Calculating mols carboxylic acid needed to nuetralize NaOH solution

[mesa]

Homework Statement

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.

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 :)

 

[Borek]

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

 

[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?

 

[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?

 

[Borek]

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

 

[mesa]

Good deal, thanks Borek