We interrogate the equilibrium....
#HOAc(aq)+H_2O(l)rightleftharpoonsH_3O^+ + ""^(-)OAc#
#HOAc=H_3C-C(=O)OH#
Now it is a fact that #K_a(HOAc)=10^(-4.76)=1.74xx10^-5#..
And so we set up the equilibrium expression...
#1.74xx10^(-5)=([H_3O^+][""^(-)OAc])/([HOAc])#
And if we put #[H_3O^+]=x#...then by stoichiometry...
#1.74xx10^(-5)=(x^2)/(0.0015-x)#..
And if #x"<<"0.0015#...then...#x~=sqrt(1.74xx10^-5xx0.0015)#
#x_1=1.60xx10^-4*mol*L^-1#...
And we can use this value for a second approximation....
#x_2=1.53xx10^-4*mol*L^-1#...
#x_3=1.53xx10^-4*mol*L^-1#...
...and since the approximations have converged, I am prepared to accept this as the true value...
And so #"% ionization"="moles of anion"/"moles of starting acid"xx100%#
#=(1.53xx10^-4*mol*L^-1)/(0.0015*mol*L^-1)xx100%=10.2%#
This #"% ionization"# is GREATER than for a more concentrated solution of weak acid .... at increasing dilution #"% ionization"# becomes greater for weak acids....
