#pH=-log_10[H_3O^+]#.
And we know that ammonium salts are weakly acidic, and undergo the acid-base reaction:
#NH_4^(+) + H_2O(l) rightleftharpoonsNH_3(aq) + H_3O^+#
For ammonium ion, #K_a=1.80xx10^-5#
And if there were a #1.0*mol*L^-1# solution of ammonium ion, we would solve the equilibrium expression in the usual way.
i.e. #K_a=1.80xx10^-5=([H_3O^+][NH_3(aq)])/([NH_4^(+)])#
And if #x*mol*L^-1# ammonium ion ionizes, then.........
#K_a=1.80xx10^-5=x^2/((1.0-x)*mol*L^-1)#, a quadratic in #x#, which may be simplified given the reasonable assumption that #1.0">>"x#, and thus #1.0-x~=1.0#.
And thus #x~=sqrt(1.80^-5)#,
so #x_1=4.2xx10^-3#, and recycling this approximation of #x# back into the expression:
#x_2=4.2xx10^-3#, and since #x# has converged, we may accept this value.
Thus #[H_3O^+]=4.2xx10^-3*mol*L^-1#, and #pH=-log_10[H_3O^+]=-log_10(4.2xx10^-3)=2.37#.
