Hi, I did the acid-base titration lab with HCl and 0.5M NaOH.

Question

I need to calculate the moles of NaOH from molarity of NaOH and the average volume used.

My English isn't good but is it asking me to find the moles of NaOH using the molarity of NaOH and the average volume used correct?

This is my table.
Molarity of NaOH used: Trial 1
Initial volume of NaOH (mL): 27.10
Final volume of NaOH (mL): 38.90
Volume of NaOH used (mL): 11.80

Trial 2
10.80
22.60
11.80

Trial 3
22.60
36.70
14.10

My average volume of NaOH:
11.80+14.10/2=12.95mL

Then for some reason I am really confused even though I know it should be very simple... Please help me, thank you kindly.

Answer 1

The average volume is not right because you threw away one of the 11.8 volumes.
Volumes are 11.80, 11.80 abd 14.10
Sum those and divide by 3.

Then mols NaOH = M NaOH x L NaOH = ?

Answer 2

I thought we only add the first two because the first one was purposely overshot?

0.5M NaOH × 12.56666667= 6.283333333
=6.3 moles

Thank you for replying!