Tính pH của dung dịch NH4Cl 0,10M? (biết $K_a(NH_4^+) = 5,56.10^{-10}$). Bỏ qua sự phân li của nước.

Đáp án:

Giải thích các bước giải:

$\begin{array}{l} N{H_4}Cl \to N{H_4}^ + + C{l^ - }\\ 0,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,1\\ \,\,\,\,\,\,\,\,\,\,\,\,\,N{H_4}^ + + {H_2}ON{H_4}OH + {H^ + }\,\,\,\,Ka = 5,{56.10^{ - 10}}\\ BĐ\,\,\,\,0,1\\ PL\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\\ CL\,0,1 - x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\\ Ka = \dfrac{{{\rm{[}}N{H_4}OH].{\rm{[}}{H^ + }{\rm{]}}}}{{{\rm{[}}N{H_4}^ + {\rm{]}}}}\,\,\, = \dfrac{{{x^2}}}{{0,1 - x}}\, = \,5,{56.10^{ - 10}}\\ \Rightarrow x = {\rm{[}}{H^ + }{\rm{]}} = 7,{46.10^{ - 6}}M\\ \Rightarrow pH = - \log \,\,\,{\rm{[}}{H^ + }{\rm{]}} = 5,13\,\,\,\,\,\,\,\,\,\,\, \end{array}$