Posted by Steve on Monday, October 4, 2010 at 10:17pm.
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.
- Chemistry - DrBob222, Monday, October 4, 2010 at 10:49pm
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.
- Chemistry - Steve, Monday, October 4, 2010 at 11:36pm
I'm given the molar ratio of base to acid is 1:1. Does that mean it's monoprotic?
- Chemistry - DrBob222, Monday, October 4, 2010 at 11:58pm
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.
- Chemistry - Steve, Tuesday, October 5, 2010 at 12:13am
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.
- Chemistry - Rayana, Sunday, October 10, 2010 at 6:28pm
What if the unknown acid diprotic?
- Chemistry - kris, Sunday, April 8, 2012 at 6:40pm
- Chemistry - Anonymous, Tuesday, October 30, 2012 at 1:55pm
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