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The pKa of vinegar is 4.76. If I know the pH of the vinegar, what would be the formula for determining the percentage acid base on the relationship between pH, pKa and concentration? Can you show the formula so that I only need to plug in the ph?

I am not a chemistry student and I have not done these kinds of calculations. I am searching for a convenient way for farmers in developing countries to quickly approximate the strength of vinegar. The final test would be done in a lab but I need a way for them to test vinegar other than titration because titration can seem more like witchcraft than science to the uninitiated.

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  • $\begingroup$ What is the definition of pKa? Ka is the equilibrium constant of which reaction? Try to write something. It is not difficult. $\endgroup$
    – Yomen Atassi
    Commented Feb 16, 2015 at 20:00
  • $\begingroup$ I understand that, "pKa = -log10Ka The lower the pKa value, the stronger the acid." I am not a chemistry student and I have not done these kinds of calculations. I am searching for a convenient way for farmers in developing countries to quickly approximate the strength of vinegar. The final test would be done in a lab but I need a way for them to test vinegar other than titration because titration can seem more like witchcraft than science to the uninitiated :-). $\endgroup$
    – VinegrMan
    Commented Feb 16, 2015 at 20:15

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If $pK_a = 4.76$, then $K_a = 1.74 \times 10^{-5}$

Then approximating that $K_a$ can be written in terms of concentration rather than the true definition which is in terms of activities:

$K_a = 1.74 \times 10^{-5}M = \frac{[A-][H+]}{[HA]} = \frac{[H+]^2}{c-[H+]} $

$1.74 \times 10^{-5}M = \frac{[H+]^2}{c-[H+]} $

where $c$ is the molar concentration of the vinegar

If you want pH explicitly in the equation, then approximating pH = -log[H+], then:

$1.74 \times 10^{-5}M = \frac{(10^{-pH}M)^2}{c-(10^{-pH}M)} $

So, for example if pH is 3, $10^{-3}$ is 0.001, so

$1.74 \times 10^{-5} = \frac{10^{-6}M}{c-(10^{-3}M)}$

$1.74 \times 10^{-5}c -1.74 \times 10^{-8}M = 10^{-6}M$

$1.74 \times 10^{-5}c = 1.02 \times 10^{-6}M$

$c = 0.059M$

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  • $\begingroup$ Thanks. What does the H+, the HA and the M stand for? If the ph is 3, what would the math look like? $\endgroup$
    – VinegrMan
    Commented Feb 16, 2015 at 20:42
  • $\begingroup$ pKa = -log(Ka), the Ka value calculated is not correct $\endgroup$
    – Jeanno
    Commented Feb 16, 2015 at 20:52
  • $\begingroup$ chemistry.about.com/od/chemistryglossary/g/Pka-Definition.htm What is it? $\endgroup$
    – VinegrMan
    Commented Feb 16, 2015 at 21:05
  • $\begingroup$ Hope this isn't double posting, but I am not understanding some of the symbols and I am not familiar with the conventions. What do the M, H+ and HA stand for? I am trying to come up with a formula where the only variable is the ph which the farmer would get with a ph meter, do the math and calculate the acidity of her vinegar. The Ka, according to the formula DavePhD gave, is .000174 because the pKa for vinegar is 4.76. The molar mass is 60.05196 g/mol. So assuming the ph is 3, could some write out the "long math" for this? Are other numbers needed to solve this? $\endgroup$
    – VinegrMan
    Commented Feb 16, 2015 at 22:25
  • $\begingroup$ I want to solve for concentration or percentage of acetic/ethanoic acid in the solution. $\endgroup$
    – VinegrMan
    Commented Feb 16, 2015 at 22:26
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As we know: $$\rm pH=\frac12(pKa-\log C)\quad\mid\quad C=\frac{\rm weight\;of\;acid}{\rm Mol\;wt.*Volume of sln.}\quad\mid\quad Strength=C*Mol.\;wt.$$ Calculating: $$\rm S=10^{pKa-2pH}*M$$ So finally: $$\huge\fbox{$\rm S=10^{4.75^{\#}-2pH}*60.05$}$$ #You can as well take that 4.76

Theoretically It is correct but some deviations are bound to occur in reality. I suggest finding out that in few samples first.

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