Calculate the percent ionization of hypobromous acid (HBrO) in solutions of each of the following concentrations (Ka = 2.5e-09.)
(a) 0.271 M %
(b) 0.388 M %
(c) 0.794 M %
a)
HBrO dissociates as:
HBrO -----> H+ + BrO-
0.271 0 0
0.271-x x x
Ka = [H+][BrO-]/[HBrO]
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 ((2.5*10^-9)*0.271) = 2.603*10^-5
since c is much greater than x, our assumption is correct
so, x = 2.603*10^-5 M
% dissociation = (x*100)/c
= 2.603*10^-5*100/0.271
= 9.605*10^-3 %
Answer: 9.60*10^-3 %
b)
HBrO dissociates as:
HBrO -----> H+ + BrO-
0.388 0 0
0.388-x x x
Ka = [H+][BrO-]/[HBrO]
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 ((2.5*10^-9)*0.388) = 3.114*10^-5
since c is much greater than x, our assumption is correct
so, x = 3.114*10^-5 M
% dissociation = (x*100)/c
= 3.114*10^-5*100/0.388
= 8.027*10^-3 %
Answer: 8.03*10^-3 %
c)
HBrO dissociates as:
HBrO -----> H+ + BrO-
0.794 0 0
0.794-x x x
Ka = [H+][BrO-]/[HBrO]
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 ((2.5*10^-9)*0.794) = 4.455*10^-5
since c is much greater than x, our assumption is correct
so, x = 4.455*10^-5 M
% dissociation = (x*100)/c
= 4.455*10^-5*100/0.794
= 5.611*10^-3 %
Answer: 5.61*10^-3 %
Calculate the percent ionization of hypobromous acid (HBrO) in solutions of each of the following concentrations...
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