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.