Calculating mols carboxylic acid needed to nuetralize NaOH solution

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Discussion Overview

The discussion revolves around calculating the number of moles of carboxylic acid needed to neutralize a NaOH solution in a specific chemical reaction context. The focus is on stoichiometry and the implications of acid dissociation constants in neutralization reactions.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents a calculation involving the moles of NaOH and HCl to determine the moles of carboxylic acid required for neutralization.
  • Another participant questions the relevance of the acid dissociation constant (Ka) in the context of simple neutralization stoichiometry.
  • A participant expresses confusion about the role of Ka and its effect on hydronium ion concentration from acetic acid.
  • There is a proposal that the carboxylic acid neutralizes NaOH in a 1:1 ratio, leading to a calculation of .0132 moles of carboxylic acid needed.
  • One participant tentatively agrees with the proposed calculation but does not verify the numbers.

Areas of Agreement / Disagreement

Participants express differing views on the relevance of Ka in this context, with some suggesting it is unnecessary for the stoichiometric calculations while others believe it may have an impact. The discussion does not reach a consensus on the necessity of including Ka in the calculations.

Contextual Notes

There are unresolved aspects regarding the application of Ka in the calculations and the assumptions made about the neutralization process. The calculations presented are based on the stoichiometry of the reaction without a definitive resolution on the role of equilibrium constants.

mesa
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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 :)
 
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What do you use Ka for? This is about simple neutralization stoichiometry, no equilibrium calculations needed.
 
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?
 
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?
 
I have not checked the numbers, but it sounds right and fits what you wrote earlier.
 
Good deal, thanks Borek
 

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