Is the Final pH Calculation Correct?

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The discussion focuses on calculating the final pH after adding 1/20 mL of 1.0M HCl to a 10mL solution of 0.1M acetic acid (CH3COOH). The user calculated the moles of H+ from both the acetic acid and HCl, arriving at a total of 9.024922359 x 10^-5 moles of H+. The resulting pH calculated is 2.52, but the user expresses uncertainty about the correctness of this answer. Additionally, there is a request for guidance on applying the common-ion effect and using an ICE table for the acetic acid in this context. The inquiry highlights the need for clarification on the pH calculation process involving weak acids and strong acids.
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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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