[AntonioVerde]

Can you explain and give me the solution of this exercise?
"Calculate the pH of a solution containing 7×10⁻² g of CH₃COOH (Ka = 1.8×10⁻⁵) in 250 mL of water. Calculate the pH of the solution when 130 mL of water containing 30 mg of HCl is added."

The first answer (the pH before adding HCl) I think it should be 3,53, correct me if I'm wrong.
But I can't understand how to do the second part of this exercise.

[Hunter2]

pH 3.53 is ok.
Now you add a strong acid (HCl) .
What happens with the equilibrium of acetic acid.

[AntonioVerde]

This was the process for the first part of the exercise, please tell me if I did wrong:

I calculated the number of moles of acetic acid, n = (7×10⁻²)/60 = 1.16×10⁻³ mol
Then I calculated the concentration [CH₃COOH] = n/V = 1.16×10⁻³/0.25 = 4.66×10⁻³

Since the acid is weak, the formula [H⁺] = Sqrt(Ka × [CH₃COOH]) = 2.89×10⁻⁴ applies.
In the end pH = -log[H⁺] = 3.53


How do I even start to do the second part? I don't even know what formulas I have to use

[Hunter2]

Correct.
You have CH3COOH => CH3COO- + H+
So what happen you add strong Acid means more H+.
What is pH if you have only 30 mg HCl .
For HCl is acetic acid a base.

[Borek]

It is not about plugging numbers into formulas, it is about finding the equilibrium using a general approach. Start with the dissociation reaction equation, write the formula for Ka, think what happens to concentrations of species involved when the equilibrium shifts (this is mostly a simple stoichiometry).

Do you know what ICE table is? It makes solving this type of problems much easier, not because it gives a magic formula, but it adds structure to the solving process, making it easier to follow.

[AntonioVerde]

Correct.
You have CH3COOH => CH3COO- + H+
So what happen you add strong Acid means more H+.
What is pH if you have only 30 mg HCl .
For HCl is acetic acid a base.

I had thought of finding the concentration [HCl], and since HCl is a strong acid it dissociates completely and therefore [HCl] = [H+ from HCl]
Then I would have calculated the concentration [CH3COOH] = nmoles/Total volume = 1.16*10^-3/(0.25+0.13).
Then I would have found [H+ from CH3COOH] = sqrt( Ka*[CH3COOH])
I would then have added the two concentrations of H+ found, and with that result I would have calculated the pH, is it wrong?

Or should I use the the formulas of buffer solutions?

[Hunter2]

I think it's not Addition. Acetic acid has higher pH as hydrochloric.The H+ of hydrochloric react with the dissoziated acetate to non dissoziate acetic acid.
So the amount of free H+ is degreased.

[AntonioVerde]

Sorry, I didn't quite understand, how would you do the exercise?

[Hunter2]

Similar like here

https://socratic.org/questions/calculate-the-ph-of-a-solution-formed-by-the-addition-of-10-0ml-of-0-050m-hydroc

Or with formula

https://chemistry.stackexchange.com/questions/69307/calculation-of-the-ph-of-a-mixture-of-a-strong-acid-and-weak-acid

[Borek]

I had thought of finding the concentration [HCl], and since HCl is a strong acid it dissociates completely and therefore [HCl] = [H+ from HCl]
Then I would have calculated the concentration [CH3COOH] = nmoles/Total volume = 1.16*10^-3/(0.25+0.13).
Then I would have found [H+ from CH3COOH] = sqrt( Ka*[CH3COOH])
I would then have added the two concentrations of H+ found, and with that result I would have calculated the pH, is it wrong?

Concentration of H⁺ is not a simple sum of these two, as adding H⁺ from HCl shifts the dissociation equilibrium of the acetic acid left (you can think about it in terms of acetate working as a base and neutralizing some of the H⁺ from HCl).

I will repeat the question: have you heard about ICE tables? (and if not, I strongly suggest you google them and see how they are used, this is a perfect case to use one).