Question

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:
. n=C.V
. =(0.0989M)(0.001375L)
. =0.001375mol



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.
. HELP?!??

Gram-Molecular acid in 25mL solution?

HELPPPP ! ! ! ! :( :(

ANY AMOUNT OF HELP WILL BE GRATEFUL! PLZ N THANK YOU :D

Answer 1

See your post above.