$\rm 25\;\rm ^\circ\rm C$时，$0.1\,\text{mol}/\text{L}\,{{\text{K}}_{2}}\text{C}{{\text{r}}_{2}}{{\text{O}}_{7}}$溶液中存在以下平衡；

①$\text{C}{{\text{r}}_{2}}\text{O}_{7}^{2-}\left( \text{aq} \right)+{{\text{H}}_{2}}\text{O}\left( \text{l} \right)\rightleftharpoons 2\text{CrO}_{4}^{2-}\left( \text{aq} \right)+2{{\text{H}}^{+}}\left( \text{aq} \right)\quad {{K}_{1}}=3.27\times {{10}^{-15}}$

②$\text{C}{{\text{r}}_{2}}\text{O}_{7}^{2-}\left( \text{aq} \right)+{{\text{H}}_{2}}\text{O}\left( \text{l} \right)\rightleftharpoons 2\text{HCrO}_{4}^{-}\left( \text{aq} \right)\quad {{K}_{2}}$

③$\text{HCrO}_{4}^{-}\left( \text{aq} \right)\rightleftharpoons \text{CrO}_{4}^{2-}\left( \text{aq} \right)+{{\text{H}}^{+}}\left( \text{aq} \right)\quad {{K}_{3}}=3.3\times {{10}^{-7}}$

其中$\text{lg}\,\dfrac{c\left( \text{CrO}_{4}^{2-} \right)}{c\left( \text{C}{{\text{r}}_{2}}\text{O}_{7}^{2-} \right)}$随$\rm pH$的变化关系如图所示。下列说法错误的是$(\quad\ \ \ \ )$

$\\rm pH$增大，$\\dfrac{c\\left( {{\\text{H}}^{+}} \\right)\\cdot {{c}^{2}}\\left( \\text{CrO}_{4}^{2-} \\right)}{c\\left( \\text{C}{{\\text{r}}_{2}}\\text{O}_{7}^{2-} \\right)}$的值减小

反应③的化学平衡常数${{K}_{3}}={{\\left( \\dfrac{{{K}_{1}}}{{{K}_{2}}} \\right)}^{\\dfrac{1}{2}}}$

$\\text{pH}=9.5$时，$\\text{HCrO}_{4}^{-}$的浓度约为$1.0\\times {{10}^{-3.8}}\\,\\text{mol}/\\text{L}$

$\\rm a$点溶液中离子浓度关系；$c\\left( {{\\text{K}}^{+}} \\right)\\gt c\\left( \\text{CrO}_{4}^{2-} \\right)\\gt c\\left( \\text{C}{{\\text{r}}_{2}}\\text{O}_{7}^{2-} \\right)\\gt c\\left( {{\\text{H}}^{+}} \\right)$

$\rm A$．${{K}_{1}}=\dfrac{{{c}^{2}}\left( {{\text{H}}^{+}} \right)\cdot {{c}^{2}}\left( \text{CrO}_{4}^{2-} \right)}{c\left( \text{C}{{\text{r}}_{2}}\text{O}_{7}^{2-} \right)}$，$\dfrac{c\left( {{\text{H}}^{+}} \right)\cdot {{c}^{2}}\left( \text{CrO}_{4}^{2-} \right)}{c\left( \text{C}{{\text{r}}_{2}}\text{O}_{7}^{2-} \right)}=\dfrac{{{K}_{1}}}{c\left( {{\text{H}}^{+}} \right)}$，可知$\rm pH$增大$c\left(\mathrm{H}^{+}\right)$减小，平衡常数不变，$\dfrac{c\left( {{\text{H}}^{+}} \right)\cdot {{c}^{2}}\left( \text{CrO}_{4}^{2-} \right)}{c\left( \text{C}{{\text{r}}_{2}}\text{O}_{7}^{2-} \right)}$的值增大，$\rm A$错误；

$\rm B$．反应③$\rm =\dfrac{1}{2}\rm ($①$\rm -$②$\rm )$，则反应③的化学平衡常数${{K}_{3}}={{\left( \dfrac{{{K}_{1}}}{{{K}_{2}}} \right)}^{\dfrac{1}{2}}}$，$\rm B$正确；

$\rm C$．$\text{pH}=9.5$时，$c\left( {{\text{H}}^{+}} \right)={{10}^{-9.5}}\,\text{mol}/\text{L}$，$\dfrac{c\left( \text{CrO}_{4}^{2-} \right)}{c\left( \text{C}{{\text{r}}_{2}}\text{O}_{7}^{2-} \right)}={{10}^{5.3}}$，${{K}_{1}}=\dfrac{{{c}^{2}}\left( {{\text{H}}^{+}} \right)\cdot {{c}^{2}}\left( \text{CrO}_{4}^{2-} \right)}{c\left( \text{C}{{\text{r}}_{2}}\text{O}_{7}^{2-} \right)}=3.27\times {{10}^{-15}}$，求得$c\left( \text{CrO}_{4}^{2-} \right)=3.27\times {{10}^{-1.3}}\,\text{mol}/\text{L}$，${{K}_{3}}=\dfrac{c\left( {{\text{H}}^{+}} \right)\cdot c\left( \text{CrO}_{4}^{2-} \right)}{c\left( \text{HCrO}_{4}^{-} \right)}=\dfrac{{{10}^{-9.5}}\times 3.27\times {{10}^{-1.3}}}{c\left( \text{HCrO}_{4}^{-} \right)}=3.3\times {{10}^{-7}}$，$\text{HCrO}_{4}^{-}$的浓度约为$1.0\times {{10}^{-3.8}}\;\rm \text{mol}/\text{L}$，$\rm C$正确；

$\rm D$．$0.1\,\text{mol}/\text{L}\,{{\text{K}}_{2}}\text{C}{{\text{r}}_{2}}{{\text{O}}_{7}}$溶液中，$c\left( {{\text{K}}^{+}} \right)$最大，$\rm a$点溶液中$\dfrac{c\left( \text{CrO}_{4}^{2-} \right)}{c\left( \text{C}{{\text{r}}_{2}}\text{O}_{7}^{2-} \right)}={{10}^{5.3}}$，$c\left( \text{CrO}_{4}^{2-} \right)\gt c\left( \text{C}{{\text{r}}_{2}}\text{O}_{7}^{2-} \right)$，根据$\rm C$项数据，$c\left( \text{CrO}_{4}^{2-} \right)=3.27\times {{10}^{-1.3}}\;\rm \text{mol}/\text{L}$，$c\left( \text{C}{{\text{r}}_{2}}\text{O}_{7}^{2-} \right)=3.27\times {{10}^{-6.3}}\;\rm \text{mol}/\text{L}$，$c\left( {{\text{H}}^{+}} \right)={{10}^{-9.5}}\;\rm \text{mol}/\text{L}$，可知$\rm a$点溶液中离子浓度关系；$c\left( {{\text{K}}^{+}} \right)\gt c\left( \text{CrO}_{4}^{2-} \right)\gt c\left( \text{C}{{\text{r}}_{2}}\text{O}_{7}^{2-} \right)\gt c\left( {{\text{H}}^{+}} \right)$，$\rm D$正确；

故选：$\rm A$