Dissolving $S\ mol$ $Ag_2Cr_2O_7$ in $1\ L$ of pure water results in saturation, meaning there is no precipitate, and the solution volume remains unchanged. The following data are known (activity coefficients are not considered):

$K_{sp}(Ag_2Cr_2O_7)=1.99\times10^{-7}$

$K_{sp}(Ag_2CrO_4)=1.12\times10^{-12}$

$K_{sp}(AgOH) = 2.01\times10^{-8}$

$K_a(HCrO_4^-)=3.23\times10^{-7}$

$2HCrO_4^-=Cr_2O_7^{2-}+H_2O,\ K_f=1.14\times10^2$

$H_2CrO_4$ is considered a strong acid upon its first ionization and a weak acid upon its second ionization; $H_2Cr_2O_7$ is considered a diprotic strong acid.

Select the correct option(s) below.
将$S\ mol$ $Ag_2Cr_2O_7$溶于$1\ L$纯水中，恰好达到饱和，即恰好没有任何沉淀，溶液体积不变。已知相关数据（不考虑活度）：

$K_{sp}(Ag_2Cr_2O_7)=1.99\times10^{-7}$

$K_{sp}(Ag_2CrO_4)=1.12\times10^{-12}$

$K_{sp}(AgOH) = 2.01\times10^{-8}$

$K_a(HCrO_4^-)=3.23\times10^{-7}$

$2HCrO_4^-=Cr_2O_7^{2-}+H_2O,\ K_f=1.14\times10^2$

$H_2CrO_4$一级电离时视为强酸，二级电离时视为弱酸；$H_2Cr_2O_7$视为二元强酸。

选择以下正确的选项。
All other options are incorrect
at equilibrium, $4.20\le pH \le4.50$
At equilibrium, $0.95\le[Ag^+]^2[Cr_2O_7^{2-}]/K_{sp}(Ag_2CrO_4)\le1.05$ (Note: Not a typo, the denominator is indeed $Ag_2CrO_4$)
The mass $m(g)$ of $Ag_2Cr_2O_7$ added at exact saturation satisfies $5.5\times10^{-2}\le m\le5.8\times10^{-2}$
Adding $2S\ mol$ of $Ag_2Cr_2O_7$ to $1\ L$ of pure water, at equilibrium the precipitate contains both $Ag_2CrO_4$ and $Ag_2Cr_2O_7$.
Adding $S\ mol\ Ag_2CrO_4$ to $1\ L$ of pure water, it can still be completely dissolved.
其他选项均不正确
平衡时满足$4.20\le pH \le4.50$
平衡时满足$0.95\le[Ag^+]^2[Cr_2O_7^{2-}]/K_{sp}(Ag_2CrO_4)\le1.05$（注：没有写错，分母就是$Ag_2CrO_4$）
恰好饱和时加入$Ag_2Cr_2O_7$的质量$m(g)$满足$5.5\times10^{-2}\le m\le5.8\times10^{-2}$
向$1\ L$纯水中加入$2S\ mol$的$Ag_2Cr_2O_7$，平衡时的沉淀同时含有$Ag_2CrO_4$和$Ag_2Cr_2O_7$。
向$1\ L$纯水中加入$S\ mol\ Ag_2CrO_4$，依然可以完全溶解。
At equilibrium, the solution is obviously weakly acidic. We can first not consider $AgOH$ saturation. Then either $Ag_2Cr_2O_7$ or $Ag_2CrO_4$ is saturated, that is, one of the following two equations holds as an equality, and the other equation holds as a less than:

$[Ag^+]^2[Cr_2O_7^{2-}]=K_{sp}(Ag_2Cr_2O_7)$ (Equation 1)

$[Ag^+]^2[CrO_4^{2-}]=K_{sp}(Ag_2CrO_4)$ (Equation 2)

<Checkpoint>[One of $[Ag^+]^2[Cr_2O_7^{2-}]=K_{sp}(Ag_2Cr_2O_7)$ and $[Ag^+]^2[CrO_4^{2-}]=K_{sp}(Ag_2CrO_4)$ holds, and the other equation holds as a less than](1)</Checkpoint>

When completely dissolved, according to the conservation of atoms $[Ag]=[Cr]$, we have

$[Ag^+]=2[Cr_2O_7^{2-}]+[HCrO_4^-]+[CrO_4^{2-}]$ (Equation 3)

<Checkpoint>[$[Ag^+]=2[Cr_2O_7^{2-}]+[HCrO_4^-]+[CrO_4^{2-}]$](1)</Checkpoint>

According to charge conservation, we have

$[Ag^+]+[H^+]=2[Cr_2O_7^{2-}]+[HCrO_4^-]+2[CrO_4^{2-}]+[OH^-]$ (Equation 4, $[OH^-]$ can also be ignored)

<Checkpoint>[$[Ag^+]+[H^+]=2[Cr_2O_7^{2-}]+[HCrO_4^-]+2[CrO_4^{2-}]+[OH^-]$](1)</Checkpoint>

