(a) Calculate the percent ionization of 0.00850 M butanoic acid
(Ka = 1.5e-05).
% ionization = %
(b) Calculate the percent ionization of 0.00850 M butanoic acid in
a solution containing 0.0130 M sodium butanoate.
% ionization = %
(a)
Ka = 1.5 x 10^-5
HA (aq) --------------> H+ (aq) + A- (aq)
0.00850 0 0
0.00850-x x x
Ka = x^2 / 0.00850 - x
1.5 x 10^-5 = x^2 / 0.00850 - x
x = 3.57 x 10^-4 M
% ionization = (x / C) x 100
= (3.57 x 10^-4 / 0.00850) x 100
% ionization = 4.20 %
% ionization = %
(b)
C4H8O2 + H2O ------------> C4H8O2- + H3O+
0.0085 x 0.0130
0.0085- x x 0.0130 + x
1.5 x 10^-5 = x (0.0130 + x) / (0.0085 - x)
x = 9.807 x 10^-6
% ionization = (9.807 x 10^-6 / 0.0085) x 100
= 0.115 %
(a) Calculate the percent ionization of 0.00850 M butanoic acid (Ka = 1.5e-05). % ionization = %...
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