Answer
- The acetic acid concentration in the vinegar is 0.08931M.
Work Step by Step
1. Find the $[H_3O^+]$ value:
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 2.9}$
$[H_3O^+] = 1.259 \times 10^{- 3}$
2. Drawing the equilibrium (ICE) table we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[H_3O^+] = [Conj.Base] = x$
-$[Acetic acid] = [Acetic acid]_{initial} - x$
3. Now, use the Ka and x values and equation to find the initial concentration value.
$Ka = \frac{[H_3O^+][Conj.Base]}{ [Initial Acetic acid] - x}$
$ 1.8\times 10^{- 5}= \frac{[x^2]}{ [Initial Acetic acid] - x}$
$ 1.8\times 10^{- 5}= \frac{( 1.259\times 10^{- 3})^2}{[Initial Acetic acid] - 1.259\times 10^{- 3}}$
$[Initial Acetic acid] - 1.259\times 10^{- 3} = \frac{ 1.585\times 10^{- 6}}{ 1.8\times 10^{- 5}}$
$[Initial Acetic acid] - 1.259\times 10^{- 3} = 0.08805$
$[Initial Acetic acid] = 0.08805 + 1.259\times 10^{- 3}$
$[Initial Acetic acid] = 0.08931M$