According to $K_a$ and $K_f$, we have

$[H^+][CrO_4^{2-}]/[HCrO_4^-]=K_a(HCrO_4^-)$ (Equation 5)

<Checkpoint>[$[H^+][CrO_4^{2-}]/[HCrO_4^-]=K_a(HCrO_4^-)$](0.5)</Checkpoint>

$[Cr_2O_7^{2-}]/[HCrO_4^-]^2=K_f$ (Equation 6)

<Checkpoint>[$[Cr_2O_7^{2-}]/[HCrO_4^-]^2=K_f$](0.5)</Checkpoint>

If $Ag_2Cr_2O_7$ is saturated, then take Equations 1 and 3-6, and solve them simultaneously (units are all $mol/L$, the same below)

$[Ag^+]=9.25\times10^{-3},[H^+]=3.78\times10^{-5},[Cr_2O_7^{2-}]=2.34\times10^{-3},[HCrO_4^-]=4.53\times10^{-3}, [CrO_4^{2-}]=3.88\times10^{-5}$

But at this time $[Ag^+]^2[CrO_4^{2-}]>K_{sp}(Ag_2CrO_4)$, discard it.

<Checkpoint>[The assumption that $Ag_2Cr_2O_7$ is just saturated is not reasonable.](3)</Checkpoint>

If $Ag_2CrO_4$ is saturated, then take Equations 2-6 and solve them simultaneously

$[Ag^+]=3.36\times10^{-4},[H^+]=9.92\times10^{-6},[Cr_2O_7^{2-}]=1.06\times10^{-5},[HCrO_4^-]=3.05\times10^{-4}, [CrO_4^{2-}]=9.94\times10^{-6}$

At this time, $[Ag^+]^2[Cr_2O_7^{2-}]<K_{sp}(Ag_2Cr_2O_7)$ and $[Ag^+][OH^-]<K_{sp}(AgOH)$ are also satisfied, which is reasonable.

<Checkpoint>[When $Ag_2CrO_4$ is just saturated, there is a reasonable solution: $[Ag^+]=3.36\times10^{-4},[H^+]=9.92\times10^{-6},[Cr_2O_7^{2-}]=1.06\times10^{-5},[HCrO_4^-]=3.05\times10^{-4}, [CrO_4^{2-}]=9.94\times10^{-6}$](5)</Checkpoint>

At this time $pH=5.00$, option B is wrong.

<Checkpoint>[At saturation, $pH=5.00$](0.5)</Checkpoint>

$[Ag^+]^2[Cr_2O_7^{2-}]/K_{sp}(Ag_2CrO_4)=1.07$, option C is wrong.

<Checkpoint>[At saturation, $[Ag^+]^2[Cr_2O_7^{2-}]/K_{sp}(Ag_2CrO_4)=1.07$](0.5)</Checkpoint>

At saturation, $S=[Ag^+]/2=1.68\times10^{-4}\ mol$, the dissolved mass $m=S\cdot M_{Ag_2Cr_2O_7}=7.3\times10^{-2}\ g$, option D is wrong.

<Checkpoint>[$S=1.68\times10^{-4}\ mol$](0.5)</Checkpoint>

<Checkpoint>[$m=7.3\times10^{-2}\ g$](0.5)</Checkpoint>

If the two precipitates coexist, then solve Equations 1, 2, 4, 5, and 6 simultaneously to get

$[Ag^+]=3.46\times10^{-3},[H^+]=4.17\times10^{-2},[Cr_2O_7^{2-}]=1.66\times10^{-2},[HCrO_4^-]=1.21\times10^{-2}, [CrO_4^{2-}]=9.33\times10^{-8}$

The amount of $Cr$ in the solution $[Cr]_{aq}=2[Cr_2O_7^{2-}]+[HCrO_4^-]+[CrO_4^{2-}]=4.53\times10^{-2}>4S$, that is, the amount of dissolved $Cr$ has exceeded the amount added, which is a contradiction. Option E is wrong.

<Checkpoint>[When two precipitates coexist, $[Cr]_{aq}>4S$](4)</Checkpoint>

$Ag_2CrO_4$ is less soluble than $Ag_2Cr_2O_7$, option F is obviously wrong.

<Checkpoint>[$Ag_2CrO_4$ is less soluble than $Ag_2Cr_2O_7$](0.5)</Checkpoint>

Therefore, choose A.
平衡时溶液显然为弱酸性，可先不考虑$AgOH$饱和，则$Ag_2Cr_2O_7$和$Ag_2CrO_4$其中之一饱和，也即以下两式之一取等号，另一式取小于号：

$[Ag^+]^2[Cr_2O_7^{2-}]=K_{sp}(Ag_2Cr_2O_7)$（式1）

$[Ag^+]^2[CrO_4^{2-}]=K_{sp}(Ag_2CrO_4)$（式2）

<Checkpoint>[$[Ag^+]^2[Cr_2O_7^{2-}]=K_{sp}(Ag_2Cr_2O_7)$和$[Ag^+]^2[CrO_4^{2-}]=K_{sp}(Ag_2CrO_4)$其中之一成立，另一式取小于号](1)</Checkpoint>

