Bài 2: $NH_{3}\rightarrow {NH_{4}}^{+}+OH^{-}$ $0,1.......... 0 ........0$ $x........x........x$$(0,1-x).....x........x$$\Leftrightarrow 1,74.10^{-5}=\frac{x.x}{(0,1-x)}$$\rightarrow x\approx 1,31.10^{-3}\rightarrow pOH=2,88\rightarrow pH=11,12$
Bài 2:
$NH_{3}\rightarrow {NH_{4}}^{+}+OH^{-}$
Ban đầu: $0,1.......... 0 ........0$
Phân li: $x........x........x$
Sau p.li: $(0,1-x).....x........x$$\Leftrightarrow 1,74.10^{-5}=\frac{x.x}{(0,1-x)}$$\rightarrow x\approx 1,31.10^{-3}\rightarrow pOH=2,88\rightarrow pH=11,12$