实验室通过下列过程制取草酸亚铁晶体：

已知：$K_{\mathrm{a1}}\left( \mathrm{NH}_{3}\cdot\mathrm{H}_{2}\mathrm{O}\right)=1.8\times10^{-5}$、$K_{\mathrm{a2}}\left( \mathrm{NH}_{3}\cdot\mathrm{H}_{2}\mathrm{O}\right)=1.8\times10^{-5}$。下列说法不正确的是$(\qquad)$

$\\text{pH}=4$的${{\\text{H}}_{2}}{{\\text{C}}_{2}}{{\\text{O}}_{4}}$溶液中：$\\textit{c}(\\text{C}_{2}\\text{O}_{4}^{2-})\\gt\\textit{c}(\\text{H}\\text{C}_{2}\\text{O}_{4}^{-})$

“酸化”后的溶液中：$\\textit{c}(\\text{NH}_{4}^{+})+2\\textit{c}(\\text{F}\\text{e}^{2+})\\lt2\\textit{c}(\\text{SO}_{4}^{2-})$

“沉淀”后的上层清液中：$\\textit{c}(\\text{NH}_{4}^{+})+\\textit{c}(\\text{N}\\text{H}_{3}\\cdot\\text{H}_{2}\\text{O})=\\textit{c}(\\text{SO}_{4}^{2-})$

水洗后，再用乙醇洗涤有利于晶体快速干燥

$\rm (NH_{4})_{2}Fe(SO_{4})_{2}$加入硫酸酸化得到$\rm Fe^{2+}$，加入草酸生成$\rm FeC_{2}O_{4}$沉淀，过滤、洗涤得到草酸亚铁晶体，据此回答。

$\rm A$．$\text{pH}=4$的${{\text{H}}_{2}}{{\text{C}}_{2}}{{\text{O}}_{4}}$溶液中$\textit{c}\left( \text{H}^{+}\right)=10^{-4}\ \text{mol/L}$，$\textit{K}_{\text{a}2}(\text{H}_{2}\text{C}_{2}\text{O}_{4})=\dfrac{\textit{c}(\text{C}_{2}\text{O}_{4}^{2-})\textit{c}(\text{H}^{+})}{\textit{c}(\text{H}\text{C}_{2}\text{O}_{4}^{-})}=1.5\times10^{-4}$，$\dfrac{\textit{c}(\text{C}_{2}\text{O}_{4}^{2-})}{\textit{c}(\text{H}\text{C}_{2}\text{O}_{4}^{-})}=\dfrac{\textit{K}_{\text{a}2}(\text{H}_{2}\text{C}_{2}\text{O}_{4})}{\textit{c}(\text{H}^{+})}=\dfrac{1.5\times10^{-4}}{10^{-4}}=1.5$，即$\textit{c}(\text{C}_{2}\text{O}_{4}^{2-})=1.5\textit{c}(\text{H}\text{C}_{2}\text{O}_{4}^{-})$，即$\textit{c}(\text{C}_{2}\text{O}_{4}^{2-})\gt\textit{c}(\text{H}\text{C}_{2}\text{O}_{4}^{-})$，$\rm A$正确；

$\rm B$．根据电荷守恒：$\textit{c}(\text{NH}_{4}^{+})+2\textit{c}(\text{F}\text{e}^{2+})+\textit{c}(\text{H}^{+})=2\textit{c}(\text{SO}_{4}^{2-})+\textit{c}(\text{O}\text{H}^{-})$，溶液显酸性，$\textit{c}\left( \text{H}^{+}\right)＞\textit{c}\left( \text{O}\text{H}^{-}\right)$，则$\textit{c}(\text{NH}_{4}^{+})+2\textit{c}(\text{F}\text{e}^{2+})\lt2\textit{c}(\text{SO}_{4}^{2-})$，$\rm B$正确；

$\rm C$．“沉淀”后的上层清液中溶质为$\rm (NH_{4})_{2}SO_{4}$和过量的${{\text{H}}_{2}}{{\text{C}}_{2}}{{\text{O}}_{4}}$，由物料守恒得：$\textit{c}(\text{NH}_{4}^{+})+\textit{c}(\text{N}\text{H}_{3}\cdot\text{H}_{2}\text{O})=2\textit{c}(\text{SO}_{4}^{2-})$，$\rm C$错误；

$\rm D$．乙醇与水互溶，且乙醇易挥发，用乙醇洗可带走水分，有利于晶体快速干燥，$\rm D$正确；

故选：$\rm C$