Question

If acetic acid is the only acid that vinegar contains (ka=1.8×10−5), calculate the concentration of acetic acid in the vinegar. express your answer using two significant figures.

Answer 1

Ethanoic (Acetic) acid is a weak acid and do not dissociate fully. Therefore its equilibrium state has to be considered here.

$CH_{3}COOH \ \textless \ ---\ \textgreater \ H^{+} + CH_{3}COO^{-}$

In this case pH value of the solution is necessary to calculate the concentration but it's not given here so pH = 2.88 (looked it up)

pH = 2.88 ==> $[H^{+}]$  = $10^{-2.88}$ =  0.001 $moldm^{-3}$

The change in Concentration Δ $[CH_{3}COOH]$= 0.001 $moldm^{-3}$


                                  CH3COOH          H+           CH3COOH    
Initial  $moldm^{-3}$                      $x$           0                     0
                                                                                                                       
Change $moldm^{-3}$        -0.001            +0.001           +0.001
                                                                                                       
Equilibrium $moldm^{-3}$      $x$- 0.001      0.001             0.001
                                                                              

Since the $k_{a}$ value is so small, the assumption 
$[CH_{3}COOH]_{initial} = [CH_{3}COOH]_{equilibrium}$ can be made.

$k_{a} = [tex]= 1.8*10^{-5} = \frac{[H^{+}][CH_{3}COO^{-}]}{[CH_{3}COOH]} = \frac{0.001^{2}}{x}$

Solve for x to get the required concentration.

note: 1.)Since you need the answer in 2SF don&t round up values in the middle of the calculation like I've done here.

         2.) The ICE (Initial, Change, Equilibrium) table may come in handy if you are new to problems of this kind

Hope this helps!