There is currently an alloy primarily composed of $\mathrm{Pb}$ and $\mathrm{Zn}$, which may contain other metals. To determine the content of each component, an experimenter designed the following experimental procedure:

(1) Dissolve the $0.7800\ \mathrm{g}$ sample in an excess mixed acid, dilute to $250.0\ \mathrm{mL}$, and transfer $25.00\ \mathrm{mL}$ of the solution to a conical flask. Add $0.1\ \mathrm{g}\ \mathrm{NaOH}$ and $10\ \mathrm{mL}$ of $1:1$ ammonia buffer solution, followed by a small amount of tartaric acid and $0.2\ \mathrm{g\ KCN}$. Using Eriochrome Black T as the indicator, titrate with a standard $\mathrm{EDTA}$ solution of concentration $c_1\ (\mathrm{mol·L^{-1}})$. When the solution changes from purple-red to blue, record the reading $V_{1}(\mathrm{mL})$. Then, take another $25.00\ \mathrm{mL}$ of the solution, add $\mathrm{NaOH}$ and ammonia buffer solution, and titrate with the $\mathrm{EDTA}$ solution, recording the reading $V_{2}(\mathrm{mL})$.

In the first experiment, the experimenter prepared a standard $\mathrm{EDTA}$ solution with $c_1=0.02016\ (\mathrm{mol·L^{-1}})$. When the titration volume reached $17.52\ \mathrm{mL}$, he realized there was an issue with the experiment and abandoned it.

In light of this, he designed a second experimental procedure:

(2) Subsequently, the experimenter revised the procedure as follows: Dissolve the $0.7800 \mathrm{~g}$ sample in an excess mixed acid, dilute to $250.0 \mathrm{~mL}$, and transfer $25.00 \mathrm{~mL}$ of the solution to a conical flask. Add $0.1 \mathrm{~g}\ \mathrm{NaOH}$ and $10 \mathrm{~mL}$ of $1:1$ ammonia buffer solution, followed by a small amount of tartaric acid and $0.2 \mathrm{~g}\ \mathrm{KCN}$. Then, introduce $0.2 \mathrm{~g}$ of triethanolamine, and using Eriochrome Black T as the indicator, titrate with an $\mathrm{EDTA}$ solution of concentration $c_{1}\left(\mathrm{~mol} \cdot \mathrm{L}^{-1}\right)$ until the endpoint is reached, recording the reading $V_{3}(\mathrm{mL})$. Next, add a slight excess of $1:1$ formaldehyde solution and continue titrating until the solution just turns blue, recording the reading $V_{4}(\mathrm{L})$.

Regarding the above analysis process, which of the following statements is correct? (For options involving calculations, an error within $1\%$ is considered acceptable.)
现有一种以$\mathrm{Pb}$与$\mathrm{Zn}$为主要材料，可能含有其他金属的合金,一实验员为测定其各组分含量,设计了如下实验方案:

（1）将$0.7800\ \mathrm{g}$样品溶于过量混酸,定容至$250.0\ \mathrm{mL}$,取$25.00\ \mathrm{mL}$溶液置于锥形瓶中,加入$0.1\ \mathrm{g}\mathrm{NaOH}$与$10\ \mathrm{mL}\ 1:1$氨缓冲溶液,再加入少量酒石酸与$0.2\ \mathrm{g\ KCN}$,以铬黑T为指示剂,
用浓度为$c_1\ (\mathrm{mol·L^{-1}})$的$\mathrm{EDTA}$标准溶液滴定,当溶液从紫红色变为蓝色时,记录下读数$V_{1}(\mathrm{mL})$。再另取$25.00\ \mathrm{mL}$溶液,加入$\mathrm{NaOH}$与氨缓冲溶液,用$\mathrm{EDTA}$溶液滴定,记录下读数
$V_{2}(\mathrm{mL})$。

实验员的第一次实验配置了$c_1=0.02016\ (\mathrm{mol·L^{-1}})$的$\mathrm{EDTA}$标准溶液,在滴定到滴定体积读数为$17.52\ \mathrm{mL}$时，他意识到实验出问题了，放弃了该次实验。

