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$
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=-\log x=2,88\rightarrow pH=11,12$