hóa 11 khó

G

galaxy98adt

Thử làm xem sao! :D
$NH_3 + H_2O ---> NH_4^+ + OH^-$

Ta có: $k_{b (NH_3)} = \frac{[NH_4^+].[OH^-]}{[NH_3]}$

$NH_4^+ ---> NH_3 + H^+$

$k_{a (NH_4^+)} = \frac{[NH_3].[H^+]}{[NH_4^+]}$

\Rightarrow $k_{a (NH_4^+)}.k_{b (NH_3)} = \frac{[NH_3].[H^+]}{[NH_4^+]}.\frac{[NH_4^+].[OH^-]}{[NH_3]} = [H^+].[OH^-] = 10^{-14}$ (đpcm)
 
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