Question

A food chemist determines the concentration of acetic acid in a sample of apple vinegar by acid base titration. The density of the sample is 1.01 g/mL. The titrant is 1.024 M NaOH. The average volume of titrant required to titrate 25.00 mL subsamples of the vinegar is 20.78 mL. What is the concentration of acetic acid in the vinegar?

Answer 1

Answer:

The concentration of acetic acid in the vinegar is 7,324 (%V/V)

Explanation:

The titration equation of acetic acid with NaOH is:

NaOH + CH₃COOH → CH₃COO⁻Na⁺ + H₂O

The moles required were:

1,024M×0,02500L = 0,02560 moles NaOH. These moles are equivalent (By the titration equation) to moles of CH₃COOH. As molar mass of CH₃COOH is 60,052g/mol, the mass in these moles of CH₃COOH is:

0,02560 moles CH₃COOH×$\frac{60,052g}{1mol}$= 1,537g of CH₃COOH

As density is 1,01g/mL:

1,537g CH₃COOH×$\frac{1mL}{1,01g}$= 1,522mL of CH₃COOH

As volume of vinegar in the sample is 20,78mL, the concentration of acetic acid in the vinegar is:

$\frac{1,522mLCH_{3}COOH}{20,78mL}$×100= 7,324 (%V/V)

I hope it helps!