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....
