Hoa 11(cần gấp)

T

thupham22011998

Coi $C_M NH_3$ ban đầu =$1M$

$NH_3+H_2O<-->NH_4^++OH^-$

--> $[NH_4^+]=[OH^-]=0,041M$

-->$[NH_3]$ dư=$0,959M$

-->$K_b=1,8.10^{-3}$
 
You must log in or register to reply here.
Top Bottom