$C_MNH_3 = \frac{0,05 + 0,05}{0,1} = 1$M
$NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-$
bd $1$
CB $1-x x x$
Ta có $K_b = \frac{[NH_4^+].[OH^-]}{NH_3} = \frac{x.x}{1-x} = 5,5.10^{-5}$
=> $x = [OH^-] = 7,4.10^{-3}$
Bài 2 :
$HF \rightleftharpoons H^+ + F^-$
bd $0,01$
CB $0,01-10^{-4} 10^{-4} 10^{-4}$
Sau khi CB :
$$\sum_{}^{}$ = 0,01-10^{-4} + 10^{-4} + 10^{-4} = 0,01 + 10^{-4}$
=> sô pư và ion = $( 0,01 + 10^{-4}).6,02.10^{23} = 6,08.10^{21}$