Calculate the percent ionization of phenol (HC6H5O) in solutions of each of the following concentrations (Ka = 1.3e-10.) (a) 0.221 M % (b) 0.476 M % (c) 0.733 M %
a)
HC6H5O dissociates as:
HC6H5O
-----> H+ + C6H5O-
0.221
0 0
0.221-x
x x
Ka = [H+][C6H5O-]/[HC6H5O]
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 ((1.3*10^-10)*0.221) = 5.36*10^-6
since c is much greater than x, our assumption is correct
so, x = 5.36*10^-6 M
% dissociation = (x*100)/c
= 5.36*10^-6*100/0.221
= 2.425*10^-3 %
Answer: 2.42*10^-3 %
b)
HC6H5O dissociates as:
HC6H5O
-----> H+ + C6H5O-
0.476
0 0
0.476-x
x x
Ka = [H+][C6H5O-]/[HC6H5O]
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 ((1.3*10^-10)*0.476) = 7.866*10^-6
since c is much greater than x, our assumption is correct
so, x = 7.866*10^-6 M
% dissociation = (x*100)/c
= 7.866*10^-6*100/0.476
= 1.653*10^-3 %
Answer: 1.65*10^-3 %
c)
HC6H5O dissociates as:
HC6H5O
-----> H+ + C6H5O-
0.733
0 0
0.733-x
x x
Ka = [H+][C6H5O-]/[HC6H5O]
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 ((1.3*10^-10)*0.733) = 9.762*10^-6
since c is much greater than x, our assumption is correct
so, x = 9.762*10^-6 M
% dissociation = (x*100)/c
= 9.762*10^-6*100/0.733
= 1.332*10^-3 %
Answer: 1.33*10^-3 %
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