Question
1.639 g of an unknown diprotic acis are used to make a 100.00 mL solutiob. Then 25.00 mL of this solution is trasferred to an Erlenmeyer flask and is titrated with 0.1008 M NaOH. The titration endpoint is reaxhed after 0.00432 mol of NaOH is added using the burette. What is the molar mass of the diprotic acid? Provide your answer to the correct nuber of significat figures

1.639 g of an unknown diprotic acid are used to make a 100.00 mL solution. Then 25.00 mL of this solution is transferred to a
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Answer #1

Consider a reaction, H2A + 2 NaOH \rightarrow Na2A + 2 H2O

From reaction, 2 mol NaOH \equiv 1 mol H2A

\therefore 0.00432 mol NaOH \equiv 1 \times 0.00432 / 2 mol H2A

0.00432 mol NaOH \equiv 0.00216 mol H2A

i e 25.00 ml of diluted solution \equiv 0.00216 mol H2A .

Therefore, 100.00 ml original solution \equiv 100.00 \times 0.00216 / 25.00 mol H2A

100 ml original solution \equiv 0.00864 mol H2A

We have, No. of moles = Mass / Molar mass

Therefore, Molar mass of H2A = Mass / No. of moles = 1.639 g /0.00864 mol = 189.699 = 190 g / mol

ANSWER : Molar mass of diprotic acid = 190 g / mol

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