What is the percent dissociation of HX (pKa = 4.49) in a 0.335 M solution?
we have below equation to be used:
pKa = -log Ka
4.49 = -log Ka
log Ka = -4.49
Ka = 10^(-4.49)
Ka = 3.236*10^-5
Lets write the dissociation equation of HA
HA -----> H+ + A-
0.335 0 0
0.335-x x x
Ka = [H+][A-]/[HA]
Ka = x*x/(c-x)
Assuming small x approximation, that is lets assume that x can be ignored as compared to c
So, above expression becomes
Ka = x*x/(c)
so, x = sqrt (Ka*c)
x = sqrt ((3.236*10^-5)*0.335) = 3.292*10^-3
since c is much greater than x, our assumption is correct
so, x = 3.292*10^-3 M
% dissociation = (x*100)/c
= 3.292*10^-3*100/0.335
= 0.983 %
Answer: 0.983 %
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