Asked by Steve

I there, this is my lab worksheet, I am having trouble filling it out. Please help me with it.

Expt #1- Molecular Weight of Unknown Acid

Unknown Acid: #3
Mass of Unknown solid acid transferred:1.0g
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: 14.34 mL

Here's where I need help:
No of moles NaOH used: ? mol

Molar ratio of base to acid?

The unknown acid is mono, di, triprotic acid:?

The titration reaction was:?

No. of moles acid in 25 mL solution:?

Gram-Molecular acid in 25mL solution?

I know it's a lot but please help me as much as you can.
Answered by DrBob222
Moles NaOH used = M x L.
From your data, that is 0.0989 M x 0.01434 L = 0.001418 moles NaOH.

You have 1.0 x (25.00 mL/100 mL) = 0.25 g of the acid used in the titration but I don't see the data to show that it is mono, di, or triprotic acid.
Answered by Steve
I'm given the molar ratio of base to acid is 1:1. Does that mean it's monoprotic?
Answered by DrBob222
Yes, it does.
So grams/moles = molar mass.
Then 0.25/0.001418 = 176.3 for the molar mass.
Titration reaction is
NaOH + HX ==> NaX + H2O where X is the remainder of the molecule.

Answered by Steve
Ok, great.

But I'm still confused, sorry.

So for No. of moles acid in 25 mL solution: 0.25/0.001418 = 176.3 moles?

And for molecular weight of unknown acid?
How would you find this.
Answered by Rayana
What if the unknown acid diprotic?
Answered by kris
119.46g/mol
Answered by Anonymous
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