# Analytical Chemistry Calculation of the Feasibility of Oxalate Complex Synthesis
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Oxalate is a good ligand that can form complexes with transition metals. Since oxalate solutions are strongly alkaline, easily hydrolyzed metal ions such as $\mathrm { F e } ^ { 3 + }$ and $\mathrm { A l } ^ { 3 + }$ need to be first converted to hydroxides, then dissolved with oxalic acid, and neutralized with base to obtain the corresponding oxalate complexes. This lengthy procedure takes a long time, the difficulty of filtering the hydroxide is high, the operational losses are large, and the yield is often not ideal.

An experimental scheme has been proposed to directly prepare a complex of oxalate with ferric iron, which is shown below.

Mix $20 \ \mathrm { mL } \ 3 \ \mathrm { mol { \cdot } L ^ { - 1 } }$ potassium oxalate hot solution and $8 ~ \mathrm { m L } ~ 1 . 5 ~ \mathrm { m o l { \cdot } L ^ { - 1 } }$ ferric chloride solution, cool to room temperature to obtain green needle-like crystals of $\mathrm { K } _ { 3 } \mathrm { F e } ( \mathrm { C } _ { 2 } \mathrm { O } _ { 4 } ) _ { 3 } { \cdot } 3 \mathrm { H } _ { 2 } \mathrm { O }$. The volume of the solution at this time was measured to be $30 \ \mathrm { mL }$. 

For this scheme, we can judge its rationality and feasibility by calculating the relative amounts of the main product and by-products (mainly $\mathrm { F e ( O H ) } _ { 3 }$).

Constants involved in this question: 

$\mathrm { p } K _ { \mathrm { s p } } ( \mathrm { F e } ( \mathrm { O H } ) _ { 3 } ) = 3 8 . 55$

 $\lg K _ { \mathrm { \tiny ~ E } } ( \mathrm { F e } ( \mathrm { C _ { 2 } O _ { 4 } }  ) _ { 3 } { } ^ { 3 - } ) = 2 0 . 2$

$\mathrm { p } K _ { \mathrm { bl } } ( \mathrm { C _ { 2 } O _ { 4 } } ^ { 2 - } ) = 9 . 7 3$ . 

The saturated solubility of the product at room temperature is $4 . 9 ~ \mathrm { g \cdot 100 ~ m L ^ { - 1 } }$, and the solubility of the two reactants is very high.

Let's first assume that $\mathrm { F e ( O H ) } _ { 3 }$ and other iron-containing species are not formed. After performing this synthesis operation (without separating the solid product from the solution), according to the assumption above, we can calculate the concentrations of $\mathrm { F e } ^ { 3 + }$, $\mathrm { F e } ( \mathrm { C } _ { 2 } \mathrm { O } _ { 4 } ) _ { 3 } { } ^ { 3 -}$, $\mathrm { C _ { 2 } O _ { 4 } } ^ { 2 - }$, $\mathrm { H C _ { 2 } O _ { 4 } } ^ { - }$, and $\mathrm { OH }^ { - }$ in the solution when it is cooled to room temperature (ignoring the secondary hydrolysis of oxalate).

Based on the preliminary calculation of the above species concentrations, we can judge whether $\mathrm { F e ( O H ) } _ { 3 }$ will be formed. So, please determine whether $\mathrm { F e ( O H ) } _ { 3 }$ will be formed. If so, calculate the mass of $\mathrm { F e ( O H ) } _ { 3 }$ formed (choose the answer closest to the calculated value).
# 草酸根配合物合成可行性的分析化学计算
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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 $\mathrm { F e } ^ { 3 + }$ ， $\mathrm { A l } ^ { 3 + }$ 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 $20 \ \mathrm { mL } \ 3 \ \mathrm { mol { \cdot } L ^ { - 1 } }$ 草酸钾热溶液和 $8 ~ \mathrm { m L } ~ 1 . 5 ~ \mathrm { m o l { \cdot } L ^ { - 1 } }$ 三氯化铁溶液混合，冷却至室温得到绿色针状晶体 $\mathrm { K } _ { 3 } \mathrm { F e } ( \mathrm { C } _ { 2 } \mathrm { O } _ { 4 } ) _ { 3 } { \cdot } 3 \mathrm { H } _ { 2 } \mathrm { O }$，测得此时溶液体积为 $3 0 \ \mathrm { m L }$ 。针对该方案，我们可以通过计算分析主产物与副产物（主要为$\mathrm { F e ( O H ) } _ { 3 }$ ）相对的多少来判断其合理性与可行性。

