Homework Help: Calculating mols carboxylic acid needed to nuetralize NaOH solution

1. Oct 23, 2012

mesa

1. The problem statement, all variables and given/known data
A 10mL mixture of 2M HCl and carboxylic acid required 43.61 mL of .762 M NaOH to nuetralize it. Calculate the number of mols of carboxylic acid present.

3. The attempt at a solution
First figured out the number of mols of Naoh and subtracted the number of mols of HCl:
.0332molNaOH-.0200molHCl = .0132 mol NaOH that still need nuetralized
Divided that by total volume so .246M H3O is needed to finish the job.

With a Ka of 1.75X10^(-5) I get 3458M of acetic acid
So I think I may be a little off :)

2. Oct 23, 2012

Staff: Mentor

What do you use Ka for? This is about simple neutralization stoichiometry, no equilibrium calculations needed.

3. Oct 23, 2012

mesa

Hey Borek, okay so the approach is wrong then lol.
I thought the Ka would affect the H3O delivered by the acetic acid?

So what do I do?

4. Oct 23, 2012

mesa

Okay, so the carboxylic acid will neutralize the NaOH in a 1:1 ratio?
If that is the case then it is just .0132 mols, is that right?

5. Oct 23, 2012

Staff: Mentor

I have not checked the numbers, but it sounds right and fits what you wrote earlier.

6. Oct 23, 2012

mesa

Good deal, thanks Borek