We interrogate the reaction...
#HOAc(aq) + H_2O(l) rightleftharpoons""^(-)OAc + H_3O^+#
Now, I think, we are quoted #K_a=3xx10^-4#...
And thus...#K_a=3.0xx10^-4=([H_3O^+][""^(-)OAc])/([HOAc])#
Now #[HOAc]_"initial"=0.01*mol*L^-1#...and if #x*mol*L^-1# undergoes protonolysis….
#K_a=3.0xx10^-4=(x^2)/(0.01-x)#... a quadratic in #x#, the which can be solved exactly...be we make the approx. that #(0.01-x)~=0.01#...we will HAVE to check whether the approx. is valid...
And so #x_1=sqrt(3.0xx10^-4xx0.01)=1.73xx10^-3#...we can plug this approx. in again, and again...i.e. this is the method of successive approximation....
#x_2=1.58xx10^-3#
#x_3=1.59xx10^-3#
#x_4=1.59xx10^-3*mol*L^-1#.
Since the approximations have converged, I am prepared to accept this as the TRUE value....
And so here....#pH=2.80#...agreed...
And degree of dissociation...#alpha=(1.59xx10^-3*mol*L^-1)/(0.010*mol*L^-1)=0.159#...or #15.9%#
Note that this relatively percentage dissociation is the result of the WEAKNESS of the acid concentration. Had #[HOAc]# been #1*mol*L^-1# initially...then....
#K_a=3.0xx10^-4=(x^2)/(0.1-x)#...#x_1=5.48xx10^-3*mol*L^-1#; #x_2=5.33*mmol*L^-1#; #x_3=5.33*mmol*L^-1#..percentage dissociation drops to about 5%. This is a manifestation of #"Le Chatelier's principle.."#
