We assess the equilibrium....
#NH_3(aq) + H_2O(l) rightleftharpoonsNH_4^(+)+HO^-#
#K_b=([NH_4^+][HO^-])/([NH_3(aq)])=1.80xx10^-5#
And if #x*mol*L^-1# of ammonia undergoes reaction....
#K_b=x^2/(0.200-x)=1.80xx10^-5#
And so #x=sqrt{K_bxx(0.200-x)}#
We make the assumption that #0.200-x~=0.200#, an assumption that we must justify later....
And so #x_1~=sqrt{K_bxx(0.200)}=sqrt{1.80xx10^-5xx0.200}=1.90xx10^-3*mol*L^-1#...which is indeed small compared to #0.200#, but now we can make a second approximation for #x#...
#x_2=sqrt{(0.200-1.90xx10^-3)xx1.80xx10^-5}=1.89xx10^-3*mol*L^-1#
#x_3=1.89xx10^-3*mol*L^-1#, and this solution is the same as we would have got had we solved the quadratic equation.....
But #x_3=[HO^-]=[NH_4^+]#
#pOH=-log_10[HO^-]=-log_10(1.89xx10^-5=-(-2.72)=2.72#
And since #pH+pOH=14#, then #pH=14-2.72=11.28#, a basic solution as we would expect for a weak base.
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