Question

What is the pH of $1.5L$ of vinegar that is 3% acetic acid by mass ? $(K_a=1.8\times {10}^{-5})$

• A. 12
• B. 4.3
• C. 7
• D. 2.5

Answer 1

The correct option is D 2.5

Assume that the vinegar is really just a solution of acetic acid in water, and that density = $1gm{L}^{-1}$
So if the vinegar is 3% acetic acid by mass and the molar mass of $C{H}_{3}COOH=60.05gmo{l}^{-1}$, then
$1.5Lvinegar\times \frac{1000mL}{1L}\times \frac{1g}{1mL}\times \frac{3gaceticacid}{100gvinegar}\times \frac{1molaceticacid}{60.05gaceticacid}=0.75molC{H}_{3}COOH$
Dividing $0.75mol$ by $1.5L$ to get an initial concentration of 0.50 M,
considering the ionization of acetic acid in water into acetate ion and hydronium ion. ignoring the concentration of water (a pure liquid).
$C{H}_{3}COO{H}_{\left(aq\right)}+H_2{O}_{\left(aq\right)}\rightleftharpoons C{H}_{3}CO{O}^{-}{2\left(aq\right)}^{-}+H_3{O}_{\left(aq\right)}^{+}$
$$\begin{array}{ccccc}& C{H}_{3}COOH& H_{2}O& C{H}_{3}CO{O}^{-}& H_3{O}^{+}\\ Initial& 0.5& ...& 0& 0\\ Change& -x& ...& +x& +x\\ Equilibrium& 0.5-x& ...& x& x\end{array}$$
the equilibrium constant expression
$K_a=\frac{\left[C{H}_{3}CO{O}^{-}\right]\left[H_3{O}^{+}\right]}{\left[C{H}_{3}COOH\right]}$
$1.8\times {10}^{-5}=\frac{\left[x\right]\left[x\right]}{[0.5-x]}$
here x can be neglected because it is very small so, 0.5-x=0.5

$x=0.003M=\left[H_3{O}^{+}\right]$
put up into the formula
$pH=-\log \left[H_3{O}^{+}\right]pH=-\log \left(0.003\right)=2.5$