# Calculating the concentration of acetic acid in vinegar from pH

DavePhD was kindly helping me with this problem but the site indicated the discussion had gone on too long. Calculating vinegar strength I don't have enough units to start a chat so I am starting a new thread.

The formula above is for determining the concentration or percentage of acetic acid in a vinegar sample when the ph is known. The farmer will get the ph with a ph meter, do the math and calculate the percentage 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 someone write out the "long math" for this? I need someone to walk me through this, step by step to solve for c.

The last time I took algebra was about 50 years ago so, while I remember some things, it is probably better to assume I don't know much :-). Let me know if you need other numbers. Dave says the "M" is just a place holder

• Could you come up with a better title for this please? No one is going to find this during a search. – jonsca Feb 17 '15 at 1:40

First off $$\ce{HC_2H_3O_2}\leftrightarrow \ce{H^+} + {\ce{C_2H_3O_2}}^-\\and\\ K_a = \frac{[\ce{H^+}][{\ce{C_2H_3O_2}}^-]}{[\ce{HC_2H_3O_2}]}$$ So let's do some simplifying. The amount of acetic acid left is the orignial molarity minus the amount that's dissociated. The amount of Hydrogen ion and conjugate base left is equal to the amount of that's been dissociated from the original amount of acid. So let's let HA equal the molarity of the acid and x equal to the amount dissociated. Soooooo $$K_a = \frac{[x][x]}{[HA-x]} = \frac{x^2}{HA-x}\\\text{Now let's solve}\\\rightarrow K_a(HA-x) = x^2\\\rightarrow HA -x= \frac{x^2}{K_a}\\\text{ switching the x interms of ph}\\ HA -10^{-ph} = \frac{(10^{-ph})^2}{K_a}\\\text{ plugging in}\\HA- 10^{-3} = \frac{(10^{-3})^2}{.0000174} = .0575$$ This is the molarity of the acid at equilibrium. Add 10^{-ph} for the concentration of the acid before dissociation.