$[H^+]=10^{-5,5}$
$NH_4Cl\to NH_4^++Cl^-$
$NH_4^++H_2O\rightleftharpoons NH_4OH+H^+$
Gọi x là mol $NH_4Cl$ thêm.
$\Rightarrow C^o=\dfrac{x}{0,25}=4x (M)$
$\Rightarrow [NH_4^+]=4x-10^{-5,5}(M)$
$[NH_4OH]=[H^+]=10^{-5,5}$
$K_a=10^{-9,24}$
$\Rightarrow \dfrac{(10^{-5,5})^2}{4x-10^{-5,5}}=10^{-9,24}$
$\Leftrightarrow x=0,004$
$\to m_{NH_4Cl}=0,004.53,5=0,214g$