Acid Mixture pH Problem: Calculating Solution pH with CH3COOH, CH3COONa, and HCl

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Discussion Overview

The discussion revolves around calculating the pH of a solution formed by mixing acetic acid (CH3COOH), sodium acetate (CH3COONa), and hydrochloric acid (HCl). Participants explore the implications of buffer chemistry and equilibrium in the context of this mixture, with a focus on the contributions of each component to the overall pH.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant calculates the pH using the buffer equation and suggests that the resulting concentration does not match any of the provided options.
  • Another participant proposes that the contribution of H+ ions from the weak acid can be neglected compared to those from HCl, leading to a simplified calculation that suggests an answer close to option (c).
  • A later reply agrees with the previous calculation but expresses confusion over the explanation provided for the answer being option (b), citing a lack of clarity in the reasoning.
  • Some participants challenge the assumption that all acetate ions will react with HCl to form acetic acid, arguing that this is not entirely accurate and that not all ions will interact.
  • One participant later concedes that the answer given as (b) is correct, explaining that the presence of HCl and sodium acetate affects the equilibrium of acetic acid dissociation according to Le Chatelier's principle.

Areas of Agreement / Disagreement

Participants express differing views on the assumptions made regarding the interaction of acetate ions with HCl and the resulting pH calculation. There is no consensus on the correct approach or final answer, as some participants challenge the reasoning behind the provided solutions.

Contextual Notes

Participants note that the calculations depend on assumptions about the behavior of weak acids and their salts in solution, as well as the equilibrium dynamics involved. The discussion highlights the complexity of buffer systems and the influence of strong acids on weak acid dissociation.

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Homework Statement



30 mL CH3COOH 0.2 M is mixed with 60 ml CH3COONa 0.1 M and then added with 10 mL HCl 0.1 M, then what is the pH of the solution ? (Ka = 10-5)

A. 3
B. 5 - log 1.4
C. 5 - log 2
D. 4 - log 1.4
E. 5 + log 2

Homework Equations



[H+] = Ka * [HX] / [X- ] ---- buffer
[H+] = ([H+]V1 + [H+]V2) / (V1 + V2) ---- mixture

The Attempt at a Solution



The hydrogen molarity from the buffer is 10^-5 * 6 / 6 = 10^-5

Hydrogen concentration of the mixture = (10^-5 * 90 + 10^-1 * 10) / ( 90 + 10 ) = (9 * 10^-4 + 1 ) / 100 = 9 * 10^-6 + 10^-2

pH = - log ( 9 * 10^-6 + 10^-2 ) which doesn't appear in the options..
Please help where I got wrong
 
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Here is what I did:

First off, we can neglect the H+ ions provided by the weak acid to the solution as they are negligible compared to the H+ provided by the HCl.
Then the concentration can be replaced with the moles, because the entire mixture is in the same solution and the volume terms will get cancelled.
On doing so, we get 4 + log 6. This is pretty close to the value of option (c).
Whats the answer?
 
CrazyNinja said:
Here is what I did:

First off, we can neglect the H+ ions provided by the weak acid to the solution as they are negligible compared to the H+ provided by the HCl.
Then the concentration can be replaced with the moles, because the entire mixture is in the same solution and the volume terms will get cancelled.
On doing so, we get 4 + log 6. This is pretty close to the value of option (c).
Whats the answer?

The answer is B. 5 - log 1,4
But, the explanation of the answer is just 10^-5 * 7 / 5 which is very confusing...
I really don't get it..
 
The solution to the answer has assumed that the CH3COONa reacts with the HCl providing you with more CH3COOH. That is not completely correct. The salt will completely dissociate, but not all the CH3COO(minus) ions will interact with HCl to give CH3COOH.
 
CrazyNinja said:
The solution to the answer has assumed that the CH3COONa reacts with the HCl providing you with more CH3COOH. That is not completely correct. The salt will completely dissociate, but not all the CH3COO(minus) ions will interact with HCl to give CH3COOH.

Why not all the ions will interact? How do you do the problem?
 
OK wait. I am sorry. the answer given is correct. It is (b).

This is because there exists an equilibrium for the dissociation of CH3COOH.

CH3COOH ↔ CH3COO(-) + H(+)

Now if there was no HCl or CH3COONa, the dissociation would proceed without any hindrance. But the prescence of these two generates CH3COO(-) ions and H(+) ions. Now these will shift the equlibrium reaction (use Le-Chatlier's principle). Thus CH3COOH will be produced. Obviously in the reaction, HCl is the limiting reagent.
 
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