Oxalate is a good ligand and can form complexes with transition metals. Since oxalate solutions are strongly alkaline, metal ions that are prone to hydrolysis, such as $\mathrm{Fe^{3+}, Al^{3+}}$, need to be converted to hydroxides first, then dissolved in oxalic acid, and neutralized with a base to obtain the corresponding oxalate complexes. This lengthy procedure is time-consuming, it is difficult to filter the hydroxides, the operation results in a lot of loss, and the yield is often not ideal.

Someone proposed the following experimental scheme to directly prepare the complex of oxalate with ferric iron: mixing $\mathrm{20 ~mL\ 3 ~mol\cdot L^{-1}}$ potassium oxalate hot solution and $\mathrm{8 ~mL\ 1.5 ~mol\cdot L^{-1}}$ ferric chloride solution, cooling to room temperature to obtain green needle-shaped crystals $\mathrm{K_3Fe(C_2O_4)_3 \cdot 3H_2O}$, and measuring the volume of the solution at this time as $\mathrm{30~mL}$. Now we theoretically calculate the feasibility of this scheme.

Constants involved in this problem: $\mathrm{pK_{sp}(Fe(OH)_{3})=38.55, \lg K_{f}(Fe(C_2O_4)_3^{3-})=20.2, pK_{b1}(C_2O_4^{2-})=9.73}$. The saturated solubility of the product at room temperature is $\mathrm{4.9~g \cdot 100 ~mL^{-1}}$, and the solubility of the two reactants is very high.

The secondary hydrolysis of oxalate is ignored in the calculation of this problem.

Which of the following options regarding this experimental scheme is correct (calculation error within 1% is considered correct):
草酸根是一种良好的配体,可以与过渡金属形成配合物。由于草酸盐溶液具有较强的碱性,所以$\mathrm{Fe^{3+},Al^{3+}}$等易水解的金属离子需要先转化为氢氧化物,再用草酸溶解、加碱中和才能制得相应的草酸根配合物。这样冗长的步骤花费的时间长,过滤氢氧化物的困难大,操作带来的损失多,产率往往不理想。

有人提出如下实验方案直接制备草酸根与三价铁的配合物:将 $\mathrm{20 ~mL\ 3 ~mol\cdot L^{-1}}$ 草酸钾热溶液和 $\mathrm{8 ~mL\ 1.5 ~mol\cdot L^{-1}}$ 三氯化铁溶液混合, 冷却至室温得到绿色针状晶体 $\mathrm{K_3Fe(C_2O_4)_3 \cdot 3H_2O}$，测得此时溶液体积为$\mathrm{30~mL}$ 。下面我们在理论上计算这种方案的可行性。

本题涉及的常数: $\mathrm{pK_{sp}(Fe(OH)_{3})=38.55, \lg K_{稳}(Fe(C_2O_4)_3^{3-})=20.2, pK_{b1}(C_2O_4^{2-})=9.73}$ 。产物在室温下的饱和溶解度为 $\mathrm{4.9~g \cdot 100 ~mL^{-1}}$, 两种反应物溶解度很大。

本题目计算时忽略草酸根的二级水解。

下列关于该实验方案，正确的选项是（计算误差在1%以内即视为正确）：
When mixing potassium oxalate solution and ferric chloride solution, one can adopt the method of slowly and gradually adding the potassium oxalate solution to the ferric chloride solution, or the method of rapidly adding the ferric chloride solution to the potassium oxalate solution.
At room temperature, the expected yield of this experimental protocol is approximately $\mathrm{5.9 ~g}$.
Assuming iron hydroxide is not formed, the iron ion concentration is approximately $\mathrm{10^{-10}~mol/L}$ when the mixture is cooled to room temperature.
Assuming that iron hydroxide is not formed, when the mixture is cooled to room temperature, the concentration of hydrogen oxalate ion is approximately $\mathrm{10^{-10}~mol/L}$.
The mass of iron-containing species precipitate, excluding the product, generated by this experimental protocol is approximately $\mathrm{0.3 ~mg}$.
This experimental protocol is completely infeasible.
All of the above options are incorrect.
混合草酸钾溶液和三氯化铁溶液时，可以采用将草酸钾溶液缓慢逐渐滴加至三氯化铁溶液的方法，或将三氯化铁溶液快速滴加至草酸钾溶液的方法。
室温下，该实验方案的预期产量为约$\mathrm{5.9 ~g}$
假设未生成氢氧化铁，混合后冷却至室温时，铁离子浓度约为$\mathrm{10^{-10}~mol/L}$
假设未生成氢氧化铁，混合后冷却至室温时，草酸氢根离子浓度约为$\mathrm{10^{-10}~mol/L}$
该实验方案生成的除产物外的含铁物种沉淀的质量约为$\mathrm{0.3 ~mg}$
该实验方案完全不可行
以上选项均不正确
In this experiment, it is necessary to ensure that ferric ions do not hydrolyze to form ferric hydroxide, which would affect the yield of potassium ferrioxalate; therefore, it is necessary to ensure that oxalate is in excess or locally in excess to avoid local overconcentration of ferric ions and hydrolysis; therefore, the ferric solution needs to be slowly dripped into the oxalate solution, or the oxalate solution needs to be quickly dripped into the ferric solution, so option A is incorrect.

