Can anyone please help with this problem?

Question

joannablackwelder · Answer

https://s3.amazonaws.com/iedu-attachments-message/3a76a08bec998dcaf0befc882bf5cb02_e8b71ab418405d60688e4570bb4bd3d7.png

cuanchi · Answer

A) ~ 0.14M
B) Trial 1 = 0.1324M
    Trial 2= 0.1278 M
 average = 0.1301 M

joannablackwelder · Answer

Thank you! But how did you get that?

vera_ewing · Answer

What do you not understand about the question?

joannablackwelder · Answer

I get part A. But part B, I don't see how you relate the data given to the concentration of KOH.

aaronq · Answer

For part B, oxalic acid is a diprotic acid - indicating that for every mole of this acid, you need two moles of KOH to neutralize it. 

So for each trial, you find the moles of the acid, then it's concentration (using the initial volume). Since the difference between the initial and final volumes, is the volume of KOH solution added, you can find it's molarity.

joannablackwelder · Answer

Ah, I see! Thanks! The piece I was missing was the one to two ratio for diprotic acids. Your help is much appreciated!