The titration of 0.02500 L of a diprotic acid solution with 0.1000 M NaOH requir

Question

The titration of 0.02500 L of a diprotic acid solution with 0.1000 M NaOH requires 34.72 mL of titrant to reach the second equivalence point and 9.27 at the second equivalence point.
If the acid solution contained 0.2020 g of acid, what is the molar mass?
What is the pKa1, and pKa2 of the acid?

THANK YOU! The titration of 0.02500 L of a diprotic acid solution with 0.1000 M NaOH requires 34.72 mL of titrant to reach the second equivalence point and 9.27 at the second equivalence point.
If the acid solution contained 0.2020 g of acid, what is the molar mass?
What is the pKa1, and pKa2 of the acid?

THANK YOU! The titration of 0.02500 L of a diprotic acid solution with 0.1000 M NaOH requires 34.72 mL of titrant to reach the second equivalence point and 9.27 at the second equivalence point.
If the acid solution contained 0.2020 g of acid, what is the molar mass?
What is the pKa1, and pKa2 of the acid?

THANK YOU!

Explanation / Answer

total volume of sodium hydroxide consumed to neutralize the diprotic acid = 34.72 + 9.27= 43.99 ml

1000 ml of sodium hydroxide contains 0.1 moles of NaOH.

so 1 ml of sodium hydroxide contains 0.1/1000 = 0.0001 moles of NaOH.

so 43.99 ml of sodium hydroxide contains 0.0001 X 43.99 = 0.004399 moles of NaOH.

as the acid is diprotic, 1 mole of NaOH will neutralize 1/2 mole of acid.

hence the moles of acid neutralized by 0.004399 moles of NaOH= 0.004399/2 = 0.0021995 moles

By using the equation moles= weight (gm) x molecular mass

molecular mass= weight (gm)/ moles = 0.2020/ 0.0021995 = 91.83 gm/ mol.

The pKa1 and pKa2 can only be calculated by volume-pH data obtained from the titration curve. So above given data is insufficient.

If you want to identify the given acid by the obtained molecular mass, then is could possibly be oxalic acid, the standard pKa1 and pKa2 values of which are 1.25 and 4.27 respectively.

Thank You

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