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

water and alcohols have similar properties because water molecules contain hydroxyl groups that can form hydrogen bonds with other water molecules and with alcohol molecules, and likewise alcohol molecules can form hydrogen bonds with other alcohol molecules as well as with water.