Calculate the percent ionization of boric acid (H3BO3) in solutions of each of the following concentrations (Ka = 5.8e-10.) (
a) 0.269 M _____%
(b) 0.324 M _____%
(c) 0.737 M _____%
a)
H3BO3 dissociates as:
H3BO3
-----> H+ + H2BO3-
0.269
0 0
0.269-x
x x
Ka = [H+][H2BO3-]/[H3BO3]
Ka = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes
Ka = x*x/(c)
so, x = sqrt (Ka*c)
x = sqrt ((5.8*10^-10)*0.269) = 1.249*10^-5
since c is much greater than x, our assumption is correct
so, x = 1.249*10^-5 M
% dissociation = (x*100)/c
= 1.249*10^-5*100/0.269
= 4.643*10^-3 %
Answer: 4.64*10^-3 %
b)
H3BO3 dissociates as:
H3BO3
-----> H+ + H2BO3-
0.324
0 0
0.324-x
x x
Ka = [H+][H2BO3-]/[H3BO3]
Ka = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes
Ka = x*x/(c)
so, x = sqrt (Ka*c)
x = sqrt ((5.8*10^-10)*0.324) = 1.371*10^-5
since c is much greater than x, our assumption is correct
so, x = 1.371*10^-5 M
% dissociation = (x*100)/c
= 1.371*10^-5*100/0.324
= 4.231*10^-3 %
Answer: 4.23*10^-3 %
c)
H3BO3 dissociates as:
H3BO3
-----> H+ + H2BO3-
0.737
0 0
0.737-x
x x
Ka = [H+][H2BO3-]/[H3BO3]
Ka = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes
Ka = x*x/(c)
so, x = sqrt (Ka*c)
x = sqrt ((5.8*10^-10)*0.737) = 2.068*10^-5
since c is much greater than x, our assumption is correct
so, x = 2.068*10^-5 M
% dissociation = (x*100)/c
= 2.068*10^-5*100/0.737
= 2.805*10^-3 %
Answer: 2.81*10^-3 %
Calculate the percent ionization of boric acid (H3BO3) in solutions of each of the following concentrations...
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