Question

The flask shown here contains 0.541 g of acid and a few drops of phenolphthalein indicator dissolved in water. The buret contains 0.230 m naoh. What volume of base is needed to reach the end point of the titration? What is the molar mass of the acid (assuming it is diprotic and that the end point corresponds to the second equivalence point)? Volume of titration is 12 ml.

Answer 1

A diprotic acid reacting with NaOH will appear as:

H2A + 2NaOH --> 2H2O + Na2A

From the diagram, 25 mL of base (NaOH) has been added to reach the endpoint

From this volume, we can get the number of moles reacted using the molarity

M = mol/V

mol = V*M

mol = (0.025 L)(0.220 M)

mol = 5.5x10-3 moles NaOH

Using the ratio from the balanced equation, we can get the moles of acid from it (H2A:NaOH) which is 1:2

5.5x10-3 mol NaOH (1 mol H2A/ 2 mole NaOH) = 2.75x10-3 moles H2A

From here, we can now get the molar mass by dividing it to the given mass 0.365g

MM of H2A = (0.365 g)/(2.75x10-3 mol)

MM of H2A = 132.7272 or 132 g/mol

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