Afood 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?

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×= 1,537g of CH₃COOH

As density is 1,01g/mL:

1,537g CH₃COOH×= 1,522mL of CH₃COOH

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

×100= 7,324 (%V/V)

I hope it helps!

a  solution  is always transparent, light passes through with no scattering from solute particles which are molecule in size. the  solution  is homogeneous and does not settle out. a colloid is intermediate between asolution  and a  suspension. while asuspension  will separate out a colloid will not.

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