完全溶解时，根据原子数守恒$[Ag]=[Cr]$，有

$[Ag^+]=2[Cr_2O_7^{2-}]+[HCrO_4^-]+[CrO_4^{2-}]$（式3）

<Checkpoint>[$[Ag^+]=2[Cr_2O_7^{2-}]+[HCrO_4^-]+[CrO_4^{2-}]$](1)</Checkpoint>

根据电荷守恒，有

$[Ag^+]+[H^+]=2[Cr_2O_7^{2-}]+[HCrO_4^-]+2[CrO_4^{2-}]+[OH^-]$（式4，也可忽略$[OH^-]$）

<Checkpoint>[$[Ag^+]+[H^+]=2[Cr_2O_7^{2-}]+[HCrO_4^-]+2[CrO_4^{2-}]+[OH^-]$](1)</Checkpoint>

另根据$K_a$和$K_f$，有

$[H^+][CrO_4^{2-}]/[HCrO_4^-]=K_a(HCrO_4^-)$（式5）

<Checkpoint>[$[H^+][CrO_4^{2-}]/[HCrO_4^-]=K_a(HCrO_4^-)$](0.5)</Checkpoint>

$[Cr_2O_7^{2-}]/[HCrO_4^-]^2=K_f$（式6）

<Checkpoint>[$[Cr_2O_7^{2-}]/[HCrO_4^-]^2=K_f$](0.5)</Checkpoint>

若为$Ag_2Cr_2O_7$饱和，则取1和3-6式，联立解得（单位均为$mol/L$，下同）

$[Ag^+]=9.25\times10^{-3},[H^+]=3.78\times10^{-5},[Cr_2O_7^{2-}]=2.34\times10^{-3},[HCrO_4^-]=4.53\times10^{-3}, [CrO_4^{2-}]=3.88\times10^{-5}$

但此时$[Ag^+]^2[CrO_4^{2-}]>K_{sp}(Ag_2CrO_4)$，舍去。

<Checkpoint>[恰好达到饱和的是$Ag_2Cr_2O_7$，这一假设不合理。](3)</Checkpoint>

若为$Ag_2CrO_4$饱和，则取2-6式，联立解得

$[Ag^+]=3.36\times10^{-4},[H^+]=9.92\times10^{-6},[Cr_2O_7^{2-}]=1.06\times10^{-5},[HCrO_4^-]=3.05\times10^{-4}, [CrO_4^{2-}]=9.94\times10^{-6}$

此时也满足$[Ag^+]^2[Cr_2O_7^{2-}]<K_{sp}(Ag_2Cr_2O_7)$和$[Ag^+][OH^-]<K_{sp}(AgOH)$，是合理的。

<Checkpoint>[恰好饱和的是$Ag_2CrO_4$时有合理解：$[Ag^+]=3.36\times10^{-4},[H^+]=9.92\times10^{-6},[Cr_2O_7^{2-}]=1.06\times10^{-5},[HCrO_4^-]=3.05\times10^{-4}, [CrO_4^{2-}]=9.94\times10^{-6}$](5)</Checkpoint>

此时$pH=5.00$，选项B错误。

<Checkpoint>[饱和时$pH=5.00$](0.5)</Checkpoint>

$[Ag^+]^2[Cr_2O_7^{2-}]/K_{sp}(Ag_2CrO_4)=1.07$，选项C错误。

<Checkpoint>[饱和时$[Ag^+]^2[Cr_2O_7^{2-}]/K_{sp}(Ag_2CrO_4)=1.07$](0.5)</Checkpoint>

饱和时$S=[Ag^+]/2=1.68\times10^{-4}\ mol$，溶解质量$m=S\cdot M_{Ag_2Cr_2O_7}=7.3\times10^{-2}\ g$，选项D错误。

<Checkpoint>[$S=1.68\times10^{-4}\ mol$](0.5)</Checkpoint>

<Checkpoint>[$m=7.3\times10^{-2}\ g$](0.5)</Checkpoint>

若两种沉淀同时存在，则联立1，2，4，5，6式，解得

$[Ag^+]=3.46\times10^{-3},[H^+]=4.17\times10^{-2},[Cr_2O_7^{2-}]=1.66\times10^{-2},[HCrO_4^-]=1.21\times10^{-2}, [CrO_4^{2-}]=9.33\times10^{-8}$

溶液中的$Cr$量$[Cr]_{aq}=2[Cr_2O_7^{2-}]+[HCrO_4^-]+[CrO_4^{2-}]=4.53\times10^{-2}>4S$，即溶解的$Cr$量已经超过了投入的量，矛盾。选项E错误。

<Checkpoint>[两种沉淀同时存在时有$[Cr]_{aq}>4S$](4)</Checkpoint>

$Ag_2CrO_4$比$Ag_2Cr_2O_7$更难溶，选项F显然错误。

<Checkpoint>[$Ag_2CrO_4$比$Ag_2Cr_2O_7$更难溶](0.5)</Checkpoint>

因此选择A。