We address the equilibrium,
#HA(aq) + H_2O(l) rightleftharpoons H_3O^(+) + A^(-)#
And #K_a=([H_3O^+][A^-])/([HA])=4.67xx10^-3#
Now if #x*mol*L^-1# #HA# dissociates then..........
#4.67xx10^-3=x^2/(0.171-x)#
And this is quadratic in #x#, but instead of using the tedious quadratic equation, we assume that #0.171">>"x#, and so.......
#4.67xx10^-3~=x^2/(0.171)#
And thus #x_1=sqrt(4.67xx10^-3xx0.171)=0.0283*mol*L^-1#, and now we have an approx. for #x#, we can recycle this back into the expression:
#x_2=0.0258*mol*L^-1#
#x_3=0.0260*mol*L^-1#
#x_4=0.0260*mol*L^-1#
Since the values have converged, I am willing to accept this answer. (This same answer would have resulted from the quadratic equation, had we bothered to do it.)
And thus percentage dissociation................
#="Concentration of hydronium ion"/"Initial concentration of acid"xx100%=(0.0260*mol*L^-1)/(0.171*mol*L^-1)xx100%=15%#