) ⟶ Na₂C₂O₄(aq) + 2H₂O(l)

3) Mole ratio:

1 mol C₂O₄H₂ (aq) ; 2 mole NaOH(aq)

4) Calculate the number of moles of C₂O₄H₂ (aq)

M = n/V ⇒ n = MV = 0.01500 l × 0.1500M / 1000 = 0.00225 mol

5) Calculate the number of moles of NaOH needed using proportion with the mole ratio:

1 mol C₂O₄H₂ (aq) / 2 mole NaOH(aq) = 0.00225 mol C₂O₄H₂ (aq) / x

⇒ x = 0.0045 mol NaOH

6) Use molarity to find the volume that contains 0.0045 mol of NaOH.

M = n / V ⇒ V = n / M = 0.0045 / 0.3300 = 0.1363 liter = 13.64 ml

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