Giải bài 10 trang 20 - SGK môn Hóa học lớp 11 nâng cao

a.

 $\begin{aligned} & {{n}_{HCl}}=0,04\,mol\Rightarrow {{C}_{M\,(HCl)}}=0,1M \\ & HCl\to {{H}^{+}}+C{{l}^{-}} \\ & 0,1M\,\,\,\,\,0,1M \\ & \Rightarrow pH=1 \\ \end{aligned}$

b. 

$\begin{aligned} & {{n}_{{{H}^{+}}}}={{n}_{HCl}}=0,1\,mol;\,{{n}_{O{{H}^{-}}}}={{n}_{NaOH}}=0,15\,mol \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{H}^{+}}+O{{H}^{-}}\to {{H}_{2}}O \\ & \text{pư:}\,\,\,\,\,\,\,\text{0,1}\,\,\,\,\,\,\,\,\,\text{0,1} \\ & \text{dư:}\,\,\,\,\,\,\,\text{0}\,\,\,\,\,\,\,\,\,\,\,\,\,\text{0,05} \\ & \Rightarrow [O{{H}^{-}}]=\dfrac{0,05}{0,2}=0,25M\Rightarrow [{{H}^{+}}]=\dfrac{{{10}^{-14}}}{0,25}={{4.10}^{-14}}\Rightarrow pH\approx 13,4 \\ \end{aligned}$

Ghi nhớ:

pH = -log $[H^+]$