We interrogate the equilibrium...............
#NH_3(aq) + H_2O(l) rightleftharpoonsNH_4^(+) + HO^-#
#K_"eq"=([NH_4^+][HO^-])/([NH_3(aq)])=1.78xx10^-5#
Now we know that #[NH_3(aq)]=0.503*mol*L^-1# initially, and call the amount of association it undergoes #x#.
So at equilibrium, #[NH_4^+]=[HO^-]=x#; and #[NH_3(aq)]=(0.503-x)*mol*L^-1#.
#K_"eq"=1.78xx10^-5=x^2/(0.503-x)#
This is quadratic in #x#, which we could solve exactly, however, we make the approximation that #(0.503-x)~=0.503*mol*L^-1#.
Thus #x_1=sqrt(1.78xx10^-5xx0.503)=0.00299*mol*L^-1#
Now that we an approximation for #x#, we can we substitute this value back into the equation.
And thus..........................................................
#x_2=sqrt(1.78xx10^-5xx(0.503-0.00299))=0.00298*mol*L^-1#
Since the values have converged, we can take this value as the true value.
And so #[HO^-]=[NH_4^+]=2.98xx10^-3*mol*L^-1#, and thus #pOH=-log_(10)[HO^-]=-log_10(2.98xx10^-3)=2.53#.
But we need #pOH#, however, we know (how?) that #pH+pOH=14#.
And so #pH=14-2.53=??#
