# Chemistry

posted by on .

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

No one has answered this question yet.

• Chemistry - ,

Yes, moles NaOH = M x L; however, you didn't convert 13.9 mL correctly. It should be 0.0139 L.

0.414 x (25 mL/100 mL) is correct.

So if the acid is diprotic (I assume you were told that), then moles H2A = 1/2 moles NaOH = ??
Now molar mass acid is
n = g/molar mass and
molar mass = grams H2A/moles H2A.

• Chemistry - ,

Thats exactly what i was thinking of doing. However a friend of mine dvided the molar mass of the acid by 4 in the end. Andinsists that its the riht thing to do. I don't understand why he did that but it seems toi have gotten him the right answer :S

• Chemistry - ,

Certainly that will work. The grams of acid/mole = molar mass.
So if you use 4 times the acid, then divide the final result by 4 you get the same answer as dividing by 4 first before dividing by moles. I wonder if this is what you did on the other problem I just did. You had 1.2/moles and your answer was 4x too high which is why you had that huge molar mass of 1304 something.

• Chemistry - ,

Thanks Alot DrBob222 :D rele means alot ! ! :D

#### Related Questions

More Related Questions

Post a New Question