Question

a. Using the initial concentration of CH,COOH (from Problem 3) and the equilibrium concentration of H,0 calculated (from Problem 4), complete the reaction table for vinegar. Ignore the column under H:O() (shaded in gray). Don't forget to write a sign (+ or -) for the "Change” row entry. (9 pts: each entry is 1 pt.) Reaction Table CH.COOH (aq) + H20(0) CH,COO (aq) + H2O (ag) Initial Change Equilibrium b. By using the concentration values at equilibrium in the reaction table above, calculate the acidity constant (K) of acetic acid, Because the acidity constant is unitless, do not use the units in this calculation. Show your work. (10 pts)

Answer 1

Solution:

A)The dissociation of acetic acid (CH3COOH) is written as,

CH3COOH + H2O   = CH3COO- + H3O+

0.833 M ------------------------- 0 M -------------- 0 M (Initial)

(-3.2 x 10^-3 M) -- (+3.2 x 10^-3 M)--(+3.2 x 10^-3 M) (Change)

(0.833 - 3.2 x10^-3)-- (3.2 x 10^-3) --- (3.2 x10^-3) (Equilibrium)

B) The dissociation constant Ka is calculated as,

Ka = [CH3COO-] [H3O+] / [CH3COOH]

Ka = (3.2 x 10^-3) (3.2 x10^-3) / (0.833 - 3.2 x 10^-3)

Since, acetic acid is a weak acid, hence 3.2 x10^-3 can be neglected from denominator.

Thus,

Ka = (3.2 x 10^-3) (3.2 x10^-3) / 0.833

Ka = 1.23 x 10^-5