91Ó°ÊÓ

First, solve for the concentration of the acetic acid in 5 %acetic acid solution (w / v). Let us assume that there is 100 mLof solution so that the mass of our acetic acid would be the 5%of 100, which is 5 g. Also, note that the density of the water is 1 g/mL.

$\begin{array}{rcl}& & \text{M}\text{=}\frac{\text{mol}}{\text{L}}\\ & & \\ & & \text{=}\text{5g}×\frac{1\mathrm{mol}}{60.1\mathrm{g}}×\frac{1}{0.100\mathrm{L}}\\ & =& \text{0.833 M acetic acid}\end{array}$

Using the${\text{K}}_{\text{a}}$of the acetic acid from Appendix C, solve for the$\left[{\text{H}}_3{\text{O}}^{+}\right]$. Construct the ICE table first.

Therefore, the${\text{K}}_{\text{a}}$is:

$\begin{array}{rcl}& & {\text{K}}_{\text{a}}\text{=}\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\left[{\text{CHI}}_{\text{3}}{\text{COO}}^{\text{-}}\right]}{\left|{\text{CH}}_{\text{3}}\text{COOH}\right|}\\ & & \text{=}\frac{{\text{x}}^{\text{2}}}{\text{0.833 - x}}.\end{array}$

We know that$\text{x}\text{=}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}\left[{\text{11O}}_{\text{2}}^{\text{-}}\right]$ since the reaction produced one mole of each. Since the${\text{K}}_{\text{a}}$is small compared to the initial concentration, one can drop the x in the denominator, then solve for x.

$\begin{array}{rcl}& & {\text{K}}_{\text{a}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0.833}}\\ & & \text{x =}\sqrt{\frac{\text{1.8}×{\text{10}}^{\text{- 5}}}{\text{0.833}}}\\ & & \text{=}\text{3.87}×{\text{10}}^{\text{- 3}}\text{M.}\end{array}$

Lastly, solve for the pH. Note that$\text{x}\text{=} \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}\left[{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}\right].$

$\begin{array}{rcl}& & \text{pH =}\text{- log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ & & \text{=}\text{- log}\left(\text{3.87}×{\text{10}}^{\text{- 3}}\text{M}\right)\\ & & \text{= 2.41.}\end{array}$

Hence, the pH of the vinegar is 2.41.