$NH_3 + HCl => NH_4Cl$ bd $0,024 0,02$
sau $0,004 0,02$
$C_{NH_3} = 0,008$M ; $C_{NH_4^+} = 0,04$M
$NH_4^+ + H_2O => NH_3 + H_3O^+$
bd $0,04 0,008$
CB $0,04-x 0,008+x x$
$K_a = 10^{-14}/K_b = 5,6.10^{-10}$
ta có : $\frac{(0,008.+x).x}{0,04-x} = 5,6.10^{-10}$
=> $x = 2,8.10^{-9} => pH = 8,553$