有鉴于此，他设计了第二种实验方案：

（2）随后, 实验员修改了实验方案, 具体操作如下: 将 $0.7800 \mathrm{~g}$ 样品溶于过量混酸, 定容至 $250.0 \mathrm{~mL}$, 取 $25.00 \mathrm{~mL}$ 溶液置于雉形瓶中, 加入 $0.1 \mathrm{~g}\ \mathrm{NaOH}$ 与 $10 \mathrm{~mL}\ 1: 1$ 氨缓冲溶液,
再加入少量酒石酸与 $0.2 \mathrm{~g}\ \mathrm{KCN}$, 随后放入 $0.2 \mathrm{~g}$ 三乙醇胺, 以铬黑 $\mathrm{T}$ 为指示剂, 用浓度为
$c_{1}\left(\mathrm{mol} \cdot \mathrm{L}^{-1}\right)$ 的$\mathrm{EDTA}$溶液滴定至终点, 记录此时读数 $V_{3}(\mathrm{mL})$, 再向其中加入稍过量 $1: 1 $
甲醛溶液, 继续滴定, 滴定至溶液刚好变为蓝色, 记录下此时读数 $V_{4}(\mathrm{mL})$。

关于上述的分析过程，下列说法正确的是（涉及到计算的选项，误差在$1\%$以内即算正确）：
In Scheme 1, the mass fraction of $\mathrm{Pb}$ can be expressed as $2.478c_1V_1$
In Scheme 1, the mass fraction of $\mathrm{Zn}$ can be expressed as $-0.8583c_1(V_1-V_2)$.
The experimenter believes that the reason for the failure of Scheme 1 is that the solution turned blue prematurely when titrated to $17.52\ \mathrm{mL}$.
In Scheme 2, when formaldehyde is added, the sum of the coefficients on the product side of the ionic equation for the reaction of formaldehyde (with the coefficients in the simplest whole-number ratio) is 8.
In Scheme 2, replacing triethanolamine with ascorbic acid + potassium cyanide will not affect the titration process.
If the sample contains the metal cadmium, according to Method 2, cadmium will block the indicator and affect the measurement results.
In Scheme 2, the mass fraction of $\mathrm{Zn}$ can be expressed as $0.8983c_1(V_4-V_3)$.
None of the above statements are correct
方案1中，$\mathrm{Pb}$的质量分数可表达为$2.478c_1V_1$
方案1中，$\mathrm{Zn}$的质量分数可表达为$-0.8583c_1(V_1-V_2)$
实验员认为方案1失败的原因是在滴定至$17.52\ \mathrm{mL}$时，溶液提前变为蓝色
方案2中，加入甲醛时，甲醛发生的反应的离子方程式（产物系数为最简整数比）的产物一侧系数和为8

方案2中，三乙醇胺如果更换为抗坏血酸+氰化钾，也不会影响滴定过程
若该样品中含有金属镉，按照方案2的方法，镉会导致指示剂封闭而影响测定结果。
方案2中，$\mathrm{Zn}$的质量分数可表达为$0.8983c_1(V_4-V_3)$
以上说法均不正确
Titration process of Method 1:  
After adding potassium cyanide, $\mathrm{Zn^{2+}}$ forms $\mathrm{Zn(CN)_4^{2-}}$ and is masked. $\mathrm{EDTA}$ only titrates $\mathrm{Pb^{2+}}$, with Eriochrome Black T as the indicator. The solution changes from purple-red to blue. Under an ammonia buffer solution, both $\mathrm{Zn^{2+}}$ and $\mathrm{Pb^{2+}}$ are titrated by $\mathrm{EDTA}$.  

<Checkpoint>[After adding potassium cyanide, $\mathrm{Zn^{2+}}$ forms $\mathrm{Zn(CN)_4^{2-}}$ and is masked](1)</Checkpoint>  

Thus, $V_{1}$ represents the amount of $\mathrm{Pb^{2+}}$, and $V_{2}$ represents the total amount of $\mathrm{Zn^{2+}}$ and $\mathrm{Pb^{2+}}$.  
Therefore, the amount of $\mathrm{Zn^{2+}}$ is $V_{2}-V_{1}$.  

<Checkpoint>[$V_{1}$ represents the amount of $\mathrm{Pb^{2+}}$, and the amount of $\mathrm{Zn^{2+}}$ is $V_{2}-V_{1}$](1)</Checkpoint>  

The mass fraction expression for $\mathrm{Pb}$ should be:  

$\frac{c_{1}V_{1} \times 207.2 \times 10}{0.7800\times1000}=2.6564 c_{1}V_{1}$.  

Option A's expression is incorrect.  

<Checkpoint>[The mass fraction of $\mathrm{Pb}$ is $2.6564 c_{1}V_{1}$, and Option A is incorrect](1)</Checkpoint>  

The mass fraction of $\mathrm{Zn}$ is:  
${\frac {c_{1}(V_{2}-V_1)\times 65.38\times 10}{0.7800\times1000}}=0.8383c_{1}(V_{2}-V_1)$.  

Option B's expression is incorrect.  

