#(i)# We first write a stoichiometric equation to inform our reasoning:
#HA+H_2OrightleftharpoonsH_3O^+ + A^-#
#(ii)# And then we write the equilibrium expression:
#([H_3O^+][A^-])/([HA])=K_a=8.1xx10^-6#
#(iii)# And then we solve the problem using approximations IF APPROPRIATE.
If the amount of dissociation is #x*mol*L^-1#, then our expression becomes:
#x^2/(0.33-x)=8.1xx10^-6#
If #0.33">>>"x#, then #0.33-x~=0.33#; and we must justify this approx. later. Solving for #x=sqrt(8.1xx10^-6xx0.33)#, we gets....
#x_1=1.63xx10^-3#; a value that is indeed small compared to #0.33#. Just to belabour the point, we can input our first approx. back in the calculation, and get a 2nd approx.......
#x_2=1.63xx10^-3#
Since the approximations have converged, we are willing to accept this value as the true value (i.e. the same as if we exactly solved the quadratic in #x#).
Given #x=1.63xx10^-3*mol*L^-1=[H_3O^+]#, #pH=-log_10[H_3O^+]# #=# #-log_(10)1.63xx10^-3=2.79#.
What is #pOH# of this solution?
