11.) 0.115 g of an unknown diprotic acid is titrated with 0.095 M NaOH. The first equivalence point occurred in the titration at a volume of 8.80 mL of NaOH added; the second equivalence point occurred at a volume of 17.60 mL of NaOH added. How many moles of diprotic acid were titrated? mol diprotic acid = mol
11.) 0.115 g of an unknown diprotic acid is titrated with 0.095 M NaOH. The first equivalence point occurred in the titration at a volume of 8.80 mL of NaOH added; the second equivalence point occurred at a volume of 17.60 mL of NaOH added. How many moles of diprotic acid were titrated? mol diprotic acid = mol
Chapter9: Acids, Bases, And Salts
Section: Chapter Questions
Problem 9.155E
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![11.) 0.115 g of an unknown diprotic acid is titrated with 0.095 M NaOH. The first equivalence point occurred in the
titration at a volume of 8.80 mL of NaOH added; the second equivalence point occurred at a volume of 17.60 mL of
NaOH added.
How many moles of diprotic acid were titrated?
mol diprotic acid =
mol](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6f9ca2ce-c481-46d9-90b7-dd41b10c6161%2F933f0f98-5b63-493e-b656-5af24a3cccd3%2Fxwpbw5v.png&w=3840&q=75)
Transcribed Image Text:11.) 0.115 g of an unknown diprotic acid is titrated with 0.095 M NaOH. The first equivalence point occurred in the
titration at a volume of 8.80 mL of NaOH added; the second equivalence point occurred at a volume of 17.60 mL of
NaOH added.
How many moles of diprotic acid were titrated?
mol diprotic acid =
mol
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