本题涉及的常数： $\mathrm { p } K _ { \mathrm { s p } } ( \mathrm { F e } ( \mathrm { O H } ) _ { 3 } ) = 3 8 . 5 5$ ， $\lg K _ { \mathrm { \tiny ~ E } } ( \mathrm { F e } ( \mathrm { C _ { 2 } O _ { 4 } }  ) _ { 3 } { } ^ { 3 - } ) = 2 0 . 2$，$\mathrm { p } K _ { \mathrm { b 1 } } ( \mathrm { C _ { 2 } O _ { 4 } } ^ { 2 - } ) = 9 . 7 3$ 。产物在室温下的饱和溶解度为 $4 . 9 ~ \mathrm { g \cdot (1 0 0 ~ m L) ^ { - 1 } }$ ，两种反应物溶解度很大。

不妨先假设未生成 $\mathrm { F e ( O H ) } _ { 3 }$ 和其他含铁物种，在进行该合成操作后（尚未将固体产物与溶液分离），我们可以计算混合后冷却至室温时溶液中 $\mathrm { F e } ^ { 3 + }$ 、 $\mathrm { F e } ( \mathrm { C } _ { 2 } \mathrm { O } _ { 4 } ) _ { 3 } { } ^ { 3 -}$ 、 $\mathrm { C _ { 2 } O _ { 4 } } ^ { 2 - }$ 、 $\mathrm { H C _ { 2 } O _ { 4 } } ^ { - }$ 、$\mathrm { OH }^ { - } $的浓度（忽略草酸根的二级水解）。

根据以上物种浓度的初步计算，我们便可以判断是否会成生成 $\mathrm { F e ( O H ) } _ { 3 }$ 。那么，请判断是否会生成 $\mathrm { F e ( O H ) } _ { 3 }$。如果是，计算生成 $\mathrm { F e ( O H ) } _ { 3 }$ 的质量（选择最接近计算值的答案）。
Completely no $\mathrm { F e ( O H ) } _ { 3 }$ is generated.
$\mathrm{Fe(OH)}_3$ will be generated, with a mass of 0.13 g.
will generate $\mathrm { F e ( O H ) } _ { 3 }$, with a mass of 0.11 mg
will produce $\mathrm { F e ( O H ) } _ { 3 }$, with a mass of $3.7 {\times} {10} ^{-22} \ \mathrm { g } $
will generate $\mathrm { F e ( O H ) } _ { 3 }$, with a mass of $5.7 {\times} {10} ^{-22} \ \mathrm { g } $
will generate $\mathrm{Fe(OH)}_3$ with a mass of 0.36 mg.
will generate $\mathrm{Fe(OH)}_3$, with a mass of 0.18 mg
完全不会生成 $\mathrm { F e ( O H ) } _ { 3 }$
会生成 $\mathrm { F e ( O H ) } _ { 3 }$，质量为0.13 g
会生成 $\mathrm { F e ( O H ) } _ { 3 }$，质量为0.11 mg
会生成 $\mathrm { F e ( O H ) } _ { 3 }$，质量为$3.7 {\times} {10} ^{-22} \ \mathrm { g }  $
会生成 $\mathrm { F e ( O H ) } _ { 3 }$，质量为$5.7 {\times} {10} ^{-22} \ \mathrm { g }  $
会生成 $\mathrm { F e ( O H ) } _ { 3 }$，质量为0.36 mg
会生成 $\mathrm { F e ( O H ) } _ { 3 }$，质量为0.18 mg
Due to the large stability constant of ferrioxalate, it can be assumed that most of the iron ions exist in the form of complexes.

First, assume that iron hydroxide is not generated.

Calculate the concentration based on the volume ratio: $[  \mathrm {{ F e } ( { C } _ { 2 } \mathrm { O } _ { 4 } ) _ { 3 }}^ { 3 -} ]_{0} = 0.40 \ \mathrm { mol \cdot L}^ { -1} $

Subtract the part of oxalate consumed to form the complex: $[  \mathrm{C _ { 2 } O _ { 4 }}^ { 2-} ] = 0 . 8 \ \mathrm { mol \cdot L}^ { -1}  $
<Checkpoint>[ $[  \mathrm{C _ { 2 } O _ { 4 }}^ { 2-} ] = 0 . 8 \ \mathrm { mol \cdot L}^ { -1}  $ ](1)</Checkpoint>

