# What's the pH of vinegar containg 5% acetic acid?

Vinegar generally contains 5% acetic acid. We would expect the pH of vinegar to be approximately:
a. 0
b. 3
c. 7
d. 9
e. 12

I don't have the key for this question, so I just want to make sure that I got it right and what I got doesn't seem to be one of the answer choices.

My answer was 1.3, so I rounded it to 1 and there wasn't an answer choice. Here is what I did:

$$\log(.05/1) = -1.3010$$

How else would you do it because you were already given the final concentration from the original concentration?

Am I missing something here?

Edit: The professor said to use this formula for calculating it:

$$K_\mathrm{a} = \frac{(.05x)^2}{x}$$

and got 3, which made no sense, any way this is possible?

• What does 5% acetic acid mean, i.e. 5% of what? What quantity do you need to calculate the pH? What property of acetic acid to you need to calculate the pH? Feb 10, 2015 at 23:34
• Well we know that vinegar is mostly composed of water and acetic acid so I guess they are asking for the pH of that 5%? Feb 10, 2015 at 23:36
• Right, but you don't know the molarity of the hydrogen ion concentration. You have the concentration of acetic acid as a percentage... Feb 10, 2015 at 23:44
• Did the question say to round it?
– L.B.
Feb 11, 2015 at 0:49
• PS - the professors solution is totally wrong.
– MaxW
Jan 30, 2017 at 21:48

Consider the choices. Since it's acetic acid, you can rule out any pH values that indicate a basic or neutral solution. Knowing that acetic acid is a weak acid and that vinegar is a fairly dilute solution, you can also rule out any pH values that would indicate either a strong acid or a highly concentrated solution. Eliminating those impossible values leaves only one choice as the answer.

The trick with a question like this is to realize that it's multiple choice for a reason. It's intended to be a conceptual problem, not a calculation problem. In general, a good rule for multiple-choice questions like this is to first see if you can solve it using general concepts; only if you can't should you reach for you calculator.

• I know you are very right, but what if you were to show your work? I know that without work 3 is the most plausible answer but with work I can't mathematically support my assumptions. Feb 11, 2015 at 22:35
• To show work (or to get an exact answer in general), I'd use basically the same method that danijoo did. The wording of the question is clear that the desired answer is only approximate, so calculating that the actual pH is about 2.4 is good enough. You're not rounding to the nearest integer; you're rounding to the nearest answer choice. That said, the only assumptions required for the method I gave are that the desired answer is one of the choices given and that work isn't required. The rest follows logically from general principles of acids, bases, and pH. Feb 12, 2015 at 2:40

You are not given a concentration, but a percentage value. Assuming its mass percent (wt%), you first have to calculate the actual concentration of the solution like this:

5% Acetic acid means in a $\pu{1 kg}$ solution, you can find $0.05 \times \pu{1000 g} = \pu{50 g}$ acetic acid. With a molar mass of $\pu{60.05 g/mol}$, this is similar to

$$n = \frac{\pu{50 g}}{\pu{60.05 g/mol}} = \pu{0.83 mol}$$

The concentration $C = n/V$, therefore is

$$C = \frac{\pu{0.83 mol}}{\pu{0.95 L}} = \pu{0.874 mol/L}$$

Acetic acid is a weak acid. This means it does not dissociate completely and $\mathrm{p}K_\mathrm{a}$ has to be taken into account to calculate the pH. The $\mathrm{p}K_\mathrm{a}$ of acetic acid is about $4.76$. The pH is then defined as:

$$\mathrm{pH} = 0.5 (\mathrm{p}K_\mathrm{a} - \log C_\mathrm{a}) = 0.5 (4.76 - \log 0.874) = 2.4$$

• Which isn't one of the choices either unless you round up and most people don't round up until 2.5.
– L.B.
Feb 11, 2015 at 0:43
• True but If I didnt make a mistake, this is the correct calculation. The closest answer is a pH of 3 which is also what google says about the pH of vinegar. The answer options just suck. Feb 11, 2015 at 0:46
• I see what you're getting at there. I was just slightly concerned since the person asking wasn't able/willing to round 1.3 and get a close answer.
– L.B.
Feb 11, 2015 at 0:48
• also, I just found this: answers.yahoo.com/question/index?qid=20110109075809AAGGAwF Its the exact same solution. Feb 11, 2015 at 0:51
• Well there 'ya have it :)
– L.B.
Feb 11, 2015 at 0:53

You don't need a calculation to know the answer. You aren't likely to use a pH 0 acid to make a salad :-). 7 is neutral and therefore not acid. 9 and 12 are bases. The only answer left is 3.

