A 0.1310 g sample of an unknown diprotic acid is diluted to 100.00 mL and titrated by using 0.1910 M NaOH. If 14.20 mL of the NaOH solution is required to reach the second equivalence point, what is the molar mass of the acid? 2.

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Chapter16: Acid-base Equilibria
Section: Chapter Questions
Problem 16.88QP: A 0.400-g sample of propionic acid was dissolved in water to give 50.0 mL of solution. This solution...
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A 0.1310 g sample of an unknown diprotic acid is diluted to 100.00 mL and titrated by using 0.1910
M NaOH. If 14.20 ml of the NaOH solution is required to reach the second equivalence point, what
is the molar mass of the acid?
2.
Transcribed Image Text:A 0.1310 g sample of an unknown diprotic acid is diluted to 100.00 mL and titrated by using 0.1910 M NaOH. If 14.20 ml of the NaOH solution is required to reach the second equivalence point, what is the molar mass of the acid? 2.
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