<Checkpoint>[It is necessary to ensure that oxalate is in excess or locally in excess to avoid local overconcentration of ferric ions and hydrolysis](1)</Checkpoint>

Obviously, the reaction ratio of oxalate to ferric iron is 3:1, thus oxalate is in excess; at this time, calculate the theoretical yield based on the amount of iron. Assuming all the iron is converted into the product, the mass of the product is $\mathrm{0.008\times1.5\times491.1=5.89~g}$. However, the dissolution loss needs to be considered. Considering the theoretical yield after dissolution is $\mathrm{5.89-4.9/100\times30=4.42~g}$, so option B is incorrect.

<Checkpoint>[The theoretical yield after considering dissolution loss is $\mathrm{4.42~g}$](1)</Checkpoint>

Since the coordination stability constant of potassium ferrioxalate is relatively large, it can be approximated that most of the iron ions exist in the form of complexes;

<Checkpoint>[It can be approximated that most of the iron ions exist in the form of complexes](1)</Checkpoint>

Since the secondary hydrolysis of oxalate can be ignored, the remaining concentration after deducting the part forming the complex is:
$\mathrm{[C_2O_4^{2-}]=(20\times3-3\times8\times1.5)/30 = 0.8 ~mol/L}$

<Checkpoint>[$\mathrm{[C_2O_4^{2-}]= 0.8 ~mol/L}$](1)</Checkpoint>

Calculate the saturation concentration through the solubility of the product in water:
$\mathrm{[Fe(C_2O_4)_3^{3-}]=4.9/491.1/0.1=0.1 ~mol/L}$

<Checkpoint>[$\mathrm{[Fe(C_2O_4)_3^{3-}]=0.1 ~mol/L}$](1)</Checkpoint>

Then calculate the concentration of ferric ions through the stability constant of the complex:
$\mathrm{[Fe^{3+}]=[Fe(C_2O_4)_3^{3-}]/(K_{f}\cdot [C_2O_4^{2-}]^3)=0.1/10^{20.2}\cdot 0.8^3=10^{-20.9}~mol/L}$, the concentration is extremely small, which is approximately reasonable.

<Checkpoint>[$\mathrm{[Fe^{3+}]=10^{-20.9}~mol/L}$, approximately reasonable](1)</Checkpoint>

Ignoring the secondary hydrolysis of oxalate, we have $\mathrm{[HC_2O_4^-]=[OH^-]=\sqrt{[C_2O_4^{2-}]K_{b1}}=10^{-4.91}~mol/L}$

<Checkpoint>[$\mathrm{[HC_2O_4^-]=[OH^-]=10^{-4.91}~mol/L}$](1)</Checkpoint>

Therefore, options C and D are both incorrect.

The hydrolysis equation of ferrioxalate ion is
$\mathrm{Fe(C_2O_4)_3^{3-}+3H_2O = Fe(OH)_3 + 3HC_2O_4^-}$

The equilibrium constant of the reaction is $\mathrm{K=K_{sp}K_{b1}^3/K_{f}=10^{-10.84}>Q=[HC_2O_4^-]^3/[Fe(C_2O_4)_3^{3-}]=10^{-13.7}}$, so ferric hydroxide will be generated;

<Checkpoint>[The hydrolysis equilibrium constant of ferrioxalate ion is $\mathrm{K=10^{-10.84}>Q=10^{-13.7}}$, so ferric hydroxide will be generated](1)</Checkpoint>

According to the equilibrium constant calculation, $\mathrm{[HC_2O_4^-]'=(K[Fe(C_2O_4)_3^{3-}])^{1/3}=10^{-3.95} ~mol/L}$ at hydrolysis equilibrium

<Checkpoint>[$\mathrm{[HC_2O_4^-]'=10^{-3.95} ~mol/L}$ at hydrolysis equilibrium](1)</Checkpoint>

The amount of precipitate is $\mathrm{n(Fe(OH)_3)=([HC_2O_4^-]'-[HC_2O_4^-])\times 0.030/3=10^{-6} ~mol}$, which is calculated as $\mathrm{0.11 ~mg}$; option E is incorrect.

<Checkpoint>[The mass of ferric hydroxide generated is $\mathrm{0.11 ~mg}$](1)</Checkpoint>

Since the mass of ferric hydroxide generated is much smaller than the mass of the product, it can be considered that the experimental scheme is feasible, so option F is incorrect.