• A rehash of j_foster’s answer
– Jan
Sep 9, 2017 at 9:16

I know this question was over 2 years ago, but I've just joined and wanted to see if I could improve on the responses already given and practice the formatting of text while also perhaps helping someone else who might come along this type of problem.

Acetic acid is a weak acid, as others have mentioned. Acetic acid does not fully ionize in water. $\mathrm{p}K_\mathrm{a}$ becomes necessary here.

The $\mathrm{p}K_\mathrm{a}$ of acetic acid is $4.76$. You should have been permitted to locate this value in a table or otherwise been provided it on the test form, unless you were required to memorize acid dissociation constants, in which case I apologize for what must have been a course destined to forever char (oxidation!) your memory and appreciation for the magical world of chemistry.

From the definition of $\mathrm{p}K_\mathrm{a}$ we see that $$\mathrm{p}K_\mathrm{a} = -\lg K_\mathrm{a}$$ which implies that the $K_\mathrm{a}$ of acetic acid is $10^{-4.76}$, which is $1.74 \times 10^{-5}$.

Now, consider that,

$$K_\mathrm{a} = \frac{[\ce{H^+}][\ce{A^-}]}{[\ce{HA}]}$$

We are asked for $\mathrm{pH}$ in the problem, and we know that: $$\mathrm{pH} = -\lg [\ce{H^+}]$$

So let's work on isolating $[\ce{H^+}]$ so that we can get the $\mathrm{pH}$ for our answer.

Let's consider it a given that our acetic acid is in pure water which means, when it reaches equilibrium, it does so to equimolar portions of protons and acetate anions, viz.,

$$\ce{HA<<=>H^+ + A^-}$$ where the $K_\mathrm{a}$ of this equilibrium we determined above to be $1.74 \times 10^{-5}$.

As we mentioned above, $$K_\mathrm{a} = \frac{[\ce{H^+}][\ce{A^-}]}{[\ce{HA}]}$$

We talked about the very reasonable assumption that $[\ce{H^+}]$ and $[\ce{A^-}]$ are in equimolar proportions on dissociation of $[\ce{HA}]$. If we let $$x = [\ce{H^+}] = [\ce{A^-}]$$ then it follows that:

$$K_\mathrm{a} = \frac{x^2}{[\ce{HA}]}$$

Since the $K_\mathrm{a}$ of acetic acid is $1.74 \times 10^{-5}$, what are we left with? What do we know? That $5~\%$ is coming into play now I think. We know that per cent is "per 100," but "per 100" of what? In chemistry, in the absence of specific details otherwise, my assumption is molarity (moles/litre). Molality is used at times (mass fraction) as may be volume and weight percentages, mole fractions, etc. Given just the "$5~\%$" statement in the problem, I would take that as molarity of the acid ($[\ce{HA}]$).

Molarity is concentration! So the equations above, particularly the one that equates $K_\mathrm{a}$ to concentrations of products/reactants should work!

If $1.74 \times 10^{-5} = K_\mathrm{a}$ and $0.05 = [\ce{HA}]$, then it follows that

$$x = [\ce{H^+}] = \sqrt{1.74 \times 10^{-5} \times 0.05}$$

On my calculator, I get $x = 0.000933 = [\ce{H^+}]$.

So, the $\mathrm{pH} = -\lg([\ce{H^+}]) = -\lg(0.000933) = 3.03$

Looks pretty close to "B" to me!

• Percentage concentrations are actually well defined as $\% = 100 \times m(\text{solute}) /m(\text{solution})$. So a $5~\%$ solution of acetic acid (we assume: in water) means that $\pu{5g}$ of acetic acid are dissolved in $\pu{95g}$ of water. This is not molarity. However, please take an upvote for the rest of the answer.
– Jan
Sep 9, 2017 at 8:33
• +1 Yeah while the use of % to indicate a percentage of 1M is innovative (read unintuitive) the fact that the answer comes out so close to the multiple choice answer that is the only possible option is kind of scary. It is possible that the person preparing the question made this same assumption. Feb 27, 2018 at 16:43
• The percent could either be percent by mass, by volume, or by moles. Definitely not a quantity with dimensions such as molarity. Dec 31, 2021 at 3:01