Considering the solubility of the product in water, calculate the saturation concentration: $[ \mathrm {F e ( C _ { 2 } O _ { 4 } ) _ { 3 } }^ { 3 -} ]= 0 . 1\ \mathrm { mol \cdot L}^ { -1} $

<Checkpoint>[ $[ \mathrm {F e ( C _ { 2 } O _ { 4 } ) _ { 3 } }^ { 3 -} ]= 0 . 1\ \mathrm { mol \cdot L}^ { -1} $ ](1)</Checkpoint>

Calculate the concentration of free iron ions based on the stability constant:  $[ \mathrm { Fe  } ^ { 3 + } ] \equiv 1 0 ^ { - 2 0 . 9 } \ \mathrm { mol \cdot L}^ { -1}$

Based on the first-level hydrolysis of oxalate, we get: $\mathrm { [{ H C _ { 2 } O _ { 4 } } ^ { - } ]  = [ O H ^ { - } ] = 1 0 ^ { - 4 . 9 1 }} \ \mathrm { mol \cdot L}^ { -1}$

Next, consider the reaction of hydrolysis to form iron hydroxide:

$\mathrm { F e } ( \mathrm { C } _ { 2 } \mathrm { O } _ { 4 } ) _ { 3 } { } ^ { 3 - } + 3 \mathrm { H } _ { 2 } \mathrm { O } = \mathrm { F e } ( \mathrm { O H } ) _ { 3 } + 3 \mathrm {{ H C } _ { 2 } \mathrm { O } _ { 4 }} ^ { - }$

<Checkpoint>[$\mathrm { F e } ( \mathrm { C } _ { 2 } \mathrm { O } _ { 4 } ) _ { 3 } { } ^ { 3 - } + 3 \mathrm { H } _ { 2 } \mathrm { O } = \mathrm { F e } ( \mathrm { O H } ) _ { 3 } + 3 \mathrm {{ H C } _ { 2 } \mathrm { O } _ { 4 }} ^ { - }$](1)</Checkpoint>

$K = K _ { \mathfrak { b } } ^ { 3 } { \cdot } K _ { \mathfrak { s p } } { \cdot } K _ { E } = 1 0 ^ { - 1 0 . 8 4 } , ~ Q = 1 0 ^ { - 1 3 . 7 } $

<Checkpoint>[$K = K _ { \mathfrak { b } } ^ { 3 } { \cdot } K _ { \mathfrak { s p } } { \cdot } K _ { E } = 1 0 ^ { - 1 0 . 8 4 } , ~ Q = 1 0 ^ { - 1 3 . 7 } $](1)</Checkpoint>

$Q<K$, $\mathrm { F e ( O H ) } _ { 3 }$ will be generated.

<Checkpoint>[$Q<K$, $\mathrm { F e ( O H ) } _ { 3 }$ will be generated。](1)</Checkpoint>

At equilibrium, $[\mathrm{ {H C _ { 2 } O _ { 4}} ^{-} } ]_{\mathrm{eq}} = 10^{-3.95} \ \mathrm { mol \cdot L}^ { -1}$

Amount of precipitated substance:

$n (\mathrm {Fe(OH) _{3}} ) = ( [\mathrm{ {H C _ { 2 } O _ { 4}} ^{-} } ]_{\mathrm{eq}}  - \mathrm { [{ H C _ { 2 } O _ { 4 } } ^ { - } ]} ) \times 0.030 / 3 =10 ^ { - 6} \ \mathrm { mol \cdot L}^ { -1}$

<Checkpoint>[$n (\mathrm {Fe(OH) _{3}} ) = ( [\mathrm{ {H C _ { 2 } O _ { 4}} ^{-} } ]_{\mathrm{eq}}  - \mathrm { [{ H C _ { 2 } O _ { 4 } } ^ { - } ]} ) \times 0.030 / 3 =10 ^ { - 6} \ \mathrm { mol \cdot L}^ { -1}$](2)</Checkpoint>

That is, the mass of the precipitate is $m=0.11 \ \mathrm{mg}$
由于三草酸合铁的稳定常数很大，可以认为绝大多数铁离子以配合物形式存在。

先假设氢氧化铁不生成

根据体积比计算浓度， $[  \mathrm {{ F e } ( { C } _ { 2 } \mathrm { O } _ { 4 } ) _ { 3 }}^ { 3 -} ]_{0} = 0.40 \ \mathrm { mol \cdot L}^ { -1} $ 

