Question

The pH of a white vinegar solution is 2.11. This vinegar is an aqueous solution of acetic acid with a density of 1.09 g/mL. What is the mass percentage of acetic acid in the solution? Ka (acetic acid)= 1.8 X10-5

Mass percentage =___________ %

Answer 1

Lets find the concentration of acetic acid:

pH = 2.11

use:

pH = -log [H+]

2.11 = -log [H+]

[H+] = 7.76*10^-3 M

CH3COOH   <—> CH3COO- + H+

C           0       0   (initial)

C-x           x       x   (at equilibrium)

here,

x = [H+] = 7.76*10^-3 M

use:

Ka = x*x/(C-x)

1.8*10^-5 =(7.76*10^-3)*(7.76*10^-3)/(C-7.76*10^-3)

(C-7.76*10^-3) = 3.345

C = 3.353 M

Now take the volume of Vinegar solution to be 1 L

So, mol of CH3COOH = concentration * volume

= 3.353 M * 1L

= 3.353 mol

Molar mass of CH3COOH,

MM = 2*MM(C) + 4*MM(H) + 2*MM(O)

= 2*12 + 4*1 + 2*16

= 60 g/mol

mass of C2H4O2,

m = number of mol * molar mass

= 3.353*60

= 201.2 g

mass of solution = density * volume

= 1.09 g/mL * 1000 mL

= 1090 g

mass % of CH3COOH = mass of CH3COOH * 100 / total mass of solution

= 201.2 * 100 / 1090

= 18.5 %

Answer: 18.5 %