Question

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)?

Answer 1

Answer : 13.64 mL

Explanation : The reaction requires 13.64 mL of NaOH 

From the reaction we know that oxalic acid requires two moles of NaOH so we have to calculate the number of moles of both.

0.150 L X 0.1500 moles/L = 2.25 X $10^{-3}$

So the moles of NaOH will be 2 X (2.25 X $10^{-3}$) =  4.5 X $10^{-3}$

Now, we know the concentration of NaOH as 0.330 M

4.5 X $10^{-3}$ moles / 0.330 moles/L = 0.01364 L or 13.64 mL

So, the volume of NaOH needed to neutralize oxalic acid will be 13.64 mL

Answer 2

Answer: 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