# pH of CH3COOH + HCl solution (shouldn't be a buffer)

$$\pu{0.05 L}$$ of a solution containing $$\pu{0.3 M}$$ acetic acid $$(\ce{CH3COOH}$$, $$K_\mathrm{a} = \pu{1.8e-5})$$ are added to $$\pu{75 mL}$$ of a solution containing $$\pu{0.02 M}$$ hydrochloric acid ($$\ce{HCl}$$, a strong acid). Calculate the $$\mathrm{pH}$$.

Can I simply find the $$[\ce{H+}]$$ released by the reaction of $$\ce{CH3COOH}$$ with water and add them to the $$[\ce{H+}]$$ released by $$\ce{HCl}$$, add the two concentrations and then calculate $$\mathrm{pH}$$ that way? Or does some reaction happen that I'm not aware of?

• Yes there is a reaction, one that you can't possibly be unaware of: the dissociation reaction. Acetic acid is weak, its dissociation is incomplete (that's why you'll need Ka), and HCl affects this process more than a little. – Ivan Neretin Apr 27 at 14:50
• This is a simple case of COMMON ION EFFECT. Where you just have to take care of the Ka being constant (as isothermal process). – user73099 Apr 27 at 14:52
• In other words, if you treat the components separately, and then combine them, the following net reaction happens: $$\ce{CH3COO- + H+ -> CH3COOH}$$ – Karsten Theis Apr 27 at 14:53

Or does some reaction happen that I'm not aware of?

There are three reactions, and you are probably aware of all of them:

$$\tag{1}\ce{H2O(l) <=> H+(aq) + OH-(aq)}$$

$$\tag{2}\ce{HCl(aq) -> H+(aq) + Cl-(aq)}$$

$$\tag{3}\ce{Ch3COOH(aq) <=> H+(aq) + CH3COO-(aq)}$$

The concentration of $$\ce{H+}$$ in those three reactions is the same because it all happens in the same phase. So what happens in reactions (1) and (2) influences the equilibrium of reaction (3).

Can I simply find the [H+] released by the reaction of CH3COOH with water and add them to the [H+] released by HCl, add the two concentrations and then calculate pH that way?

No you can't because if you go back to reaction (3), you are no longer at equilibrium. To avoid trying to fix one equilibrium while messing up another, there are two strategies:

a) Put everything into a system of equations, and solve them in one swoop (preferred method if you are using an equation solver).

b) Start with the major species ignoring minor species and reactions that don't influence major species much. Then, move on to those other reactions and the minor species. This is preferred when you have to do the calculations on paper and don't need the exact solution.

After mixing and ignoring all acid dissociation reactions, the concentrations are the following:

c(acetic acid) = 50 / 125 * 0.3 M = 0.12 M

c(hydrochloric acid) = 75 / 125 * 0.2 = 0.12 M

So here are the steps:

1. Let hydrochloric acid dissociate and get hydrogen ion concentration and pH
2. Check whether acetic acid dissociates appreciably
3. Check whether water dissociated appreciably

For the first step, we get [H+] = 0.12 M, and pH = 0.92. These are tentative because we did not let the other acids (water and acetic acid) dissociate.

For the second step, we are not at equilibrium yet (no acetate yet). However, the pH is very low compared to the pKa of acetic acid, so unless the pH changes a lot, it won't dissociate much. Let's try to calculate the acetate concentration assuming that the change in acetic acid and hydrogen ion concentration is negligible.

$$[\text{acetate}] = K_a * [\text{acetic acid}] / \ce{[H+]} = \pu{1.8e−5} * 0.12 / 0.12$$

So we are not making a big mistake if we say the concentrations of acetic acid and hydrogen ions did not change much due to reaction 2. If we want, we can update the hydrogen ion concentration from 0.12 M to 0.120018 M.

For the third step, we do what we always do when the pH is substantially acidic. We just calculate the hydroxide concentration assuming the hydrogen ion concentration is not much affected by water dissociation. It comes out as $$\pu{8.3e-14}$$. If we want, we can update the hydrogen ion concentration from 0.120018 M to 0.120018000000013 M and a pH of 0.92075. (None of this makes sense because we only had 2 significant figures for the dissociation constant of acetic acid.)

Because we assumed a hydrogen ion concentration of 0.12 M for the second step and 0.120018 M for the third, even though this is not quite true, reactions (2) and (3) are not quite at equilibrium for the concentrations we calculated. The good thing is that in most cases, that does not matter.

• I'll add that the point about significant figures is VERY important. If you try to solve all three equations simultaneously you end up with a very nasty cubic equation which is impossible to solve by hand. No problem for a computer, but 32 digits of precision makes no sense when the equilibrium constants aren't define to better than 1%. // -- The gist -- In chemistry it is all about finding approximate solutions, not mathematical exactitude. So $20.0002 \ne 20$ in a math class, but in a chemistry class they are equal. – MaxW Apr 27 at 16:16