题目内容：

0.5MOL/L氯化铵的PH值是多少?给出氨水的电离常数是K1.

优质解答

c(H+) = [ Kh* c(NH4+) ] ^1/2
= [ (Kw / K1)* 0.5 ] ^1/2
= [ 10^-14/ 1.77*10^-5]^1/2
= 2.38 *10^-5
pH = 4.6