We interrogate the equilibrium.......
#NH_3(aq) + H_2O(l) rightleftharpoonsNH_4^(+) + HO^(-)#,
for which #K_b=([NH_4^+][HO^-])/([NH_3(aq)])=4.3xx10^-5#
ANd if we propose that #x*mol*L^-1#of #"ammonia"# associates, then we can rewrite the #K_b# expression to give.......
#x^2/(1.5-x)=4.3xx10^-5#, and if we make the approx., that #1.5">>"x#, then.......
#x~=sqrt(1.5xx4.3xx10^-5)#
#x_1=0.00803*mol*L^-1#
#x_2=0.00801*mol*L^-1# (we plug the first approx. back into the expression...)
#x_3=0.00801*mol*L^-1#. Since the approximations have converged
And thus #[NH_4^+]=[HO^-]=0.00801*mol*L^-1#, and #[NH_3(aq)]=1.492*mol*L^-1#.
Now #pOH=2.10#, and thus #pH=14-2.10=11.90#
