Determine the percent ionization of a 0.240 m solution of benzoic acid
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Ka of benzoic acid = 5.3 x 10^-5
C6H5COOH <=> C6H5COO- + H+
start
0.210
change
-x .. . . . . . . .. . .. . . +x. . . . ..+x
at equilibrium
0.210-x. . . . . . .. . . .x .. . . . . . x
Ka = [C6H5COO-][H+]/ [C6H5COOH]= (x^2/ 0.210-x
x = 0.00364
% ionization = 0.00364 x 100/ 0.210=1.73
C6H5COOH <=> C6H5COO- + H+
start
0.210
change
-x .. . . . . . . .. . .. . . +x. . . . ..+x
at equilibrium
0.210-x. . . . . . .. . . .x .. . . . . . x
Ka = [C6H5COO-][H+]/ [C6H5COOH]= (x^2/ 0.210-x
x = 0.00364
% ionization = 0.00364 x 100/ 0.210=1.73
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