Question

You decide to titrate the solution of NaOH using oxalic acid as a primary standard and phenolphthalein as the indicator. In the reaction of oxalic acid with NaOH, every 1 mole of oxalic acid is deprotonated by 2 moles of NaOH. The molecular weight of oxalic acid is 90.03 g/mol and you use 0.60 g in 50 mL water for the titration. You use 15.20 mL of your NaOH solution in order to reach the endpoint of the titration. What is the concentration of your NaOH solution?

Answer 1

Answer:

C = 0.875 M

Explanation:

• The concentration of the NaOH solution is equal to the number of moles of NaOH divided by the volume:

C = n/V

From the problem we know that V = 15.20 mL = 0.0152 L

So we just need to calculate the moles of NaOH:

• First we calculate the moles of oxalic acid, using its molecular weight:

0.60 g ÷ 90.03 g/mol = 6.66 *10⁻³ mol.

Every 1 mole of oxalic acid is deprotonated by 2 moles of NaOH, so the moles of NaOH are:

6.66 *10⁻³ mol Acid * $\frac{2molNaOH}{1molAcid}$ = 0.0133 mol NaOH

• The concentration of the NaOH solution is

C = n / V = 0.0133 mol / 0.0152 L = 0.875 M