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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Chapter9: Acids, Bases, And Salts
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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
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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