giải giúp mình với ạ

Đáp án:

a,

$C_5H_5N+H_2O\rightleftharpoons C_5H_6N^++OH^-$ 

b,

$C_{NH_3}=C_{C_5H_5N}=\dfrac{0,005}{0,200}=0,025M$ 

$C_5H_5N+H_2O\rightleftharpoons C_5H_5NH^+ + OH^-\quad K_{b_1}=10^{-8,82}$

$NH_3+H_2O\rightleftharpoons NH_4^++OH^-\quad K_{b_2}=10^{-4,74}$ 

$H_2O\rightleftharpoons H^++OH^-\quad K_W=10^{-14}$

Dung dịch hai base yếu có $oh>>h$ nên ta có:

$oh=[C_5H_5NH^+]+[NH_4^+]=\dfrac{K_{b_1}C_{b_1}}{oh}+\dfrac{K_{b_2}C_{b_2}}{oh}$ 

$\Rightarrow oh=\sqrt{K_{b_1}C_{b_1}+K_{b_2}C_{b_2}}=\sqrt{10^{-8,82}.0,025+10^{-4,74}.0,025}=6,75.10^{-4}M$ 

$\Rightarrow pH=10,83$

Ta có $K_{b_1}=\dfrac{[C_5H_5NH^+]oh}{C_{b_1}-[C_5H_5NH^+]}$ 

$\Rightarrow (C_{b_1}-[C_5H_5NH^+])K_{b_1}=oh[C_5H_5NH^+]$ 

$\Rightarrow [C_5H_5NH^+]=\dfrac{C_{b_1}K_{b_1}}{oh+K_{b_1}}=\dfrac{0,025.10^{-8,82}}{6,75.10^{-4}+10^{-8,82}}=5,61.10^{-8}M$