A bottle of vinegar is found to have a pH of 2.90. Calculate the molar concentration of acetic acid in this vinegar bottle (assume that acetic acid is the only acid present in this aqueous solution). The pKaof acetic acid is 4.76.

Answer 1

so you know the pK and the pH, looking for Molarity M.
The amount of hydrogen ion (x) is the original Molarity M minus the amount dissociate.
Ka=x^2/(M-x)
or Ka(HA-x)=x^2 or HA-x=x^2/Ka
now, lets swithc to logs in the x term..
HA-10^^-pH = (10^^-pH)^^2< /K
now with some algebra, solve for HA. You know pK=4.76, so you then know K= 10^(-4.76)== 1.73780083e-5
HA= (10^^-pH)^^2< /K + 10^^-2pH
That is the equalibirum cocentration
original cocentration M then Add 10^{-ph} for the concentration of the acid before dissociation.

Answer 2

check my math, it is a pain to type in HTML to get superscripts, then proof it.