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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?

User Skyhan
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1 Answer

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Answer:

C = 0.875 M

Step-by-step 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 *
(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

User Valene
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