A certain weak acid, HA, has a Ka value of 9.1×10−7. Part A Calculate the percent dissociation of HA in a 0.10 M solution.
A weak acid will form the next equilibrium
HA <--> H+ and A-
This is descirbed by Keq
K = [H+][A-]/[HA]
K is given as 9.1*10^-7
HA initial = 0.10 M
BUT this is in equilibrium so:
[H+] = x (assume a variable)
[A-] = [H+] due to stoichiometry so [A-] = x
[HA] = (0.1-x) since we need to account the dissociated substances
Substitute all in KA expression
K = [H+][A-]/[HA]
9.1*10^-7 = x*x / (0.1-x)
solve for x
x = 3.01*10^-4
[H+] = 3.01*10^-4
We want % of dissociation so:
% dissociation = [H+] in solution / [HA initial] * 100%
% dissociation = 3.01*10^-4 / (0.1) *100 = 0.301%
% dissociation = 0.301%
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