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?




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.


 18

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:


$$\mathrm{pH} = 0.5 (\mathrm{p}K\_\mathrm{a} - \log C\_\mathrm{a}) = 0.5 (4.76 - \log 0.874) = 2.4$$


 12

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. 


 4

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!


 2

“EXPECT” is the key word here. It has nothing to do with an equation. The PH values are quite funny 0? 7? 9? 12? Really??? I haven’t been in a chemistry class in almost forty years…. For vinegar to be shelf stable it needs an acetic acid concentration around 5% and a Ph under 4 (mine have run anywhere from 2.4 to 3.6). I’m sure the equations listed here work for acetic acid solution of 5%. However vinegar’s PH is not created by its acetic acid percentage (exclusively I should say). Type of fruit, sugars when harvested, which yeasts were used etc etc


 0