Question

2. Use the literature value of K, for acetic acid to calculate the percent ionization of a 0.40M solution.

Answer 1

Answer:

The percent ionization is 0,66%.

Explanation:

K for acetic acid = 1,74x10*-5

CH3COOH (ac) + H2O(l) ⇄  H3O+ (ac)   +  CH3COO- (ac)

initial.  0,40 M                           ----                  ------

react.      X                                 X                       X

eq.     (0,4 -X)                             X                       X

This is the expression for K:

Ka = ( [H3O+] . [CH3COO-] )  / [CH3COOH]

1,74x10*-5 = (X . X) / 0,40 - X

Since the X in the numerator is going to be a very small number, since my Ka is in the order of -5, I can despise it in order to avoid the quadratic equation. You can't always do that!.. Be careful and pay attention to the value of Ka.

$\sqrt{1,74x10*-5x0,40}$ = X

X = 2,64x10*-3

% ionization is  (X /acid concentration) x 100

(2,64x10*-3 / 0,4) x 100 = 0,66%