常温下，向$0.1\,\text{mol}/\text{L}$的稀氨水中通入氯化氢气体$\rm ($忽略溶液体积变化及溶液中$\text{N}{{\text{H}}_{3}}$的存在$\rm )$，溶液中$\text{lg}\,\dfrac{n\left( \text{Cl} \right)}{n\left( \text{N} \right)}$随$\text{pH}$的变化如图所示。已知：$n\left( \ \text{N} \right)=n\left( \text{N}{{\text{H}}_{3}}\cdot {{\text{H}}_{2}}\text{O} \right)+n\left( \text{NH}_{4}^{+} \right)$；${{K}_{\text{b}}}\left( \text{N}{{\text{H}}_{3}}\cdot {{\text{H}}_{2}}\text{O} \right)=2.0\times {{10}^{-5}}$，$\text{lg}\,2=0.30$。则下列说法正确的是$(\quad\ \ \ \ )$

$\\rm A$点：$c\\left( \\text{C}{{\\text{l}}^{-}} \\right)=2\\left[ c\\left( \\text{N}{{\\text{H}}_{3}}\\cdot {{\\text{H}}_{2}}\\text{O} \\right)+c\\left( \\text{NH}_{4}^{+} \\right) \\right]$

$\\rm B$点：$c\\left( \\text{C}{{\\text{l}}^{-}} \\right)+c\\left( {{\\text{H}}^{+}} \\right)=2c\\left( \\text{N}{{\\text{H}}_{3}}\\cdot {{\\text{H}}_{2}}\\text{O} \\right)+c\\left( \\text{NH}_{4}^{+} \\right)+c\\left( \\text{O}{{\\text{H}}^{-}} \\right)$且$\\rm y$约等于$\\rm 5.15$

$\\text{C}$点：$c\\left( \\text{N}{{\\text{H}}_{3}}\\cdot {{\\text{H}}_{2}}\\text{O} \\right)=\\dfrac{20}{201}\\,\\,\\text{mol}/\\text{L}$

水的电离程度$\\text{B}\\gt \\text{A}\\gt \\text{C}$

根据图像知，氨水中通入$\rm HCl$气体，曲线由$\rm C$向$\rm A$变化，$\rm B$点$\text{lg}\,\dfrac{c\text{(Cl)}}{c\text{(N)}}=0$，$n\left( \text{Cl} \right)=n\left( \text{N} \right)$，恰好中和，溶质只有$\rm NH_{4}Cl$，$\rm B$点左侧$\rm HCl$过量，$\rm B$点右侧$\text{N}{{\text{H}}_{\text{3}}}\cdot {{\text{H}}_{\text{2}}}\text{O}$过量，据此解答。

$\rm A$．当溶液中溶质为浓度均为$\rm 0.1$ $\rm mol/L$的$\rm NH_{4}Cl$和$\rm HCl$时溶液中物料守恒为$c\left( \text{C}{{\text{l}}^{-}} \right)=2[c\left( \text{N}{{\text{H}}_{\text{3}}}\cdot {{\text{H}}_{\text{2}}}\text{O} \right)+c(\text{NH}_{4}^{+})]$，$\rm A$点$\rm pH=0.5$，$c\left( {{\text{H}}^{+}} \right)={{10}^{-0.5}}\,\text{mol/L}\gt 0.1\,\text{mol/L}$，$\rm A$点溶液中$\rm HCl$的浓度大于$\rm NH_{4}Cl$的浓度，则$c\left( \text{C}{{\text{l}}^{-}} \right)\gt 2[c\left( \text{N}{{\text{H}}_{\text{3}}}\cdot {{\text{H}}_{\text{2}}}\text{O} \right)+c(\text{NH}_{4}^{+})]$，故$\rm A$错误；

