12.5 mL sample of vinegar, containing acid, HC2H3O2, was titrated using 0.504 M NaOH solution.?

Question

The titration required 20.65 mL of base

a) What was the molar concentrationof acetic acid in vinegar?

b.) Assuming the density of vinegar to be 1.0g/mL, what was the precent by mass of acetic acid in vinegar?

Answer 1

The most useful equation is M1V1=M2V2 using this you can solve for whatever the unknown is. I also believe you need to convert the ml to L(1000mL in 1L). so .0125L(M1)=.504M(.02065L) so using simple algebra the answer is 0.833M.

For part 2 I will only give you a hint use the known that
density = mass/volume and since we can figure out the total volume of the solution it should be simple.

Answer 2

The molar concentration of acetic acid in vinegar is 0.833M

Answer 3

a) 0.504x20.65/12.5 = 0.833 M

b) Mass of CH3COOH = 0.833 x 60 = 49.98 g

Which is 4.998 %

Answer 4

Look in your textbook, stop being lazy.