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.
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.
check my math, it is a pain to type in HTML to get superscripts, then proof it.
To calculate the molar concentration of acetic acid, we need to use the equation relating pH and pKa:
pH = pKa + log([A-]/[HA])
Where:
pH is the given pH value (2.90)
pKa is the given pKa value (4.76)
[A-] is the concentration of the conjugate base (acetate ion, C2H3O2-) of acetic acid
[HA] is the concentration of acetic acid (C2H4O2)
First, let's rearrange the equation:
pH - pKa = log([A-]/[HA])
Now, we can substitute the given values into the equation:
2.90 - 4.76 = log([A-]/[HA])
-1.86 = log([A-]/[HA])
Next, we need to convert the logarithmic equation into an exponential equation:
10^(-1.86) = [A-]/[HA]
Now, we can calculate the value on the left side of the equation:
10^(-1.86) = 0.01584893192
Next, let's assume that the concentration of acetic acid ([HA]) is equal to x. Since acetic acid dissociates into one hydrogen ion (H+) and one acetate ion (C2H3O2-), the concentration of acetate ion ([A-]) is also x.
So, [A-] = x and [HA] = x.
Substituting these values into the equation:
0.01584893192 = x/x
Simplifying the equation, we get:
0.01584893192 = 1
This equation indicates that the concentrations of acetate ion and acetic acid are equal, both being 0.01584893192 M.
Therefore, the molar concentration of acetic acid in the vinegar bottle is approximately 0.0158 M.