扣除形成配合物消耗的部分草酸根， $[  \mathrm{C _ { 2 } O _ { 4 }}^ { 2-} ] = 0 . 8 \ \mathrm { mol \cdot L}^ { -1}  $ 
<Checkpoint>[ $[  \mathrm{C _ { 2 } O _ { 4 }}^ { 2-} ] = 0 . 8 \ \mathrm { mol \cdot L}^ { -1}  $ ](1)</Checkpoint>

考虑到产物在水中的溶解度，计算出饱和浓度， $[ \mathrm {F e ( C _ { 2 } O _ { 4 } ) _ { 3 } }^ { 3 -} ]= 0 . 1\ \mathrm { mol \cdot L}^ { -1} $ 

<Checkpoint>[ $[ \mathrm {F e ( C _ { 2 } O _ { 4 } ) _ { 3 } }^ { 3 -} ]= 0 . 1\ \mathrm { mol \cdot L}^ { -1} $ ](1)</Checkpoint>

根据稳定常数计算出游离铁离子浓度，  $[ \mathrm { Fe  } ^ { 3 + } ] \equiv 1 0 ^ { - 2 0 . 9 } \ \mathrm { mol \cdot L}^ { -1}$ 

根据草酸根的一级水解得到， $\mathrm { [{ H C _ { 2 } O _ { 4 } } ^ { - } ]  = [ O H ^ { - } ] = 1 0 ^ { - 4 . 9 1 }} \ \mathrm { mol \cdot L}^ { -1}$ 

之后，考虑水解生成氢氧化铁的反应

$\mathrm { F e } ( \mathrm { C } _ { 2 } \mathrm { O } _ { 4 } ) _ { 3 } { } ^ { 3 - } + 3 \mathrm { H } _ { 2 } \mathrm { O } = \mathrm { F e } ( \mathrm { O H } ) _ { 3 } + 3 \mathrm {{ H C } _ { 2 } \mathrm { O } _ { 4 }} ^ { - }$ 

<Checkpoint>[$\mathrm { F e } ( \mathrm { C } _ { 2 } \mathrm { O } _ { 4 } ) _ { 3 } { } ^ { 3 - } + 3 \mathrm { H } _ { 2 } \mathrm { O } = \mathrm { F e } ( \mathrm { O H } ) _ { 3 } + 3 \mathrm {{ H C } _ { 2 } \mathrm { O } _ { 4 }} ^ { - }$](1)</Checkpoint>

$K = K _ { \mathfrak { b } } ^ { 3 } { \cdot } K _ { \mathfrak { s p } } { \cdot } K _ { E } = 1 0 ^ { - 1 0 . 8 4 } , ~ Q = 1 0 ^ { - 1 3 . 7 } $

<Checkpoint>[$K = K _ { \mathfrak { b } } ^ { 3 } { \cdot } K _ { \mathfrak { s p } } { \cdot } K _ { E } = 1 0 ^ { - 1 0 . 8 4 } , ~ Q = 1 0 ^ { - 1 3 . 7 } $](1)</Checkpoint>

$Q<K$， 会生成$\mathrm { F e ( O H ) } _ { 3 }$。

<Checkpoint>[$Q<K$， 会生成$\mathrm { F e ( O H ) } _ { 3 }$。](1)</Checkpoint>

平衡时，$[\mathrm{ {H C _ { 2 } O _ { 4}} ^{-} } ]_{\mathrm{eq}} = 10^{-3.95} \ \mathrm { mol \cdot L}^ { -1}$

沉淀物质的量

$n (\mathrm {Fe(OH) _{3}} ) = ( [\mathrm{ {H C _ { 2 } O _ { 4}} ^{-} } ]_{\mathrm{eq}}  - \mathrm { [{ H C _ { 2 } O _ { 4 } } ^ { - } ]} ) \times 0.030 / 3 =10 ^ { - 6} \ \mathrm { mol}$

<Checkpoint>[$n (\mathrm {Fe(OH) _{3}} ) = ( [\mathrm{ {H C _ { 2 } O _ { 4}} ^{-} } ]_{\mathrm{eq}}  - \mathrm { [{ H C _ { 2 } O _ { 4 } } ^ { - } ]} ) \times 0.030 / 3 =10 ^ { - 6} \ \mathrm { mol }$](2)</Checkpoint>

亦即沉淀质量 $m=0.11 \ \mathrm{mg}$