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Topic: Acid-Base Titration  (Read 2184 times)

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Offline Lanochka

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Acid-Base Titration
« on: November 04, 2009, 10:57:26 PM »
I have this lab assigned, but I was absent for the lesson due to the flu. I'm trying to work my way through it but I simply don't understand where to start.



For the first trial in the top table, would the calculation be:

(25.89 mL/1) x (1 L/1000 mL) x (0.1024 mol HCL/1 L) = 0.002651 mol/L

or would it be

(23.23 mL/1) x (1 L/1000 mL) x (0.1024 mol NaOH/1 L) = 0.002379 mol/L


I feel like I'm doing it all incorrectly.

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