#NH_3(aq) + H_2O(l) rightleftharpoonsNH_4^+ + HO^-#
And so #K_b=1.8xx10^-5=([NH_4^+][HO^-])/([NH_3(aq)])#
We put in some numbers, and we assume that #x*mol*L^-1# of ammonia ASSOCIATES, and thus at equilibrium, we know that #x=[NH_4^+]=[HO^-]#, and #[NH_3]=(0.10-x)*mol*L^-1#.
And we plug in the numbers to our equilibrium expression..
#K_b=1.8xx10^-5=x^2/(0.10-x)#
And this is a quadratic in #x#, which we could solve EXACTLY if we were so-minded, but because chemists are workshy, we make the approximation that #0.10-x~=0.10#. Note that we must justify this approx. later.
And thus #x_1=sqrt(1.8xx10^-5xx0.10)=1.34xx10^-3*mol*L^-1#.
And now that we have an approximation for #x#, we may use #x_1# in our expression to get a second approximation........
#x_2=1.33xx10^-3*mol*L^-1#
#x_3=1.33xx10^-3*mol*L^-1#
This method is as exact as if we used the quadratic equation, and so.....
#x_3=[HO^-]=1.33xx10^-3*mol*L^-1#; #pOH=-log_10([HO^-])=2.88#; and #pH=14-pOH=11.1.#
We know that in aqueous solution, #14=pOH+pH#, and so if you knows #[HO^-]# youze also knows #[H_3O^+]#.
