9. A researcher has produced an extract from a tropical plant that contains a di

ID: 906883 • Letter: 9

Question

9. A researcher has produced an extract from a tropical plant that contains a diprotic acid.

The researcher titrates 100.0 mL of the plant extract with 0.100 M NaOH at 25°C. The first

equivalence point of the titration is reached when 25.6 mL of NaOH have been added. The

pH at the first half equivalence point was 3.58.

In a different experiment at 25°C with the same plant extract, the researcher determines that

there is only 0.123 g of the diprotic acid in every 100.0 mL of the plant extract. What is the

molar mass of this acid?

a. 48.0 g/mol

b. 24.0 g/mol

c. 12.3 g/mol

d. Not enough information is provided.

Volume of 0.100 M NaOH required to reach the 1st equivalent point = 25.6 mL

therefore concentration of the acid in 100 mL extract = 25.6 x 0.100/ 100 = 0.0256 M

so number of moles of the acid present in 100 mL = 100 x 0.0256 / 1000 = 0.00256 moles

Now 0.00256 moles has mass 0.123 g

therefore 1 mole has a mass = 0.123 g/ 0.00256 moles = 48.0 g/ moles

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