(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?