Is the Final pH Calculation Correct?

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Discussion Overview

The discussion revolves around the calculation of the final pH of a solution after the addition of hydrochloric acid (HCl) to a solution of acetic acid (CH3COOH). Participants explore the implications of the common-ion effect and the use of an ICE table in this context.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Conceptual clarification

Main Points Raised

  • The initial poster calculated the number of moles of H+ from both the acetic acid and the added HCl, arriving at a final pH of 2.52 but expressed uncertainty about the correctness of the solution.
  • One participant inquired about the application of the common-ion effect in this scenario.
  • Another participant suggested using an ICE table for acetic acid, treating the H+ from HCl as an initial concentration.

Areas of Agreement / Disagreement

The discussion does not appear to reach a consensus, as participants are exploring different methods and concepts without confirming the correctness of the initial calculation.

Contextual Notes

Participants have not fully resolved the assumptions regarding the ionization of acetic acid and the impact of the added HCl on the pH calculation.

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Homework Statement


What will be the final pH of the solution if a drop (approximately equal to 1/20 mL) of 1.0M HCl is added to a 10mL 0.1 M CH3COOH sol'n?


Homework Equations


pH = -log [H+]


The Attempt at a Solution


I first determined the number of moles of H+ from the partial ionization of CH3COOH (CH3COOH <--> CH3COO- + H+) using the ICE table. The number of moles H+ from my computation is 4.024922359 x 10^-5.

Then I also determined the number of moles of H+ from HCl--> H+ + Cl-. What I got is 5 x 10^-5 moles.

I added the two values to get the total number of moles H+.

The pH I got is 2.52.


Is my answer correct?
I'm not quite sure about how I solved it.
Need help. Thanks. :smile:
 
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How do I apply the common-ion effect for this problem?
 
Do an ICE table for acetic acid treating H+ from HCl as initial concentration.
 
Thanks much. I'll do that.
 

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