常温下，向$\text{N}{{\text{H}}_{4}}\text{F}$溶液中滴加$\rm NaOH$溶液，混合溶液中$\text{lg}X\rm (X$表示$\dfrac{c\left( {{\text{F}}^{-}} \right)}{c(\text{HF})}$或$\dfrac{c\left( \text{NH}_{4}^{+} \right)}{c\left( \text{N}{{\text{H}}_{3}}\cdot {{\text{H}}_{2}}\text{O} \right)}\rm )$随$\rm pH$变化如图，已知${{K}_{\text{a}}}(\text{HF})={{10}^{-4.5}}$，Ⅰ线和Ⅱ线交点横坐标为$\rm 6.6$，下列说法错误的是$(\quad\ \ \ \ )$

Ⅰ线代表$\\lg \\dfrac{c\\left( {{\\text{F}}^{-}} \\right)}{c(\\text{HF})}$

Ⅱ线和横轴交点坐标为$\\left( 8.7,0 \\right)$

$\\text{N}{{\\text{H}}_{4}}\\text{F}$溶液加水稀释过程中$\\rm pH$值逐渐减小

$\\text{N}{{\\text{H}}_{4}}\\text{F}$溶液中存在：$c\\left( \\text{NH}_{4}^{+} \\right)+c\\left( \\text{O}{{\\text{H}}^{-}} \\right)+2c\\left( \\text{N}{{\\text{H}}_{3}}\\cdot {{\\text{H}}_{2}}\\text{O} \\right)=c\\left( {{\\text{H}}^{+}} \\right)+c\\left( {{\\text{F}}^{-}} \\right)+2c(\\text{HF})$

分析：向$\text{N}{{\text{H}}_{4}}\text{F}$溶液中滴加$\rm NaOH$溶液，随着$\rm PH$增大，$c\left( {{\text{F}}^{-}} \right)$增大$c\left( \text{H}{{\text{F}}^{{}}} \right)$减小，$\text{lg}\dfrac{c\left( {{\text{F}}^{-}} \right)}{c(\text{HF})}$增大， $c\left( \text{NH}_{4}^{+} \right)$减小$c\left( \text{N}{{\text{H}}_{3}}\cdot {{\text{H}}_{2}}\text{O} \right)$增大，$\text{lg}\dfrac{c\left( \text{NH}_{4}^{+} \right)}{c\left( \text{N}{{\text{H}}_{3}}\cdot {{\text{H}}_{2}}\text{O} \right)}$减小。

$\rm A$．Ⅰ线代表$\lg \dfrac{c\left( {{\text{F}}^{-}} \right)}{c(\text{HF})}$，$\rm A$正确；

$\rm B$．Ⅰ线和Ⅱ线交点横坐标为$\rm 6.6$可得$\dfrac{c\left( {{\text{F}}^{-}} \right)}{c\text{(HF)}}=\dfrac{{{\text{K}}_{\text{a}}}}{c\text{(}{{\text{H}}^{+}}\text{)}}=\dfrac{{{K}_{\text{b}}}}{c\text{(O}{{\text{H}}^{-}}\text{)}}=\dfrac{c\left( \text{NH}_{4}^{+} \right)}{c\left( \text{N}{{\text{H}}_{3}}\cdot {{\text{H}}_{2}}\text{O} \right)}\Rightarrow \dfrac{\text{1}{{\text{0}}^{\text{-4}\text{.5}}}}{\text{1}{{\text{0}}^{\text{-6}\text{.6}}}}=\dfrac{{{K}_{\text{b}}}}{\text{1}{{\text{0}}^{\text{6}\text{.6}}}^{-14}}\Rightarrow {{K}_{\text{b}}}={{10}^{-5.3}}$，横轴交点说明$\dfrac{c\left( \text{NH}_{4}^{+} \right)}{c\left( \text{N}{{\text{H}}_{3}}\cdot {{\text{H}}_{2}}\text{O} \right)}\rm =1$则$\dfrac{{{K}_{\text{b}}}}{c\text{(O}{{\text{H}}^{-}}\text{)}}=\dfrac{c\left( \text{NH}_{4}^{+} \right)}{c\left( \text{N}{{\text{H}}_{3}}\cdot {{\text{H}}_{2}}\text{O} \right)}\Rightarrow c\text{(O}{{\text{H}}^{-}}\text{)}={{10}^{-5.3}}$，$\text{pH}=8.7$，$\rm B$正确；

$\rm C$．${{K}_{\text{a}}}(\text{HF})={{10}^{-4.5}}$，$\rm F^{-}$的水解平衡常数${{K}_{\text{h}}}=\dfrac{{{K}_{\text{w}}}}{{{K}_{\text{a}}}}=\dfrac{{{10}^{-14}}}{{{10}^{-4.5}}}={{10}^{-9.5}}$，$\text{NH}_{\text{4}}^{+}$的水解平衡常数${{K}_{\text{h}}}=\dfrac{{{K}_{\text{w}}}}{{{K}_{\text{a}}}}=\dfrac{{{10}^{-14}}}{{{10}^{-5.3}}}={{10}^{-8.7}}$，$\text{N}{{\text{H}}_{4}}\text{F}$溶液显酸性，加水稀释过程中水解程度增大，但氢离子浓度减小，$\rm pH$值逐渐增大，$\rm C$错误；

$\rm D$．$\text{N}{{\text{H}}_{4}}\text{F}$溶液中存在：电荷平衡$c\left( \text{NH}_{4}^{+} \right)+c\left( {{\text{H}}^{+}} \right)=c\left( {{\text{F}}^{-}} \right)+c\left( \text{O}{{\text{H}}^{-}} \right)$，物料守恒

$c\left( \text{NH}_{4}^{+} \right)+c\left( \text{N}{{\text{H}}_{3}}\cdot {{\text{H}}_{2}}\text{O} \right)=c\left( {{\text{F}}^{-}} \right)+c(\text{HF})$，两式相减得$c\left( \text{O}{{\text{H}}^{-}} \right)+c\left( \text{N}{{\text{H}}_{3}}\cdot {{\text{H}}_{2}}\text{O} \right)=c\left( {{\text{H}}^{+}} \right)+c(\text{HF})$，与$c\left( \text{NH}_{4}^{+} \right)+c\left( \text{N}{{\text{H}}_{3}}\cdot {{\text{H}}_{2}}\text{O} \right)=c\left( {{\text{F}}^{-}} \right)+c(\text{HF})$相加得$c\left( \text{NH}_{4}^{+} \right)+c\left( \text{O}{{\text{H}}^{-}} \right)+2c\left( \text{N}{{\text{H}}_{3}}\cdot {{\text{H}}_{2}}\text{O} \right)=c\left( {{\text{H}}^{+}} \right)+c\left( {{\text{F}}^{-}} \right)+2c(\text{HF})$，$\rm D$正确；

故选：$\rm C$