In addition to the acid-dissociation constant, K a, another measure of the stren
ID: 899126 • Letter: I
Question
In addition to the acid-dissociation constant, Ka, another measure of the strength of an acid is percent dissociation, determined by the following formula:
Percent dissociation=[HA] dissociated[HA] initial×100%
Percent dissociation increases with increasing Ka. Strong acids, for which Ka is very large, dissociate completely(100%). For weak acids, the percent dissociation changes with concentration. The more diluted the acid is, the greater percent dissociation.
A certain weak acid, HA, has a Ka value of 4.7×107.
Part A
Calculate the percent dissociation of HA in a 0.10 M solution.
Part B
Calculate the percent dissociation of HA in a 0.010 M solution.
Express your answer as a percent using two significant figures.
Explanation / Answer
Part A :
Let a be the dissociation of the weak acid
HA <---> H + + A-
initial conc. c 0 0
Equb. conc. c(1-a) ca ca
Dissociation constant , Ka = ca x ca / ( c(1-a)
= c a2 / (1-a)
In the case of weak acids is very small so 1-a is taken as 1
So Ka = ca2
==> a = ( Ka / c )
Given Ka = 4.7x10-7
c = concentration = 0.10 M
Plug the values we get a = ( Ka / c )
= ((4.7x10-7)/ 0.10)
= 2.17 x10-3
% dissociation = 2.17x10-3 x 100 = 0.216
Part B :
Let a be the dissociation of the weak acid
HA <---> H + + A-
initial conc. c 0 0
Equb. conc. c(1-a) ca ca
Dissociation constant , Ka = ca x ca / ( c(1-a)
= c a2 / (1-a)
In the case of weak acids is very small so 1-a is taken as 1
So Ka = ca2
==> a = ( Ka / c )
Given Ka = 4.7x10-7
c = concentration = 0.010 M
Plug the values we get a = ( Ka / c )
= ((4.7x10-7)/ 0.010)
= 6.85 x10-3
% dissociation = 6.85x10-3 x 100 = 0.68
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