<Checkpoint>[The mass fraction of $\mathrm{Zn}$ is $0.8383c_{1}(V_{2}-V_1)$, and Option B is incorrect](1)</Checkpoint>  

Analysis of the first failed experiment: For the first titration of lead content, assuming the sample consists entirely of lead, the theoretical endpoint should be $0.7800/207.2/0.02016\times1000=18.67\ mL$. The experimenter titrated up to $17.52\ mL$, only about 1 mL short of the endpoint. At this point, each drop of titrant added should cause localized color change to the endpoint color due to excessive local concentration, reverting to the pre-endpoint color upon mixing. The experimenter likely observed no color change at all and concluded the experiment failed, so Option C is incorrect. In this case, the failure was caused by indicator僵化.  

<Checkpoint>[Excessive local concentration causes localized color change to the endpoint color, reverting to the pre-endpoint color upon mixing](0.5)</Checkpoint>  

<Checkpoint>[No color change was observed, leading to the conclusion that the experiment failed, so Option C is incorrect.](1)</Checkpoint>  

<Checkpoint>[The failure was caused by indicator僵化](0.5)</Checkpoint>  

Method 2 differs from Method 1 by adding triethanolamine, which is commonly used to mask $\mathrm{Fe^{3+},Al^{3+}}$, thereby preventing僵化 of Eriochrome Black T. Since the failed experiment cannot determine which ion caused the indicator僵化, the possibility of both $\mathrm{Fe^{3+},Al^{3+}}$ being present must be considered. Therefore, triethanolamine is added instead of the ascorbic acid + potassium cyanide combination, which only masks $\mathrm{Fe^{3+}}$, making Option E incorrect.  

<Checkpoint>[Triethanolamine is used to mask $\mathrm{Fe^{3+},Al^{3+}}$](0.5)</Checkpoint>  

<Checkpoint>[The possibility of both $\mathrm{Fe^{3+},Al^{3+}}$ being present must be considered](0.5)</Checkpoint>  

<Checkpoint>[Triethanolamine is added instead of the ascorbic acid + potassium cyanide combination, which only masks $\mathrm{Fe^{3+}}$, making Option E incorrect](1)</Checkpoint>  

The purpose of adding formaldehyde is to demask the masked $\mathrm{Zn(CN)_4^{2-}}$ for continuous titration. The ionic equation is:  

<Checkpoint>[Formaldehyde demasks $\mathrm{Zn(CN)_4^{2-}}$](1)</Checkpoint>  

$\mathrm{Zn(CN)_4^{2-}+4HCHO+4H_2O=Zn^{2+}+4HOCH_2CN+4OH^-}$  

The sum of the product coefficients is 9, making Option D incorrect.  

<Checkpoint>[$\mathrm{Zn(CN)_4^{2-}+4HCHO+4H_2O=Zn^{2+}+4HOCH_2CN+4OH^-}$  The sum of the product coefficients is 9, making Option D incorrect.](1)</Checkpoint>  

$\mathrm{Cd^{2+}}$ formed from metallic cadmium is masked by the added potassium cyanide and does not block the indicator, making Option F incorrect.  

<Checkpoint>[$\mathrm{Cd^{2+}}$ is masked by the added potassium cyanide and does not block the indicator, making Option F incorrect.](1)</Checkpoint>  

Method 2 follows the same principle as Method 1, involving masking followed by back-titration. Thus, the volume of titrant corresponding to $\mathrm{Zn^{2+}}$ is $V_4-V_3$, and the mass fraction is $0.8383c_{1}(V_{4}-V_3)$, making Option G incorrect.  

<Checkpoint>[The mass fraction of $\mathrm{Zn}$ is $0.8383c_{1}(V_{4}-V_3)$, making Option G incorrect.](1)</Checkpoint>  

In conclusion, Options A-G are all incorrect, and Option H is correct.  

<Checkpoint>[Options A-G are all incorrect, and Option H is correct.](1)</Checkpoint>
方案1的滴定过程:
加入氰化钾后，$\mathrm{Zn^{2+}}$形成$\mathrm{Zn(CN)_4^{2-}}$而被掩蔽， $\mathrm{EDTA}$只滴定$\mathrm{Pb^{2+}}$,铬黑T指示终点, 溶液从紫红色变为蓝色。而在氨缓溶液下，$\mathrm{Zn^{2+}}$和$\mathrm{Pb^{2+}}$均被
$\mathrm{EDTA}$滴定。

<Checkpoint>[加入氰化钾后，$\mathrm{Zn^{2+}}$形成$\mathrm{Zn(CN)_4^{2-}}$而被掩蔽](1)</Checkpoint>

从而$V_{1}$为$\mathrm{Pb^{2+}}$的量, $V_{2}$ 为$\mathrm{Zn^{2+}}$和$\mathrm{Pb^{2+}}$的总量,
因此$\mathrm{Zn^{2+}}$的量为 $V_{2}-V_{1}$ 。

<Checkpoint>[$V_{1}$为$\mathrm{Pb^{2+}}$的量，$\mathrm{Zn^{2+}}$的量为 $V_{2}-V_{1}$](1)</Checkpoint>

$\mathrm{Pb}$的质量分数表达式应为:

$\frac{c_{1}V_{1} \times 207.2 \times 10}{0.7800\times1000}=2.6564 c_{1}V_{1}$.

