Question: Over time, as their free fatty acid (FFA) content increases, edible fats and oils become rancid. To measure rancidity, the fat or oil is dissolved in ethanol, and any FFA present is titrated with KOH dissolved in ethanol. In a series of tests on olive oil, a stock solution of 0.050 M ethanolic KOH was prepared at 25⁰C , stored at 0^oC, and then placed in a 100-mL buret to titrate any oleic acid $\left[{\mathbf{CH}}_{\mathbf{3}}{\left({\mathbf{CH}}_{\mathbf{2}}\right)}_{\mathbf{7}}\mathbf{CH}\mathbf{=}\mathbf{CH}{\left({\mathbf{CH}}_{\mathbf{2}}\right)}_{\mathbf{7}}\mathbf{COOH}\right]$present in the oil. Each of four 10.00-g samples of oil took several minutes to titrate: the first required 19.60 mL, the second 19.80 mL, and the third and fourth 20.00 mL of the ethanolic KOH.

(a) What is the apparent acidity of each sample, in terms of mass % of oleic acid? (Note: As the ethanolic KOH warms in the buret, its volume increases by a factor of .)

(b) Is the variation in acidity a random or systematic error? Explain.

(c) What is the actual acidity? How would you demonstrate this?

The reaction is,

$C_{17}{H}_{33}COOH+KOH\to C_{17}{H}_{33}COOK+H_{2}O$

For first sample,

Number of moles of KOH are,

$$\begin{gathered} Moles=volume\times molarity \\ =0.0196L\times 0.050M \\ =9.8\times {10}^{-4}mol \end{gathered}$$

Mass of oleic acid is,

$$\begin{gathered} Mass=9.8\times {10}^{-4}mol\times \frac{1mol C_{17}{H}_{33}COOH}{1mol KOH}\times \frac{282.47g C_{17}{H}_{33}COOH }{1mol C_{17}{H}_{33}COOH} \\ =2768.21\times {10}^{-4}g \end{gathered}$$

Now, mass percent of oleic acid is,

$$\begin{gathered} Mass\%=\frac{mass of C_{17}{H}_{33}COOH}{total mass}\times 100 \\ =\frac{2768.21\times {10}^{-4}g}{10.0g}\times 100 \\ =2.8\% \end{gathered}$$

For second sample,

Number of moles of KOH are,

$$\begin{gathered} Moles=volume\times molarity \\ =0.0198L\times 0.050M \\ =9.9\times {10}^{-4}mol \end{gathered}$$

Mass of oleic acid is,

$$\begin{gathered} Mass=9.9\times {10}^{-4}mol\times \frac{1mol C_{17}{H}_{33}COOH}{1mol KOH}\times \frac{282.47g C_{17}{H}_{33}COOH }{1mol C_{17}{H}_{33}COOH} \\ =2796.45\times {10}^{-4}g \end{gathered}$$

Now, mass percent of oleic acid is,

$$\begin{gathered} Mass\%=\frac{mass of C_{17}{H}_{33}COOH}{total mass}\times 100 \\ =\frac{2796.45\times {10}^{-4}g}{10.0g}\times 100 \\ =2.8\% \end{gathered}$$

For third sample,

Number of moles of KOH are,

$$\begin{gathered} Moles=volume\times molarity \\ =0.0200L\times 0.050M \\ =10.0\times {10}^{-4}mol \end{gathered}$$

Mass of oleic acid is,

$$\begin{gathered} Mass=10.0\times {10}^{-4}mol\times \frac{1mol C_{17}{H}_{33}COOH}{1mol KOH}\times \frac{282.47g C_{17}{H}_{33}COOH }{1mol C_{17}{H}_{33}COOH} \\ =2824.7\times {10}^{-4}g \end{gathered}$$

Now, mass percent of oleic acid is,

$$\begin{gathered} Mass\%=\frac{mass of C_{17}{H}_{33}COOH}{total mass}\times 100 \\ =\frac{2824.7\times {10}^{-4}g}{10.0g}\times 100 \\ =2.8\% \end{gathered}$$

The variation in the acidity can be considered as systematic error. It is given in the problem that KOH is warmed in the burette and its volume increased.

Thus, the acidity is increased.

The actual acidity in third and fourth sample is 2.82%.

By performing the titration multiple times with ethanol, the actual acidity can be easily demonstrated.