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?

C = 0.875 M

Explanation:

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:

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 * = 0.0133 mol NaOH

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

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