We address the equilibrium....
#underbrace(HOAc(aq))_"acetic acid" + H_2O(l)rightleftharpoons""^(-)OAc +H_3O^+#
And of course, while this will give us #H_3O^+#, and thus #pH#, #pOH=14-pH#...now given the equilibrium expression, and if we put #[""^(-)OAc]=[H_3O^+]=x#...then....
#K_a=([AcO^(-)][H_3O^+])/([HOAc])=x^2/(0.50*mol*L^-1-x)=1.75xx10^-5#.
This is a quadratic in #x# that COULD be solved EXACTLY....but chemists are lazy....and so we make the approximation that #0.50">>"x#...and thus #0.50-x~=x#....you know the usual story, approximate, then justify, then re-approximate using the first approximation....
And so #x_1=sqrt{0.50xx1.75xx10^-5}=2.96xx10^-3*mol*L^-1#...the which value is indeed small compared to #0.50*mol*L^-1#...but given an approximation for #x#, we resubstitute this back into the expression...
#x_2=sqrt{(0.50-2.96xx10^-3)xx1.75xx10^-5}=2.95xx10^-3*mol*L^-1#
#x_3=sqrt{(0.50-2.95xx10^-3)xx1.75xx10^-5}=2.95xx10^-3*mol*L^-1#
...and since the approximations have converged, I am prepared to accept this as the TRUE value...the same as if we solved the quadratic equation...
But #x=[H_3O^+]=[""^(-)OAc]# by our definition....and #[HOAc]=0.50-2.95xx10^-3)*mol*L^-1=0.497*mol*L^-1#...
And so #[H_3O^+]=[""^(-)OAc]=2.95xx10^-3*mol*L^-1#...
#pH=-log_10{2.95xx10^-3}=-(-2.53)=2.53#
And since...
#K_w=[H_3O^+][HO^-]=10^(-14)#...
And we can take #log_10# of both sides to give....
#log_(10)K_w=log_(10)10^(-14)=log_10[H_3O^+]+log_10[HO^-]#.
And thus.... #-14=log_(10)[H_3O^+]+log_(10)[HO^-]#
Or.....
#14=-log_(10)[H_3O^+]-log_(10)[HO^-]#
#14=underbrace(-log_10[H^+])_(pH)underbrace(-log_10[OH^-])_(pOH)#
#14=pH+pOH#
By definition, #-log_10[H^+]=pH#, #-log_10[HO^-]=pOH#
#pOH=14-pH=14-2.53=11.47#...for the given solution under standard conditions.....got all that?
