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Two 23.0 mL samples, one 0.200 MKOH and the other 0.200 M CH3NH2, were titrated with...

Two 23.0 mL samples, one 0.200 MKOH and the other 0.200 M CH3NH2, were titrated with 0.100 MHI. Answer each of the following questions regarding these two titrations.

a) What is the volume of added acid at the equivalence point for KOH?

b) What is the volume of added acid at the equivalence point for CH3NH2?

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Answer #1

No. of moles = Molarity Volume of solution in litres

moles KOH = 0.023 L x 0.200 M = 4.6 10-3
moles HI required = 4.6 10-3
Volume HI added at equivalence point = (4.6 10-3) / 0.100 M = 0.046 L = 46.0 mL

the moles of CH3NH2 are the same, so the volume of HI added at equivalence point is the same : 46.0 mL

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