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
See your post above.
To determine the number of moles of acid in the 25 mL solution, you need to use the molar mass of the acid. First, calculate the moles of acid used in the titration:
Moles of acid = (mass of acid transferred) / (molar mass of acid)
You mentioned that the mass of acid transferred is 0.414 g. However, you didn't provide the molar mass of the acid, so we cannot calculate the moles of acid without this information.
To calculate the gram-molecular weight of the acid in the 25 mL solution, you will need to use the concentration and volume of the NaOH solution used in the titration. First, calculate the moles of NaOH used:
Moles of NaOH = (concentration of NaOH) x (volume of NaOH used)
In your case, the concentration of NaOH is 0.0989 M and the volume used is 13.9 mL (convert to L by dividing by 1000):
Moles of NaOH = (0.0989 M) x (13.9 mL / 1000 mL/L)
Next, determine the molar ratio between NaOH and the acid. Since the acid is diprotic (H2A), the molar ratio is 2:1. This means that for every 2 moles of NaOH, 1 mole of acid is involved in the reaction.
Now, using the molar ratio, you can calculate the moles of acid in the titration:
Moles of acid = (moles of NaOH) / 2
Once you have the moles of acid in the titration, you can use this to calculate the gram-molecular weight of the acid in the 25 mL solution. However, since we don't have the molar mass of the acid, it is not possible to provide you with the final answer.
To complete the calculation, you would use the equation:
Gram-Molecular acid in 25 mL solution = (moles of acid) x (molar mass of acid)
Please provide the molar mass of the acid to proceed with the calculation.