<Checkpoint>[The mass of ferric hydroxide generated is much smaller than the mass of the product, so it can be considered that the experimental scheme is feasible](1)</Checkpoint>

In summary, options A-F are all incorrect, and option G is correct.
本实验，要保证三价铁离子不水解生成氢氧化铁从而影响三草酸合铁的产率；因此需要保证草酸根过量或者局部过量，避免三价铁离子局部过浓而水解；因此需要将三价铁溶液缓慢滴加入草酸根溶液或者将草酸根溶液快速滴加入三价铁溶液，选项A错误。

<Checkpoint>[需要保证草酸根过量或者局部过量，避免三价铁离子局部过浓而水解](1)</Checkpoint>

很明显草酸根与三价铁为3：1反应，从而草酸根过量；此时以铁的量计算理论产量，假设铁全都转化为产物，此时产物质量为$\mathrm{0.008\times1.5\times491.1=5.89~g}$，但此时需要考虑溶解损失，考虑溶解后的理论产量为 $\mathrm{5.89-4.9/100\times30=4.42~g}$，选项B错误。

<Checkpoint>[考虑溶解损失后的理论产量为 $\mathrm{4.42~g}$](1)</Checkpoint>

由于三草酸合铁的配位稳定常数比较大，可以近似绝大多数铁离子以配合物形式存在；

<Checkpoint>[可以近似绝大多数铁离子以配合物形式存在](1)</Checkpoint>

由于可忽略草酸根二级水解，扣除形成配合物的部分，剩余浓度为：
$\mathrm{[C_2O_4^{2-}]=(20\times3-3\times8\times1.5)/30 = 0.8 ~mol/L}$

<Checkpoint>[$\mathrm{[C_2O_4^{2-}]= 0.8 ~mol/L}$](1)</Checkpoint>

通过产物在水中的溶解度计算饱和浓度：
$\mathrm{[Fe(C_2O_4)_3^{3-}]=4.9/491.1/0.1=0.1 ~mol/L}$

<Checkpoint>[$\mathrm{[Fe(C_2O_4)_3^{3-}]=0.1 ~mol/L}$](1)</Checkpoint>

再通过配合物的稳定常数计算三价铁离子的浓度：
$\mathrm{[Fe^{3+}]=[Fe(C_2O_4)_3^{3-}]/(K_{稳}\cdot [C_2O_4^{2-}]^3)=0.1/10^{20.2}\cdot 0.8^3=10^{-20.9}~mol/L}$，浓度极小，近似合理。

<Checkpoint>[$\mathrm{[Fe^{3+}]=10^{-20.9}~mol/L}$，近似合理](1)</Checkpoint>

忽略草酸根二级水解，有$\mathrm{[HC_2O_4^-]=[OH^-]=\sqrt{[C_2O_4^{2-}]K_{b1}}=10^{-4.91}~mol/L}$

<Checkpoint>[$\mathrm{[HC_2O_4^-]=[OH^-]=10^{-4.91}~mol/L}$](1)</Checkpoint>

从而选项C,D均错误。

三草酸铁离子水解的方程式为
$\mathrm{Fe(C_2O_4)_3^{3-}+3H_2O = Fe(OH)_3 + 3HC_2O_4^-}$

反应的平衡常数$\mathrm{K=K_{sp}^{-1}K_{b1}^3K_{稳}^{-1}=10^{-10.84}>Q=[HC_2O_4^-]^3/[Fe(C_2O_4)_3^{3-}]=10^{-13.7}}$，从而会生成氢氧化铁；

<Checkpoint>[三草酸铁离子水解平衡常数$\mathrm{K=10^{-10.84}>Q=10^{-13.7}}$，从而会生成氢氧化铁](1)</Checkpoint>

根据平衡常数计算，水解平衡时的$\mathrm{[HC_2O_4^-]'=(K[Fe(C_2O_4)_3^{3-}])^{1/3}=10^{-3.95} ~mol/L}$

<Checkpoint>[水解平衡时的$\mathrm{[HC_2O_4^-]'=10^{-3.95} ~mol/L}$](1)</Checkpoint>

沉淀的物质的量$\mathrm{n(Fe(OH)_3)=([HC_2O_4^-]'-[HC_2O_4^-])\times 0.030/3=10^{-6} ~mol}$，计算为$\mathrm{0.11 ~mg}$；选项E错误。

<Checkpoint>[生成氢氧化铁质量为$\mathrm{0.11 ~mg}$](1)</Checkpoint>

由于生成的氢氧化铁质量远小于产物质量，可以认为该实验方案可行，选项F错误。

<Checkpoint>[生成的氢氧化铁质量远小于产物质量，可以认为该实验方案可行](1)</Checkpoint>

综上，选项A-F均错误，选项G正确。