Question

Calculate the percent ionization of a 0.484 M solution of benzoic acid, C₆H₅COOH.

% Ionization = ___%

Answer 1

Ka of benzoic acid = 5.3 x 10^-5

C6H5COOH <=> C6H5COO- + H+
start
0.484                          0                      0
change
-x .. . . . . . . .. . .. . . +x. . . . ..+x
at equilibrium
0.210-x. . . . . . .. . . .x .. . . . . . x

Ka = [C6H5COO-][H+]/ [C6H5COOH]= (x^2/ 0.484-x           5.3 x 10^-5 =x^2/0.484           (if x is small 0.484-x= 0.484)

x = 0.0056 M

% ionization = 0.0056 x 100/ 0.484=1.04 %