Q165CP - Chemistry: Molecular Nature Of Matter And Change

Question

What is the pH of a vinegar with  5.0 % (w/v) acetic acid in water?

Step-by-Step Solution

Answer

The pH of the vinegar is 2.41.

Step 1: To the pH of a vinegar

First, solve for the concentration of the acetic acid in 5 % acetic acid solution (w / v). Let us assume that there is 100 mL of 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 .

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

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.

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

Step 2: Solve the equation

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

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

Hence, the pH of the vinegar is 2.41.