选项A的表达式错误。

<Checkpoint>[$\mathrm{Pb}$的质量分数为$2.6564 c_{1}V_{1}$，选项A错误](1)</Checkpoint>

$\mathrm{Zn}$的质量分数为
${\frac {c_{1}(V_{2}-V_1)\times 65.38\times 10}{0.7800\times1000}}=0.8383c_{1}(V_{2}-V_1)$ 。

选项B的表达式错误。

<Checkpoint>[$\mathrm{Zn}$的质量分数为$0.8383c_{1}(V_{2}-V_1)$，选项B错误](1)</Checkpoint>

对第一次实验失败进行分析：第一次滴定铅的含量，假设如果样品全为铅，则理论的化学终点应当为$0.7800/207.2/0.02016\times1000=18.67\ mL$；实验员已经滴定到了$17.52\ mL$，距离终点仅差约1毫升，按理说此时每一滴滴定剂加入后，由于局部浓度过大，溶液都会局部变色为终点颜色，摇匀后才回到滴定终点前的颜色；实验员应当是看到了溶液完全不变色，才判断实验失败，故选项C说法错误。此时，应当是指示剂僵化导致这一实验失败。

<Checkpoint>[局部浓度过大，溶液会局部变色为终点颜色，摇匀后才回到滴定终点前的颜色](0.5)</Checkpoint>

<Checkpoint>[溶液完全不变色，才判断实验失败，故选项C说法错误。](1)</Checkpoint>

<Checkpoint>[指示剂僵化导致这一实验失败](0.5)</Checkpoint>

方案2的过程相比方案1加入了三乙醇胺，其经常用于掩蔽$\mathrm{Fe^{3+},Al^{3+}}$，从而防止指示剂铬黑T僵化。由于失败的实验并不能判断是哪种离子导致指示剂僵化，故需要考虑$\mathrm{Fe^{3+},Al^{3+}}$均存在的可能性，所以需要加入三乙醇胺而不是只能掩蔽$\mathrm{Fe^{3+}}$的抗坏血酸+氰化钾组合，选项E错误。

<Checkpoint>[三乙醇胺用于掩蔽$\mathrm{Fe^{3+},Al^{3+}}$](0.5)</Checkpoint>

<Checkpoint>[需要考虑$\mathrm{Fe^{3+},Al^{3+}}$均存在的可能性](0.5)</Checkpoint>

<Checkpoint>[需要加入三乙醇胺而不是只能掩蔽$\mathrm{Fe^{3+}}$的抗坏血酸+氰化钾组合，选项E错误](1)</Checkpoint>

加入甲醛的目的是为了使得被掩蔽的$\mathrm{Zn(CN)_4^{2-}}$解掩蔽从而连续滴定，其离子方程式为

<Checkpoint>[甲醛使$\mathrm{Zn(CN)_4^{2-}}$解掩蔽](1)</Checkpoint>

$\mathrm{Zn(CN)_4^{2-}+4HCHO+4H_2O=Zn^{2+}+4HOCH_2CN+4OH^-}$

方程式右边系数和为9，选项D错误。

<Checkpoint>[$\mathrm{Zn(CN)_4^{2-}+4HCHO+4H_2O=Zn^{2+}+4HOCH_2CN+4OH^-}$  产物系数和为9，选项D错误。](1)</Checkpoint>

金属镉形成的$\mathrm{Cd^{2+}}$会被加入的氰化钾掩蔽，不会封闭指示剂，选项F错误。

<Checkpoint>[$\mathrm{Cd^{2+}}$会被加入的氰化钾掩蔽，不会封闭指示剂，选项F错误。](1)</Checkpoint>

方案2与方案1的原理相同，都是先掩蔽后返滴定，从而$\mathrm{Zn^{2+}}$对应的滴定剂体积为$V_4-V_3$，质量分数为$0.8383c_{1}(V_{4}-V_3)$，选项G错误。

<Checkpoint>[$\mathrm{Zn}$质量分数为$0.8383c_{1}(V_{4}-V_3)$，选项G错误。](1)</Checkpoint>

综上，选项A-G均不正确，选项H正确。

<Checkpoint>[选项A-G均不正确，选项H正确。](1)</Checkpoint>