$\rm B$．根据分析可知，$\rm B$点溶液中$\text{lg}\,\dfrac{n\left( \text{Cl} \right)}{n\left( \text{N} \right)}=0$，溶液为$\rm 0.1$ $\rm mol/L$ $\rm NH_{4}Cl$溶液，溶液中电荷守恒$c\left( \text{C}{{\text{l}}^{-}} \right)+c\left( \text{O}{{\text{H}}^{-}} \right)=c\left( \text{NH}_{4}^{+} \right)+c\left( {{\text{H}}^{+}} \right)$，物料守恒为$c\left( \text{C}{{\text{l}}^{-}} \right)=c\left( \text{N}{{\text{H}}_{\text{3}}}\cdot {{\text{H}}_{\text{2}}}\text{O} \right)+c(\text{NH}_{4}^{+})$，两式整理得$c\left( \text{C}{{\text{l}}^{-}} \right)+c\left( {{\text{H}}^{+}} \right)=2c\left( \text{N}{{\text{H}}_{\text{3}}}\cdot {{\text{H}}_{\text{2}}}\text{O} \right)+c\left( \text{NH}_{\text{4}}^{+} \right)+c\left( \text{O}{{\text{H}}^{-}} \right)$，若设水解的$c\left( \text{NH}_{4}^{+} \right)=a\,\text{mol/L}$，则： $\begin{matrix} {} & \text{NH}_{4}^{+} & + & {{\text{H}}_{\text{2}}}\text{O} & \overset{{}}{\mathop{\rightleftharpoons }}\, & \text{N}{{\text{H}}_{3}}\cdot {{\text{H}}_{2}}\text{O} & + & {{\text{H}}^{+}} \\起始量 \text{mol/L} & 0.1 & {} & {} & {} & 0 & {} & 0 \\转化量 \text{mol/L} & a & {} & {} & {} & a & {} & a \\ 平衡量\text{mol/L} & 0.1-a\approx 0.1 & {} & {} & {} & a & {} & a \\ \end{matrix}$，由${{K}_{\text{h}}}=\dfrac{{{K}_{\text{w}}}}{{{K}_{\text{b}}}}$，得$\dfrac{{{a}^{2}}}{0.1}=\dfrac{{{10}^{-14}}}{2\times {{10}^{-5}}}$，解得$a={{\left( \dfrac{1}{2} \right)}^{\dfrac{1}{2}}}\times {{10}^{-5}}$，$\rm pH=5.15$，即$y=5.15$，故$\rm B$正确；

$\rm C$．由$\rm c$点$\rm pH=7$，结合电荷守恒得：$c\left( \text{NH}_{\text{4}}^{+} \right)=c\left( \text{C}{{\text{l}}^{-}} \right)$，又知$c(\text{N}{{\text{H}}_{3}}\cdot {{\text{H}}_{\text{2}}}\text{O})+c(\text{NH}_{4}^{+})=0.1\,\text{mol/L}$，${{K}_{\text{b}}}\left( \text{N}{{\text{H}}_{3}}\cdot {{\text{H}}_{\text{2}}}\text{O} \right)=\dfrac{c\left( \text{NH}_{\text{4}}^{+} \right)c\left( \text{O}{{\text{H}}^{-}} \right)}{c\left( \text{N}{{\text{H}}_{\text{3}}}\cdot {{\text{H}}_{\text{2}}}\text{O} \right)}=\dfrac{c\left( \text{NH}_{\text{4}}^{+} \right)\times {{10}^{-7}}}{0.1-c\left( \text{NH}_{\text{4}}^{+} \right)}=2.0\times {{10}^{-5}}$，得$c\left( \text{NH}_{4}^{+} \right)=\dfrac{20}{201}\,\,\text{mol}/\text{L}$，$c\left( \text{N}{{\text{H}}_{3}}\cdot {{\text{H}}_{2}}\text{O} \right)=\dfrac{0.1}{201}\,\,\text{mol}/\text{L}$，故$\rm C$错误；

$\rm D$．$\rm B$点溶质只有$\rm NH_{4}Cl$，水的电离程度最大，$\rm C$点显中性，$\text{N}{{\text{H}}_{\text{3}}}\cdot {{\text{H}}_{\text{2}}}\text{O}$稍过量，溶质为$\rm NH_{4}Cl$和少量$\text{N}{{\text{H}}_{\text{3}}}\cdot {{\text{H}}_{\text{2}}}\text{O}$，$\rm A$点显强酸性，溶质为$\rm NH_{4}Cl$和大量$\rm HCl$，$\rm A$点水的电离程度小于$\rm C$点，综上得水的电离程度：$\rm B\gt C\gt A$，故$\rm D$错误；

故选：$\rm B$