posted by Rayana .
Expt #1- Molecular Weight of Unknown Acid
Unknown Acid: #2
Mass of Unknown solid acid transferred:0.414g
Volume of volumetric flask: 100.00 mL
Concentration of NaOH: 0.0989 M
Aliqot of acid titrated with NaOH: 25.00 mL
Average volume of Naoh from titration: 13.9 mL
Here's where I need help:
No of moles NaOH used: ? mol
--This is wat i did:
Molar ratio of base to acid?
. Diprotic acid. so 2:1
The unknown acid is mono, di, triprotic acid:? Diprotic
The titration reaction was:?
. H2A + 2NaOH --> 2H2O +Na2H
. where A is the unknown acid compound
No. of moles acid in 25 mL solution:?
. HELP >.<' I don't understand the
. question :P But here is wat i think
. of it. (0.414g)x(25mL/100ml)
. =0.1035 g of acid used in
. the titration.
Gram-Molecular acid in 25mL solution?
HELPPPP ! ! ! ! :( :(
ANY AMOUNT OF HELP WILL BE GRATEFUL! PLZ N THANK YOU :D
See your post above.